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values | name stringlengths 2 58 | type stringclasses 2
values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522
values | time-limit stringclasses 8
values | memory-limit stringclasses 8
values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3
values | verdict stringclasses 14
values | testset stringclasses 12
values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
922 | A | Cloning Toys | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Imp likes his plush toy a lot.
Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies.
Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies. | The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one). | Print "Yes", if the desired configuration is possible, and "No" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"6 3\n",
"4 2\n",
"1000 1001\n"
] | [
"Yes\n",
"No\n",
"Yes\n"
] | In the first example, Imp has to apply the machine twice to original toys and then twice to copies. | 500 | [
{
"input": "6 3",
"output": "Yes"
},
{
"input": "4 2",
"output": "No"
},
{
"input": "1000 1001",
"output": "Yes"
},
{
"input": "1000000000 999999999",
"output": "Yes"
},
{
"input": "81452244 81452247",
"output": "No"
},
{
"input": "188032448 86524683",... | 1,606,968,638 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | x,y = map(int, input().split())
if y == 0:
print("No")
elif y == 1 and x > 0:
print("No")
elif x < y-1:
print("No")
else:
no_of_original_toys = y
no_of_copies = y-1
if no_of_copies % 2 == x % 2
print("Yes")
else:
print("No") | Title: Cloning Toys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imp likes his plush toy a lot.
Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies.
Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies.
Input Specification:
The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one).
Output Specification:
Print "Yes", if the desired configuration is possible, and "No" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['6 3\n', '4 2\n', '1000 1001\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
In the first example, Imp has to apply the machine twice to original toys and then twice to copies. | ```python
x,y = map(int, input().split())
if y == 0:
print("No")
elif y == 1 and x > 0:
print("No")
elif x < y-1:
print("No")
else:
no_of_original_toys = y
no_of_copies = y-1
if no_of_copies % 2 == x % 2
print("Yes")
else:
print("No")
``` | -1 | |
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
... | 1,693,244,322 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 92 | 0 | n=int(input())
x=n-10
if(x==0):
print(0)
elif(1<=x<=9):
print(4)
elif(x==10):
print(15)
elif(x==11):
print(4)
| Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
n=int(input())
x=n-10
if(x==0):
print(0)
elif(1<=x<=9):
print(4)
elif(x==10):
print(15)
elif(x==11):
print(4)
``` | 0 |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,650,014,774 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n=6
m=6
a=4
if m%a==0:
x=m//a
else:
x=m//a + 1
if n%a==0:
y=n//a
else:
y=m//a + 1
nf=x*y
print("number of flugstone",nf)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n=6
m=6
a=4
if m%a==0:
x=m//a
else:
x=m//a + 1
if n%a==0:
y=n//a
else:
y=m//a + 1
nf=x*y
print("number of flugstone",nf)
``` | 0 |
32 | B | Borze | PROGRAMMING | 800 | [
"expression parsing",
"implementation"
] | B. Borze | 2 | 256 | Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. | The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). | Output the decoded ternary number. It can have leading zeroes. | [
".-.--\n",
"--.\n",
"-..-.--\n"
] | [
"012",
"20",
"1012"
] | none | 1,000 | [
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "1000022... | 1,675,076,506 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | string = input().strip().split(".")
ans = ''
for i in string:
if(len(i) == 0):
ans += '0'
elif(len(i) == 1):
ans += '1'
elif(len(i) == 2):
ans += '2'
print(ans) | Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none | ```python
string = input().strip().split(".")
ans = ''
for i in string:
if(len(i) == 0):
ans += '0'
elif(len(i) == 1):
ans += '1'
elif(len(i) == 2):
ans += '2'
print(ans)
``` | 0 |
217 | A | Ice Skating | PROGRAMMING | 1,200 | [
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates. | The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct. | Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one. | [
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] | [
"1\n",
"0\n"
] | none | 500 | [
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182... | 1,587,516,547 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 216 | 307,200 | n = int(input())
li = []
for i in range(n):
x, y = map(int, input().split())
li.append((x, y))
cnt = 0
vis = {}
for i in range(n):
for j in range(n):
if i == j:
continue
if li[i] not in vis or li[j] not in vis:
if li[i][0] == li[j][0] or li[i][1] == li[j][1]:
cnt += 1
vis[li[i]], vis[li[j]] = True, True
print(min(n - len(vis), n - 1))
| Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none | ```python
n = int(input())
li = []
for i in range(n):
x, y = map(int, input().split())
li.append((x, y))
cnt = 0
vis = {}
for i in range(n):
for j in range(n):
if i == j:
continue
if li[i] not in vis or li[j] not in vis:
if li[i][0] == li[j][0] or li[i][1] == li[j][1]:
cnt += 1
vis[li[i]], vis[li[j]] = True, True
print(min(n - len(vis), n - 1))
``` | 0 | |
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "... | 1,574,152,829 | 2,429 | Python 3 | OK | TESTS | 25 | 248 | 0 | n,m=map(int,input().split())
s,k=0,0
for i in range(1,n+1):
s+=i
while m>s:
m-=s
for i in range(1,n+1):
if m>=i:
m-=i
else:
break
print(m)
| Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
n,m=map(int,input().split())
s,k=0,0
for i in range(1,n+1):
s+=i
while m>s:
m-=s
for i in range(1,n+1):
if m>=i:
m-=i
else:
break
print(m)
``` | 3.938 |
978 | A | Remove Duplicates | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. | The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. | In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. | [
"6\n1 5 5 1 6 1\n",
"5\n2 4 2 4 4\n",
"5\n6 6 6 6 6\n"
] | [
"3\n5 6 1 \n",
"2\n2 4 \n",
"1\n6 \n"
] | In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | 0 | [
{
"input": "6\n1 5 5 1 6 1",
"output": "3\n5 6 1 "
},
{
"input": "5\n2 4 2 4 4",
"output": "2\n2 4 "
},
{
"input": "5\n6 6 6 6 6",
"output": "1\n6 "
},
{
"input": "7\n1 2 3 4 2 2 3",
"output": "4\n1 4 2 3 "
},
{
"input": "9\n100 100 100 99 99 99 100 100 100",
... | 1,681,492,102 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | def main():
size = int(input())
numbers = tuple(map(int, input().split()))
for i in range(size - 1):
if numbers[i + 1:].count(numbers[i]) == 0:
print(numbers[i], end=' ')
print(numbers[-1])
if __name__ == '__main__':
main() | Title: Remove Duplicates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.
Output Specification:
In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
Demo Input:
['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n']
Demo Output:
['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n']
Note:
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | ```python
def main():
size = int(input())
numbers = tuple(map(int, input().split()))
for i in range(size - 1):
if numbers[i + 1:].count(numbers[i]) == 0:
print(numbers[i], end=' ')
print(numbers[-1])
if __name__ == '__main__':
main()
``` | 0 | |
400 | C | Inna and Huge Candy Matrix | PROGRAMMING | 1,500 | [
"implementation",
"math"
] | null | null | Inna and Dima decided to surprise Sereja. They brought a really huge candy matrix, it's big even for Sereja! Let's number the rows of the giant matrix from 1 to *n* from top to bottom and the columns — from 1 to *m*, from left to right. We'll represent the cell on the intersection of the *i*-th row and *j*-th column as (*i*,<=*j*). Just as is expected, some cells of the giant candy matrix contain candies. Overall the matrix has *p* candies: the *k*-th candy is at cell (*x**k*,<=*y**k*).
The time moved closer to dinner and Inna was already going to eat *p* of her favourite sweets from the matrix, when suddenly Sereja (for the reason he didn't share with anyone) rotated the matrix *x* times clockwise by 90 degrees. Then he performed the horizontal rotate of the matrix *y* times. And then he rotated the matrix *z* times counterclockwise by 90 degrees. The figure below shows how the rotates of the matrix looks like.
Inna got really upset, but Duma suddenly understood two things: the candies didn't get damaged and he remembered which cells contained Inna's favourite sweets before Sereja's strange actions. Help guys to find the new coordinates in the candy matrix after the transformation Sereja made! | The first line of the input contains fix integers *n*, *m*, *x*, *y*, *z*, *p* (1<=≤<=*n*,<=*m*<=≤<=109; 0<=≤<=*x*,<=*y*,<=*z*<=≤<=109; 1<=≤<=*p*<=≤<=105).
Each of the following *p* lines contains two integers *x**k*, *y**k* (1<=≤<=*x**k*<=≤<=*n*; 1<=≤<=*y**k*<=≤<=*m*) — the initial coordinates of the *k*-th candy. Two candies can lie on the same cell. | For each of the *p* candies, print on a single line its space-separated new coordinates. | [
"3 3 3 1 1 9\n1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n3 3\n"
] | [
"1 3\n1 2\n1 1\n2 3\n2 2\n2 1\n3 3\n3 2\n3 1\n"
] | Just for clarity. Horizontal rotating is like a mirroring of the matrix. For matrix: | 1,500 | [
{
"input": "3 3 3 1 1 9\n1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n3 3",
"output": "1 3\n1 2\n1 1\n2 3\n2 2\n2 1\n3 3\n3 2\n3 1"
},
{
"input": "5 5 0 0 0 1\n1 4",
"output": "1 4"
},
{
"input": "14 76 376219315 550904689 16684615 24\n11 21\n1 65\n5 25\n14 63\n11 30\n1 19\n5 7\n9 51\n2 49\n1... | 1,638,455,716 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 102,400 | import sys
import collections
import bisect
input = sys.stdin.readline
ip = lambda: input().rstrip('\n')
IP = lambda: list(map(int, ip().split()))
def horizontal_flip(x):
global n, m
x[1] = m+1 - x[1]
return x
def vertical_flip(x):
global n, m
x[0] = n+1 - x[0]
return x
def rotate(x, clockwise = 0):
x[0], x[1] = x[1], x[0]
if clockwise:
x = horizontal_flip(x)
else:
x = vertical_flip(x)
return x
n, m, x, y, z, p = IP()
P = []
for i in range(p):
P.append(IP())
for i in range(p):
for cnt in range(x % 4):
P[i] = rotate(P[i])
for cnt in range(y % 2):
P[i] = horizontal_flip(P[i])
for cnt in range(z % 4):
P[i] = rotate(P[i], 1)
for i in range(p):
print(*P[i]) | Title: Inna and Huge Candy Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Inna and Dima decided to surprise Sereja. They brought a really huge candy matrix, it's big even for Sereja! Let's number the rows of the giant matrix from 1 to *n* from top to bottom and the columns — from 1 to *m*, from left to right. We'll represent the cell on the intersection of the *i*-th row and *j*-th column as (*i*,<=*j*). Just as is expected, some cells of the giant candy matrix contain candies. Overall the matrix has *p* candies: the *k*-th candy is at cell (*x**k*,<=*y**k*).
The time moved closer to dinner and Inna was already going to eat *p* of her favourite sweets from the matrix, when suddenly Sereja (for the reason he didn't share with anyone) rotated the matrix *x* times clockwise by 90 degrees. Then he performed the horizontal rotate of the matrix *y* times. And then he rotated the matrix *z* times counterclockwise by 90 degrees. The figure below shows how the rotates of the matrix looks like.
Inna got really upset, but Duma suddenly understood two things: the candies didn't get damaged and he remembered which cells contained Inna's favourite sweets before Sereja's strange actions. Help guys to find the new coordinates in the candy matrix after the transformation Sereja made!
Input Specification:
The first line of the input contains fix integers *n*, *m*, *x*, *y*, *z*, *p* (1<=≤<=*n*,<=*m*<=≤<=109; 0<=≤<=*x*,<=*y*,<=*z*<=≤<=109; 1<=≤<=*p*<=≤<=105).
Each of the following *p* lines contains two integers *x**k*, *y**k* (1<=≤<=*x**k*<=≤<=*n*; 1<=≤<=*y**k*<=≤<=*m*) — the initial coordinates of the *k*-th candy. Two candies can lie on the same cell.
Output Specification:
For each of the *p* candies, print on a single line its space-separated new coordinates.
Demo Input:
['3 3 3 1 1 9\n1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n3 3\n']
Demo Output:
['1 3\n1 2\n1 1\n2 3\n2 2\n2 1\n3 3\n3 2\n3 1\n']
Note:
Just for clarity. Horizontal rotating is like a mirroring of the matrix. For matrix: | ```python
import sys
import collections
import bisect
input = sys.stdin.readline
ip = lambda: input().rstrip('\n')
IP = lambda: list(map(int, ip().split()))
def horizontal_flip(x):
global n, m
x[1] = m+1 - x[1]
return x
def vertical_flip(x):
global n, m
x[0] = n+1 - x[0]
return x
def rotate(x, clockwise = 0):
x[0], x[1] = x[1], x[0]
if clockwise:
x = horizontal_flip(x)
else:
x = vertical_flip(x)
return x
n, m, x, y, z, p = IP()
P = []
for i in range(p):
P.append(IP())
for i in range(p):
for cnt in range(x % 4):
P[i] = rotate(P[i])
for cnt in range(y % 2):
P[i] = horizontal_flip(P[i])
for cnt in range(z % 4):
P[i] = rotate(P[i], 1)
for i in range(p):
print(*P[i])
``` | 0 | |
908 | A | New Year and Counting Cards | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true. | The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit. | Print a single integer, the minimum number of cards you must turn over to verify your claim. | [
"ee\n",
"z\n",
"0ay1\n"
] | [
"2\n",
"0\n",
"2\n"
] | In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | 500 | [
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01... | 1,624,393,339 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | s = list(input())
ans = 0
v = ['a','e','i','o','u']
n = ['0', '2', '4', '6', '8']
for i in s:
if i in v or i in n:
ans += 1
print(ans)
| Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | ```python
s = list(input())
ans = 0
v = ['a','e','i','o','u']
n = ['0', '2', '4', '6', '8']
for i in s:
if i in v or i in n:
ans += 1
print(ans)
``` | 0 | |
108 | A | Palindromic Times | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | A. Palindromic Times | 2 | 256 | Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him. | The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits. | Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time. | [
"12:21\n",
"23:59\n"
] | [
"13:31\n",
"00:00\n"
] | none | 500 | [
{
"input": "12:21",
"output": "13:31"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "15:51",
"output": "20:02"
},
{
"input": "10:44",
"output": "11:11"
},
{
"input": "04:02",
"output": "04:40"
},
{
"input": "02:11",
"output": "02:20"
},
... | 1,632,385,345 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 124 | 0 | s=input("")
l=s.split(":")
rev=l[0][::-1]
nrev,ntime=int(rev),int(l[1])
if(int(l[0])<=23):
if(nrev<60 and ntime+10<=60):
if(nrev<=ntime):
out=str(int(l[0])+1)+":"+str(nrev+10)
if(len(out)==5):
print(out)
else:
print("0"+out)
else:
print(l[0]+":"+rev)
else:
if(int(l[0])>=15):
print("20:02")
else:
print("10:01")
else:
print("00:00")
| Title: Palindromic Times
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
Input Specification:
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
Output Specification:
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
Demo Input:
['12:21\n', '23:59\n']
Demo Output:
['13:31\n', '00:00\n']
Note:
none | ```python
s=input("")
l=s.split(":")
rev=l[0][::-1]
nrev,ntime=int(rev),int(l[1])
if(int(l[0])<=23):
if(nrev<60 and ntime+10<=60):
if(nrev<=ntime):
out=str(int(l[0])+1)+":"+str(nrev+10)
if(len(out)==5):
print(out)
else:
print("0"+out)
else:
print(l[0]+":"+rev)
else:
if(int(l[0])>=15):
print("20:02")
else:
print("10:01")
else:
print("00:00")
``` | 0 |
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"inpu... | 1,664,376,060 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 92 | 0 | # from math import gcd
a, b, n = map( int, input().split() )
# def nod(x, y):
# while(y):
# x, x = y, x % y
# return x
def gcd_rem_division(num1, num2):
while num1 != 0 and num2 != 0:
if num1 >= num2:
num1 %= num2
else:
num2 %= num1
return num1 or num2
last_move = True
while n > 0:
n -= gcd_rem_division(a, n) if last_move else gcd_rem_division(b, n)
last_move = False if last_move else True
print(1) if last_move else print(0)
| Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
# from math import gcd
a, b, n = map( int, input().split() )
# def nod(x, y):
# while(y):
# x, x = y, x % y
# return x
def gcd_rem_division(num1, num2):
while num1 != 0 and num2 != 0:
if num1 >= num2:
num1 %= num2
else:
num2 %= num1
return num1 or num2
last_move = True
while n > 0:
n -= gcd_rem_division(a, n) if last_move else gcd_rem_division(b, n)
last_move = False if last_move else True
print(1) if last_move else print(0)
``` | 3 | |
895 | A | Pizza Separation | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360. | Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya. | [
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] | [
"0\n",
"40\n",
"360\n",
"0\n"
] | In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | 500 | [
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 4... | 1,514,994,855 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 1,000 | 53,964,800 | def main():
x = int(input())
a = list(map(int,input().split(" ")))
while x != 4:
a.append(0)
x = x + 1
a = sorted(a)
print(abs(a[0]+a[3]-a[1]-a[2]))
main()
| Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | ```python
def main():
x = int(input())
a = list(map(int,input().split(" ")))
while x != 4:
a.append(0)
x = x + 1
a = sorted(a)
print(abs(a[0]+a[3]-a[1]-a[2]))
main()
``` | 0 | |
114 | A | Cifera | PROGRAMMING | 1,000 | [
"math"
] | null | null | When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it. | The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1). | You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*. | [
"5\n25\n",
"3\n8\n"
] | [
"YES\n1\n",
"NO\n"
] | none | 500 | [
{
"input": "5\n25",
"output": "YES\n1"
},
{
"input": "3\n8",
"output": "NO"
},
{
"input": "123\n123",
"output": "YES\n0"
},
{
"input": "99\n970300",
"output": "NO"
},
{
"input": "1000\n6666666",
"output": "NO"
},
{
"input": "59\n3571",
"output": "N... | 1,673,862,497 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 64 | 124 | 0 | k = int(input())
l = int(input())
p = k
importance = 0
cifiera = False
if k == 1:
cifiera = (k == l)
else:
while p < 2 ** 31:
if p == l:
cifiera = True
break
p *= k
importance += 1
if cifiera:
print('YES')
print(importance)
else:
print('NO')
| Title: Cifera
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Petya went to school, he got interested in large numbers and what they were called in ancient times. For instance, he learned that the Russian word "tma" (which now means "too much to be counted") used to stand for a thousand and "tma tmyschaya" (which literally means "the tma of tmas") used to stand for a million.
Petya wanted to modernize the words we use for numbers and invented a word petricium that represents number *k*. Moreover, petricium la petricium stands for number *k*2, petricium la petricium la petricium stands for *k*3 and so on. All numbers of this form are called petriciumus cifera, and the number's importance is the number of articles la in its title.
Petya's invention brought on a challenge that needed to be solved quickly: does some number *l* belong to the set petriciumus cifera? As Petya is a very busy schoolboy he needs to automate the process, he asked you to solve it.
Input Specification:
The first input line contains integer number *k*, the second line contains integer number *l* (2<=≤<=*k*,<=*l*<=≤<=231<=-<=1).
Output Specification:
You should print in the first line of the output "YES", if the number belongs to the set petriciumus cifera and otherwise print "NO". If the number belongs to the set, then print on the seconds line the only number — the importance of number *l*.
Demo Input:
['5\n25\n', '3\n8\n']
Demo Output:
['YES\n1\n', 'NO\n']
Note:
none | ```python
k = int(input())
l = int(input())
p = k
importance = 0
cifiera = False
if k == 1:
cifiera = (k == l)
else:
while p < 2 ** 31:
if p == l:
cifiera = True
break
p *= k
importance += 1
if cifiera:
print('YES')
print(importance)
else:
print('NO')
``` | 3 | |
680 | A | Bear and Five Cards | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | A little bear Limak plays a game. He has five cards. There is one number written on each card. Each number is a positive integer.
Limak can discard (throw out) some cards. His goal is to minimize the sum of numbers written on remaining (not discarded) cards.
He is allowed to at most once discard two or three cards with the same number. Of course, he won't discard cards if it's impossible to choose two or three cards with the same number.
Given five numbers written on cards, cay you find the minimum sum of numbers on remaining cards? | The only line of the input contains five integers *t*1, *t*2, *t*3, *t*4 and *t*5 (1<=≤<=*t**i*<=≤<=100) — numbers written on cards. | Print the minimum possible sum of numbers written on remaining cards. | [
"7 3 7 3 20\n",
"7 9 3 1 8\n",
"10 10 10 10 10\n"
] | [
"26\n",
"28\n",
"20\n"
] | In the first sample, Limak has cards with numbers 7, 3, 7, 3 and 20. Limak can do one of the following.
- Do nothing and the sum would be 7 + 3 + 7 + 3 + 20 = 40. - Remove two cards with a number 7. The remaining sum would be 3 + 3 + 20 = 26. - Remove two cards with a number 3. The remaining sum would be 7 + 7 + 20 = 34.
You are asked to minimize the sum so the answer is 26.
In the second sample, it's impossible to find two or three cards with the same number. Hence, Limak does nothing and the sum is 7 + 9 + 1 + 3 + 8 = 28.
In the third sample, all cards have the same number. It's optimal to discard any three cards. The sum of two remaining numbers is 10 + 10 = 20. | 500 | [
{
"input": "7 3 7 3 20",
"output": "26"
},
{
"input": "7 9 3 1 8",
"output": "28"
},
{
"input": "10 10 10 10 10",
"output": "20"
},
{
"input": "8 7 1 8 7",
"output": "15"
},
{
"input": "7 7 7 8 8",
"output": "16"
},
{
"input": "8 8 8 2 2",
"output"... | 1,600,349,121 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 139 | 0 | import math
def main():
a = list(map(int, input().split()))
if len(a) == len(set(a)):
print(sum(a))
else:
s = sum(a)
b = set(a)
b = list(b)
b.sort()
b.reverse()
for x in b:
if a.count(x) >= 3:
s -= x * 3
break
elif a.count(x) == 2:
s -= 2 * x
break
print(s)
if __name__ == '__main__':
main()
| Title: Bear and Five Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A little bear Limak plays a game. He has five cards. There is one number written on each card. Each number is a positive integer.
Limak can discard (throw out) some cards. His goal is to minimize the sum of numbers written on remaining (not discarded) cards.
He is allowed to at most once discard two or three cards with the same number. Of course, he won't discard cards if it's impossible to choose two or three cards with the same number.
Given five numbers written on cards, cay you find the minimum sum of numbers on remaining cards?
Input Specification:
The only line of the input contains five integers *t*1, *t*2, *t*3, *t*4 and *t*5 (1<=≤<=*t**i*<=≤<=100) — numbers written on cards.
Output Specification:
Print the minimum possible sum of numbers written on remaining cards.
Demo Input:
['7 3 7 3 20\n', '7 9 3 1 8\n', '10 10 10 10 10\n']
Demo Output:
['26\n', '28\n', '20\n']
Note:
In the first sample, Limak has cards with numbers 7, 3, 7, 3 and 20. Limak can do one of the following.
- Do nothing and the sum would be 7 + 3 + 7 + 3 + 20 = 40. - Remove two cards with a number 7. The remaining sum would be 3 + 3 + 20 = 26. - Remove two cards with a number 3. The remaining sum would be 7 + 7 + 20 = 34.
You are asked to minimize the sum so the answer is 26.
In the second sample, it's impossible to find two or three cards with the same number. Hence, Limak does nothing and the sum is 7 + 9 + 1 + 3 + 8 = 28.
In the third sample, all cards have the same number. It's optimal to discard any three cards. The sum of two remaining numbers is 10 + 10 = 20. | ```python
import math
def main():
a = list(map(int, input().split()))
if len(a) == len(set(a)):
print(sum(a))
else:
s = sum(a)
b = set(a)
b = list(b)
b.sort()
b.reverse()
for x in b:
if a.count(x) >= 3:
s -= x * 3
break
elif a.count(x) == 2:
s -= 2 * x
break
print(s)
if __name__ == '__main__':
main()
``` | 0 | |
672 | B | Different is Good | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation",
"strings"
] | null | null | A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters. | If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes. | [
"2\naa\n",
"4\nkoko\n",
"5\nmurat\n"
] | [
"1\n",
"2\n",
"0\n"
] | In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko". | 1,000 | [
{
"input": "2\naa",
"output": "1"
},
{
"input": "4\nkoko",
"output": "2"
},
{
"input": "5\nmurat",
"output": "0"
},
{
"input": "6\nacbead",
"output": "1"
},
{
"input": "7\ncdaadad",
"output": "4"
},
{
"input": "25\npeoaicnbisdocqofsqdpgobpn",
"outp... | 1,563,368,919 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 124 | 204,800 | n=int(input())
s=input()
freq=[0]*26
for i in s:
freq[ord(i)-ord('a')]+=1
change=0
for i in range(26):
if freq[i]:
change+=freq[i]-1
print(change)
| Title: Different is Good
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
Output Specification:
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
Demo Input:
['2\naa\n', '4\nkoko\n', '5\nmurat\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko". | ```python
n=int(input())
s=input()
freq=[0]*26
for i in s:
freq[ord(i)-ord('a')]+=1
change=0
for i in range(26):
if freq[i]:
change+=freq[i]-1
print(change)
``` | 0 | |
287 | A | IQ Test | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color. | Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black. | Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise. | [
"####\n.#..\n####\n....\n",
"####\n....\n####\n....\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | 500 | [
{
"input": "###.\n...#\n###.\n...#",
"output": "NO"
},
{
"input": ".##.\n#..#\n.##.\n#..#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.#.",
"output": "NO"
},
{
"input": "##..\n..##\n##..\n..##",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n.#.#\n.#.#",
"ou... | 1,584,817,558 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 124 | 307,200 | n = 4
vois = 1
def nb_same_square(mat,x,y):
nb_sign1 = 0
for dx in range(vois + 1):
for dy in range(vois + 1):
elem = mat[x + dx][y + dy]
nb_sign1 += (1 if elem == '#' else 0)
nb_sign2 = 4 - nb_sign1
max_same = max(nb_sign1,nb_sign2)
return max_same
def solve(mat):
for i in range(n - 1):
for j in range(n - 1):
if nb_same_square(mat,i,j) >= 3:
print('YES')
return
print('NO')
mat = [[None for i in range(n)] for j in range(n)]
for i in range(n):
s = input()
for j in range(n):
mat[i][j] = s[j]
solve(mat) | Title: IQ Test
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.
Input Specification:
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
Output Specification:
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
Demo Input:
['####\n.#..\n####\n....\n', '####\n....\n####\n....\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | ```python
n = 4
vois = 1
def nb_same_square(mat,x,y):
nb_sign1 = 0
for dx in range(vois + 1):
for dy in range(vois + 1):
elem = mat[x + dx][y + dy]
nb_sign1 += (1 if elem == '#' else 0)
nb_sign2 = 4 - nb_sign1
max_same = max(nb_sign1,nb_sign2)
return max_same
def solve(mat):
for i in range(n - 1):
for j in range(n - 1):
if nb_same_square(mat,i,j) >= 3:
print('YES')
return
print('NO')
mat = [[None for i in range(n)] for j in range(n)]
for i in range(n):
s = input()
for j in range(n):
mat[i][j] = s[j]
solve(mat)
``` | 3 | |
913 | A | Modular Exponentiation | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108). | Output a single integer — the value of . | [
"4\n42\n",
"1\n58\n",
"98765432\n23456789\n"
] | [
"10\n",
"0\n",
"23456789\n"
] | In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | 500 | [
{
"input": "4\n42",
"output": "10"
},
{
"input": "1\n58",
"output": "0"
},
{
"input": "98765432\n23456789",
"output": "23456789"
},
{
"input": "8\n88127381",
"output": "149"
},
{
"input": "32\n92831989",
"output": "92831989"
},
{
"input": "92831989\n25... | 1,638,953,205 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 1,000 | 45,875,200 |
# from time import time as t
n, m = int(input()), int(input())
# s = t()
print(pow(m, 1, pow(2, n)))
# print(t()-s) | Title: Modular Exponentiation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108).
Output Specification:
Output a single integer — the value of .
Demo Input:
['4\n42\n', '1\n58\n', '98765432\n23456789\n']
Demo Output:
['10\n', '0\n', '23456789\n']
Note:
In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | ```python
# from time import time as t
n, m = int(input()), int(input())
# s = t()
print(pow(m, 1, pow(2, n)))
# print(t()-s)
``` | 0 | |
444 | C | DZY Loves Colors | PROGRAMMING | 2,400 | [
"data structures"
] | null | null | DZY loves colors, and he enjoys painting.
On a colorful day, DZY gets a colorful ribbon, which consists of *n* units (they are numbered from 1 to *n* from left to right). The color of the *i*-th unit of the ribbon is *i* at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.
DZY loves painting, we know. He takes up a paintbrush with color *x* and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit *i* currently is *y*. When it is painted by this paintbrush, the color of the unit becomes *x*, and the colorfulness of the unit increases by |*x*<=-<=*y*|.
DZY wants to perform *m* operations, each operation can be one of the following:
1. Paint all the units with numbers between *l* and *r* (both inclusive) with color *x*. 1. Ask the sum of colorfulness of the units between *l* and *r* (both inclusive).
Can you help DZY? | The first line contains two space-separated integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105).
Each of the next *m* lines begins with a integer *type* (1<=≤<=*type*<=≤<=2), which represents the type of this operation.
If *type*<==<=1, there will be 3 more integers *l*,<=*r*,<=*x* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*x*<=≤<=108) in this line, describing an operation 1.
If *type*<==<=2, there will be 2 more integers *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) in this line, describing an operation 2. | For each operation 2, print a line containing the answer — sum of colorfulness. | [
"3 3\n1 1 2 4\n1 2 3 5\n2 1 3\n",
"3 4\n1 1 3 4\n2 1 1\n2 2 2\n2 3 3\n",
"10 6\n1 1 5 3\n1 2 7 9\n1 10 10 11\n1 3 8 12\n1 1 10 3\n2 1 10\n"
] | [
"8\n",
"3\n2\n1\n",
"129\n"
] | In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].
After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].
After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].
So the answer to the only operation of type 2 is 8. | 2,000 | [
{
"input": "3 3\n1 1 2 4\n1 2 3 5\n2 1 3",
"output": "8"
},
{
"input": "3 4\n1 1 3 4\n2 1 1\n2 2 2\n2 3 3",
"output": "3\n2\n1"
},
{
"input": "10 6\n1 1 5 3\n1 2 7 9\n1 10 10 11\n1 3 8 12\n1 1 10 3\n2 1 10",
"output": "129"
},
{
"input": "3 3\n1 2 2 31844623\n1 1 2 37662529\n... | 1,689,251,538 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1689251538.034106")# 1689251538.034152 | Title: DZY Loves Colors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves colors, and he enjoys painting.
On a colorful day, DZY gets a colorful ribbon, which consists of *n* units (they are numbered from 1 to *n* from left to right). The color of the *i*-th unit of the ribbon is *i* at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.
DZY loves painting, we know. He takes up a paintbrush with color *x* and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit *i* currently is *y*. When it is painted by this paintbrush, the color of the unit becomes *x*, and the colorfulness of the unit increases by |*x*<=-<=*y*|.
DZY wants to perform *m* operations, each operation can be one of the following:
1. Paint all the units with numbers between *l* and *r* (both inclusive) with color *x*. 1. Ask the sum of colorfulness of the units between *l* and *r* (both inclusive).
Can you help DZY?
Input Specification:
The first line contains two space-separated integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105).
Each of the next *m* lines begins with a integer *type* (1<=≤<=*type*<=≤<=2), which represents the type of this operation.
If *type*<==<=1, there will be 3 more integers *l*,<=*r*,<=*x* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*x*<=≤<=108) in this line, describing an operation 1.
If *type*<==<=2, there will be 2 more integers *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) in this line, describing an operation 2.
Output Specification:
For each operation 2, print a line containing the answer — sum of colorfulness.
Demo Input:
['3 3\n1 1 2 4\n1 2 3 5\n2 1 3\n', '3 4\n1 1 3 4\n2 1 1\n2 2 2\n2 3 3\n', '10 6\n1 1 5 3\n1 2 7 9\n1 10 10 11\n1 3 8 12\n1 1 10 3\n2 1 10\n']
Demo Output:
['8\n', '3\n2\n1\n', '129\n']
Note:
In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].
After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].
After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].
So the answer to the only operation of type 2 is 8. | ```python
print("_RANDOM_GUESS_1689251538.034106")# 1689251538.034152
``` | 0 | |
508 | A | Pasha and Pixels | PROGRAMMING | 1,100 | [
"brute force"
] | null | null | Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed. | The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move. | If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0. | [
"2 2 4\n1 1\n1 2\n2 1\n2 2\n",
"2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n",
"5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n"
] | [
"4\n",
"5\n",
"0\n"
] | none | 500 | [
{
"input": "2 2 4\n1 1\n1 2\n2 1\n2 2",
"output": "4"
},
{
"input": "2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1",
"output": "5"
},
{
"input": "5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2",
"output": "0"
},
{
"input": "3 3 11\n2 1\n3 1\n1 1\n1 3\n1 2\n2 3\n3 3\n3 2\n2 2\n1 3\n3 3",
... | 1,697,698,268 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 0 | # -*- coding: utf-8 -*-
"""
Created on Wed Oct 18 16:18:28 2023
@author: masiyu004
"""
def y(a,b,x):
for i in [(1,1),(1,-1),(-1,1),(-1,-1)]:
nexta=a+i[0]
nextb=b+i[1]
if 0<=nexta<=n-1 and 0<=nextb<=m-1:
test1=x[nexta][nextb]
test2=x[nexta][b]
test3=x[a][nextb]
if (test1,test2,test3)==(1,1,1):
return True
return False
import numpy
n,m,k=map(int,input().split())
x=numpy.zeros(shape=(n+2,m+2))
s=0
for i in range(k):
a,b=map(int,input().split())
x[a-1][b-1]=1
if s==0 and y(a,b,x):
s=i+1
print(s)
| Title: Pasha and Pixels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
Input Specification:
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
Output Specification:
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
Demo Input:
['2 2 4\n1 1\n1 2\n2 1\n2 2\n', '2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n', '5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n']
Demo Output:
['4\n', '5\n', '0\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 18 16:18:28 2023
@author: masiyu004
"""
def y(a,b,x):
for i in [(1,1),(1,-1),(-1,1),(-1,-1)]:
nexta=a+i[0]
nextb=b+i[1]
if 0<=nexta<=n-1 and 0<=nextb<=m-1:
test1=x[nexta][nextb]
test2=x[nexta][b]
test3=x[a][nextb]
if (test1,test2,test3)==(1,1,1):
return True
return False
import numpy
n,m,k=map(int,input().split())
x=numpy.zeros(shape=(n+2,m+2))
s=0
for i in range(k):
a,b=map(int,input().split())
x[a-1][b-1]=1
if s==0 and y(a,b,x):
s=i+1
print(s)
``` | -1 | |
137 | A | Postcards and photos | PROGRAMMING | 900 | [
"implementation"
] | null | null | Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items? | The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100 characters. If the *i*-th character in the string is the letter "С", that means that the *i*-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the *i*-th character is the letter "P", than the *i*-th object on the wall is a photo. | Print the only number — the minimum number of times Polycarpus has to visit the closet. | [
"CPCPCPC\n",
"CCCCCCPPPPPP\n",
"CCCCCCPPCPPPPPPPPPP\n",
"CCCCCCCCCC\n"
] | [
"7\n",
"4\n",
"6\n",
"2\n"
] | In the first sample Polycarpus needs to take one item to the closet 7 times.
In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.
In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.
In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go). | 500 | [
{
"input": "CPCPCPC",
"output": "7"
},
{
"input": "CCCCCCPPPPPP",
"output": "4"
},
{
"input": "CCCCCCPPCPPPPPPPPPP",
"output": "6"
},
{
"input": "CCCCCCCCCC",
"output": "2"
},
{
"input": "CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC... | 1,531,914,531 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 248 | 0 | import sys
import math
s=str(input())
c,p,cona=0,0,0
for i in range(len(s)):
if(s[i]=='C'):
c+=1
if(p!=0):
cona+=p//5
if(p%5!=0): cona+=1
p=0
else:
p+=1
if(c!=0):
cona+=c//5
if (c%5!=0): cona+=1
c=0
if(c!=0 or p!=0):
cona+=c//5+p//5
if(c%5!=0 or p%5!=0): cona+=1
print(cona)
| Title: Postcards and photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items?
Input Specification:
The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100 characters. If the *i*-th character in the string is the letter "С", that means that the *i*-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the *i*-th character is the letter "P", than the *i*-th object on the wall is a photo.
Output Specification:
Print the only number — the minimum number of times Polycarpus has to visit the closet.
Demo Input:
['CPCPCPC\n', 'CCCCCCPPPPPP\n', 'CCCCCCPPCPPPPPPPPPP\n', 'CCCCCCCCCC\n']
Demo Output:
['7\n', '4\n', '6\n', '2\n']
Note:
In the first sample Polycarpus needs to take one item to the closet 7 times.
In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.
In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.
In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go). | ```python
import sys
import math
s=str(input())
c,p,cona=0,0,0
for i in range(len(s)):
if(s[i]=='C'):
c+=1
if(p!=0):
cona+=p//5
if(p%5!=0): cona+=1
p=0
else:
p+=1
if(c!=0):
cona+=c//5
if (c%5!=0): cona+=1
c=0
if(c!=0 or p!=0):
cona+=c//5+p//5
if(c%5!=0 or p%5!=0): cona+=1
print(cona)
``` | 3 | |
855 | A | Tom Riddle's Diary | PROGRAMMING | 800 | [
"brute force",
"implementation",
"strings"
] | null | null | Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes). | First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100. | Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower). | [
"6\ntom\nlucius\nginny\nharry\nginny\nharry\n",
"3\na\na\na\n"
] | [
"NO\nNO\nNO\nNO\nYES\nYES\n",
"NO\nYES\nYES\n"
] | In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 is "YES". | 500 | [
{
"input": "6\ntom\nlucius\nginny\nharry\nginny\nharry",
"output": "NO\nNO\nNO\nNO\nYES\nYES"
},
{
"input": "3\na\na\na",
"output": "NO\nYES\nYES"
},
{
"input": "1\nzn",
"output": "NO"
},
{
"input": "9\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnb... | 1,651,937,711 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 62 | 102,400 |
from sys import stdin
from math import log, floor, ceil, gcd
from collections import defaultdict as dd
#from bisect import bisect_left as bl,bisect_right as br
input = stdin.readline
def inp(): return int(stdin.readline())
def rs(): return stdin.readline().strip()
def ra(typ): return list(map(typ, stdin.readline().split()))
def rv(typ): return map(typ, stdin.readline().split())
#mod = 1000000007
def main():
x = set()
for _ in range(inp()):
s = rs()
if s in x:
print("YES")
continue
x.add(s)
print("NO")
main()
| Title: Tom Riddle's Diary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes).
Input Specification:
First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100.
Output Specification:
Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower).
Demo Input:
['6\ntom\nlucius\nginny\nharry\nginny\nharry\n', '3\na\na\na\n']
Demo Output:
['NO\nNO\nNO\nNO\nYES\nYES\n', 'NO\nYES\nYES\n']
Note:
In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 is "YES". | ```python
from sys import stdin
from math import log, floor, ceil, gcd
from collections import defaultdict as dd
#from bisect import bisect_left as bl,bisect_right as br
input = stdin.readline
def inp(): return int(stdin.readline())
def rs(): return stdin.readline().strip()
def ra(typ): return list(map(typ, stdin.readline().split()))
def rv(typ): return map(typ, stdin.readline().split())
#mod = 1000000007
def main():
x = set()
for _ in range(inp()):
s = rs()
if s in x:
print("YES")
continue
x.add(s)
print("NO")
main()
``` | 3 | |
267 | A | Subtractions | PROGRAMMING | 900 | [
"math",
"number theory"
] | null | null | You've got two numbers. As long as they are both larger than zero, they go through the same operation: subtract the lesser number from the larger one. If they equal substract one number from the another. For example, one operation transforms pair (4,17) to pair (4,13), it transforms (5,5) to (0,5).
You've got some number of pairs (*a**i*,<=*b**i*). How many operations will be performed for each of them? | The first line contains the number of pairs *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000). Then follow *n* lines, each line contains a pair of positive integers *a**i*,<=*b**i* (1<=<=≤<=<=*a**i*,<=<=*b**i*<=<=≤<=<=109). | Print the sought number of operations for each pair on a single line. | [
"2\n4 17\n7 987654321\n"
] | [
"8\n141093479\n"
] | none | 500 | [
{
"input": "2\n4 17\n7 987654321",
"output": "8\n141093479"
},
{
"input": "10\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321",
"output": "141093479\n141093479\n141093479\n141093479\n141093479\n141093479\n141093479... | 1,683,454,550 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 1,000 | 0 | n=int(input())
for i in range(0,n):
a=list(map(int,input().split()))
a.sort()
x=a[0]
y=a[1]
c=0
while(x!=0 and y!=0):
y=y-x
if x>=y:
t=x
x=y
y=t
c=c+1
print(c) | Title: Subtractions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got two numbers. As long as they are both larger than zero, they go through the same operation: subtract the lesser number from the larger one. If they equal substract one number from the another. For example, one operation transforms pair (4,17) to pair (4,13), it transforms (5,5) to (0,5).
You've got some number of pairs (*a**i*,<=*b**i*). How many operations will be performed for each of them?
Input Specification:
The first line contains the number of pairs *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000). Then follow *n* lines, each line contains a pair of positive integers *a**i*,<=*b**i* (1<=<=≤<=<=*a**i*,<=<=*b**i*<=<=≤<=<=109).
Output Specification:
Print the sought number of operations for each pair on a single line.
Demo Input:
['2\n4 17\n7 987654321\n']
Demo Output:
['8\n141093479\n']
Note:
none | ```python
n=int(input())
for i in range(0,n):
a=list(map(int,input().split()))
a.sort()
x=a[0]
y=a[1]
c=0
while(x!=0 and y!=0):
y=y-x
if x>=y:
t=x
x=y
y=t
c=c+1
print(c)
``` | 0 | |
855 | B | Marvolo Gaunt's Ring | PROGRAMMING | 1,500 | [
"brute force",
"data structures",
"dp"
] | null | null | Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative. | First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). | Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. | [
"5 1 2 3\n1 2 3 4 5\n",
"5 1 2 -3\n-1 -2 -3 -4 -5\n"
] | [
"30\n",
"12\n"
] | In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | 1,000 | [
{
"input": "5 1 2 3\n1 2 3 4 5",
"output": "30"
},
{
"input": "5 1 2 -3\n-1 -2 -3 -4 -5",
"output": "12"
},
{
"input": "5 886327859 82309257 -68295239\n-731225382 354766539 -48222231 -474691998 360965777",
"output": "376059240645059046"
},
{
"input": "4 -96405765 -495906217 6... | 1,620,935,812 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 4 | 202 | 30,617,600 | from sys import stdin,stdout
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
NI=float('-inf')
def fn(pos,state):
if pos>=n:
if state in [0,1,2]:return NI
return 0
if (pos,state) in dp:return dp[pos,state]
if state==0:
ans=max(fn(pos+1,0),a[pos]*p+fn(pos,1))
elif state==1:
ans=max(fn(pos+1,1),a[pos]*q+fn(pos,2))
else:
ans=max(fn(pos+1,2),a[pos]*r)
dp[pos,state]=ans
return ans
for _ in range(1):#nmbr()):
dp={}
n,p,q,r=lst()
a=lst()
print(fn(0,0)) | Title: Marvolo Gaunt's Ring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative.
Input Specification:
First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*.
Demo Input:
['5 1 2 3\n1 2 3 4 5\n', '5 1 2 -3\n-1 -2 -3 -4 -5\n']
Demo Output:
['30\n', '12\n']
Note:
In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | ```python
from sys import stdin,stdout
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
NI=float('-inf')
def fn(pos,state):
if pos>=n:
if state in [0,1,2]:return NI
return 0
if (pos,state) in dp:return dp[pos,state]
if state==0:
ans=max(fn(pos+1,0),a[pos]*p+fn(pos,1))
elif state==1:
ans=max(fn(pos+1,1),a[pos]*q+fn(pos,2))
else:
ans=max(fn(pos+1,2),a[pos]*r)
dp[pos,state]=ans
return ans
for _ in range(1):#nmbr()):
dp={}
n,p,q,r=lst()
a=lst()
print(fn(0,0))
``` | -1 | |
343 | B | Alternating Current | PROGRAMMING | 1,600 | [
"data structures",
"greedy",
"implementation"
] | null | null | Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples. | The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise. | Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled. | [
"-++-\n",
"+-\n",
"++\n",
"-\n"
] | [
"Yes\n",
"No\n",
"Yes\n",
"No\n"
] | The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself: | 1,000 | [
{
"input": "-++-",
"output": "Yes"
},
{
"input": "+-",
"output": "No"
},
{
"input": "++",
"output": "Yes"
},
{
"input": "-",
"output": "No"
},
{
"input": "+-+-",
"output": "No"
},
{
"input": "-+-",
"output": "No"
},
{
"input": "-++-+--+",
... | 1,591,812,733 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 216 | 0 | string = input()
if len(string)==1:
print('No')
else:
if string[0] == string[-1]:
plus = 0
minus = 0
for i in range(len(string)):
if string[i]=='+':
plus+=1
else:
minus+=1
if plus%2 == 0 and minus%2==0:
print('Yes')
else:
print('No')
else:
print('NO') | Title: Alternating Current
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
Input Specification:
The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
Output Specification:
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
Demo Input:
['-++-\n', '+-\n', '++\n', '-\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n', 'No\n']
Note:
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself: | ```python
string = input()
if len(string)==1:
print('No')
else:
if string[0] == string[-1]:
plus = 0
minus = 0
for i in range(len(string)):
if string[i]=='+':
plus+=1
else:
minus+=1
if plus%2 == 0 and minus%2==0:
print('Yes')
else:
print('No')
else:
print('NO')
``` | 0 | |
386 | A | Second-Price Auction | PROGRAMMING | 800 | [
"implementation"
] | null | null | In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder. | The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based. | [
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] | [
"2 5\n",
"1 8\n",
"6 9\n"
] | none | 500 | [
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{... | 1,606,475,011 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 109 | 307,200 | n = int(input())
nums = [int(j) for j in input().split()]
ref = 10000 * [-1]
for j in range(n):
ref[nums[j] - 1] = j + 1
index, pointer = -1, 9999
while pointer >= 0:
if ref[pointer] != -1:
index = ref[pointer]
pointer -= 1
break
pointer -= 1
value = -1
while pointer >= 0:
if ref[pointer] != -1:
value = pointer + 1
break
pointer -= 1
print(index, value)
| Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none | ```python
n = int(input())
nums = [int(j) for j in input().split()]
ref = 10000 * [-1]
for j in range(n):
ref[nums[j] - 1] = j + 1
index, pointer = -1, 9999
while pointer >= 0:
if ref[pointer] != -1:
index = ref[pointer]
pointer -= 1
break
pointer -= 1
value = -1
while pointer >= 0:
if ref[pointer] != -1:
value = pointer + 1
break
pointer -= 1
print(index, value)
``` | 3 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4... | 1,641,905,340 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 13,516,800 | n = int(input())
numbers = [x for x in map(int, input().split())]
srebmun = numbers[::-1]
m = int(input())
searches = [x for x in map(int, input().split())]
a = 0
b = 0
for item in searches:
a += numbers.index(item) + 1
b += srebmun.index(item) + 1
print (a, b) | Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
n = int(input())
numbers = [x for x in map(int, input().split())]
srebmun = numbers[::-1]
m = int(input())
searches = [x for x in map(int, input().split())]
a = 0
b = 0
for item in searches:
a += numbers.index(item) + 1
b += srebmun.index(item) + 1
print (a, b)
``` | 0 | |
4 | A | Watermelon | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Watermelon | 1 | 64 | One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. | The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. | Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. | [
"8\n"
] | [
"YES\n"
] | For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | 0 | [
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"... | 1,688,741,619 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 92 | 0 | n = int(input())
if n%2 == 0:
x = n - 2
y = n - x
if x % 2 == 0 and y % 2 == 0 :
print('YES')
else:
print('NO')
else:
print('NO')
| Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | ```python
n = int(input())
if n%2 == 0:
x = n - 2
y = n - x
if x % 2 == 0 and y % 2 == 0 :
print('YES')
else:
print('NO')
else:
print('NO')
``` | 0 |
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
... | 1,684,710,006 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 124 | 0 | def main():
original_points = 10
needed = int(input())
dict = {1: 4, 11: 4, 10: 15, 2: 4, 3: 4, 4: 4, 5: 4, 6: 4, 7: 4, 8: 4, 9: 4}
to_find = needed - original_points
if to_find not in dict:
print(0)
else:
print(dict[to_find])
main()
| Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
def main():
original_points = 10
needed = int(input())
dict = {1: 4, 11: 4, 10: 15, 2: 4, 3: 4, 4: 4, 5: 4, 6: 4, 7: 4, 8: 4, 9: 4}
to_find = needed - original_points
if to_find not in dict:
print(0)
else:
print(dict[to_find])
main()
``` | 3.969 |
965 | A | Paper Airplanes | PROGRAMMING | 800 | [
"math"
] | null | null | To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy? | The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively. | Print a single integer — the minimum number of packs they should buy. | [
"5 3 2 3\n",
"5 3 100 1\n"
] | [
"4\n",
"5\n"
] | In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets. | 500 | [
{
"input": "5 3 2 3",
"output": "4"
},
{
"input": "5 3 100 1",
"output": "5"
},
{
"input": "10000 10000 1 1",
"output": "100000000"
},
{
"input": "1 1 10000 10000",
"output": "1"
},
{
"input": "300 300 21 23",
"output": "196"
},
{
"input": "300 2 37 51... | 1,579,607,186 | 2,147,483,647 | Python 3 | OK | TESTS | 18 | 109 | 307,200 | import math
if __name__ == '__main__':
k, n, s, p = map(int, input().split())
ans = math.ceil(math.ceil(n / s) * k / p)
print(ans)
| Title: Paper Airplanes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
Input Specification:
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
Output Specification:
Print a single integer — the minimum number of packs they should buy.
Demo Input:
['5 3 2 3\n', '5 3 100 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets. | ```python
import math
if __name__ == '__main__':
k, n, s, p = map(int, input().split())
ans = math.ceil(math.ceil(n / s) * k / p)
print(ans)
``` | 3 | |
479 | C | Exams | PROGRAMMING | 1,400 | [
"greedy",
"sortings"
] | null | null | Student Valera is an undergraduate student at the University. His end of term exams are approaching and he is to pass exactly *n* exams. Valera is a smart guy, so he will be able to pass any exam he takes on his first try. Besides, he can take several exams on one day, and in any order.
According to the schedule, a student can take the exam for the *i*-th subject on the day number *a**i*. However, Valera has made an arrangement with each teacher and the teacher of the *i*-th subject allowed him to take an exam before the schedule time on day *b**i* (*b**i*<=<<=*a**i*). Thus, Valera can take an exam for the *i*-th subject either on day *a**i*, or on day *b**i*. All the teachers put the record of the exam in the student's record book on the day of the actual exam and write down the date of the mark as number *a**i*.
Valera believes that it would be rather strange if the entries in the record book did not go in the order of non-decreasing date. Therefore Valera asks you to help him. Find the minimum possible value of the day when Valera can take the final exam if he takes exams so that all the records in his record book go in the order of non-decreasing date. | The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=5000) — the number of exams Valera will take.
Each of the next *n* lines contains two positive space-separated integers *a**i* and *b**i* (1<=≤<=*b**i*<=<<=*a**i*<=≤<=109) — the date of the exam in the schedule and the early date of passing the *i*-th exam, correspondingly. | Print a single integer — the minimum possible number of the day when Valera can take the last exam if he takes all the exams so that all the records in his record book go in the order of non-decreasing date. | [
"3\n5 2\n3 1\n4 2\n",
"3\n6 1\n5 2\n4 3\n"
] | [
"2\n",
"6\n"
] | In the first sample Valera first takes an exam in the second subject on the first day (the teacher writes down the schedule date that is 3). On the next day he takes an exam in the third subject (the teacher writes down the schedule date, 4), then he takes an exam in the first subject (the teacher writes down the mark with date 5). Thus, Valera takes the last exam on the second day and the dates will go in the non-decreasing order: 3, 4, 5.
In the second sample Valera first takes an exam in the third subject on the fourth day. Then he takes an exam in the second subject on the fifth day. After that on the sixth day Valera takes an exam in the first subject. | 1,500 | [
{
"input": "3\n5 2\n3 1\n4 2",
"output": "2"
},
{
"input": "3\n6 1\n5 2\n4 3",
"output": "6"
},
{
"input": "1\n1000000000 999999999",
"output": "999999999"
},
{
"input": "1\n2 1",
"output": "1"
},
{
"input": "2\n3 2\n3 2",
"output": "2"
},
{
"input": "... | 1,694,682,432 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 46 | 716,800 | n=int(input())
l=[]
for i in range(n):
a, b= map(int, input().split())
l.append([a,b])
l.sort()
prev=0
for a, b in l:
if b>= prev:
prev=b
else:
prev=a
print(prev) | Title: Exams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Student Valera is an undergraduate student at the University. His end of term exams are approaching and he is to pass exactly *n* exams. Valera is a smart guy, so he will be able to pass any exam he takes on his first try. Besides, he can take several exams on one day, and in any order.
According to the schedule, a student can take the exam for the *i*-th subject on the day number *a**i*. However, Valera has made an arrangement with each teacher and the teacher of the *i*-th subject allowed him to take an exam before the schedule time on day *b**i* (*b**i*<=<<=*a**i*). Thus, Valera can take an exam for the *i*-th subject either on day *a**i*, or on day *b**i*. All the teachers put the record of the exam in the student's record book on the day of the actual exam and write down the date of the mark as number *a**i*.
Valera believes that it would be rather strange if the entries in the record book did not go in the order of non-decreasing date. Therefore Valera asks you to help him. Find the minimum possible value of the day when Valera can take the final exam if he takes exams so that all the records in his record book go in the order of non-decreasing date.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=5000) — the number of exams Valera will take.
Each of the next *n* lines contains two positive space-separated integers *a**i* and *b**i* (1<=≤<=*b**i*<=<<=*a**i*<=≤<=109) — the date of the exam in the schedule and the early date of passing the *i*-th exam, correspondingly.
Output Specification:
Print a single integer — the minimum possible number of the day when Valera can take the last exam if he takes all the exams so that all the records in his record book go in the order of non-decreasing date.
Demo Input:
['3\n5 2\n3 1\n4 2\n', '3\n6 1\n5 2\n4 3\n']
Demo Output:
['2\n', '6\n']
Note:
In the first sample Valera first takes an exam in the second subject on the first day (the teacher writes down the schedule date that is 3). On the next day he takes an exam in the third subject (the teacher writes down the schedule date, 4), then he takes an exam in the first subject (the teacher writes down the mark with date 5). Thus, Valera takes the last exam on the second day and the dates will go in the non-decreasing order: 3, 4, 5.
In the second sample Valera first takes an exam in the third subject on the fourth day. Then he takes an exam in the second subject on the fifth day. After that on the sixth day Valera takes an exam in the first subject. | ```python
n=int(input())
l=[]
for i in range(n):
a, b= map(int, input().split())
l.append([a,b])
l.sort()
prev=0
for a, b in l:
if b>= prev:
prev=b
else:
prev=a
print(prev)
``` | 3 | |
26 | B | Regular Bracket Sequence | PROGRAMMING | 1,400 | [
"greedy"
] | B. Regular Bracket Sequence | 5 | 256 | A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained? | Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106. | Output the maximum possible length of a regular bracket sequence. | [
"(()))(\n",
"((()())\n"
] | [
"4\n",
"6\n"
] | none | 1,000 | [
{
"input": "(()))(",
"output": "4"
},
{
"input": "((()())",
"output": "6"
},
{
"input": "(",
"output": "0"
},
{
"input": ")",
"output": "0"
},
{
"input": ")(()(",
"output": "2"
},
{
"input": "))))))(",
"output": "0"
},
{
"input": "()()(()((... | 1,666,594,525 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 26 | 5,000 | 3,174,400 | a=input()
b=0
if a.count("(") != 0 and a.count(')') != 0:
if a.count("(") == 1 and a.count(')') ==1:
b = 1
else:
while '()' in a:
b+=a.count('()')
a=a.replace('()','')
print(b*2) | Title: Regular Bracket Sequence
Time Limit: 5 seconds
Memory Limit: 256 megabytes
Problem Description:
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input Specification:
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output Specification:
Output the maximum possible length of a regular bracket sequence.
Demo Input:
['(()))(\n', '((()())\n']
Demo Output:
['4\n', '6\n']
Note:
none | ```python
a=input()
b=0
if a.count("(") != 0 and a.count(')') != 0:
if a.count("(") == 1 and a.count(')') ==1:
b = 1
else:
while '()' in a:
b+=a.count('()')
a=a.replace('()','')
print(b*2)
``` | 0 |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
... | 1,699,365,782 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 21 | 216 | 2,048,000 | ans=0
for i in range(int(input())):
task = sum(map(int, input().split()))
if task>=2:
ans+=1
print(ans) | Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
ans=0
for i in range(int(input())):
task = sum(map(int, input().split()))
if task>=2:
ans+=1
print(ans)
``` | 3 | |
875 | A | Classroom Watch | PROGRAMMING | 1,200 | [
"brute force",
"math"
] | null | null | Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109). | In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order. | [
"21\n",
"20\n"
] | [
"1\n15\n",
"0\n"
] | In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | 500 | [
{
"input": "21",
"output": "1\n15"
},
{
"input": "20",
"output": "0"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "3",
"output": "0"
},
{
"input": "100000001",
"output": "2\n99999937\n100000000"
},
{
"i... | 1,508,152,236 | 336 | Python 3 | OK | TESTS | 18 | 61 | 5,529,600 | n = int(input())
a = []
for i in range(min(n, 82), -1, -1):
c = n - i
c += sum([int(j) for j in str(c)])
if c == n:
a.append(n - i)
print(len(a))
for i in a:
print(i) | Title: Classroom Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order.
Demo Input:
['21\n', '20\n']
Demo Output:
['1\n15\n', '0\n']
Note:
In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | ```python
n = int(input())
a = []
for i in range(min(n, 82), -1, -1):
c = n - i
c += sum([int(j) for j in str(c)])
if c == n:
a.append(n - i)
print(len(a))
for i in a:
print(i)
``` | 3 | |
451 | A | Game With Sticks | PROGRAMMING | 900 | [
"implementation"
] | null | null | After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game? | The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). | Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game. | [
"2 2\n",
"2 3\n",
"3 3\n"
] | [
"Malvika\n",
"Malvika\n",
"Akshat\n"
] | Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks. | 500 | [
{
"input": "2 2",
"output": "Malvika"
},
{
"input": "2 3",
"output": "Malvika"
},
{
"input": "3 3",
"output": "Akshat"
},
{
"input": "20 68",
"output": "Malvika"
},
{
"input": "1 1",
"output": "Akshat"
},
{
"input": "1 2",
"output": "Akshat"
},
... | 1,690,872,428 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 46 | 0 | n,m = map(int, input().split())
a = min(n,m)
if a%2==0:
print('Malvika')
else:
print('Akshat') | Title: Game With Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game?
Input Specification:
The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Output Specification:
Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game.
Demo Input:
['2 2\n', '2 3\n', '3 3\n']
Demo Output:
['Malvika\n', 'Malvika\n', 'Akshat\n']
Note:
Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks. | ```python
n,m = map(int, input().split())
a = min(n,m)
if a%2==0:
print('Malvika')
else:
print('Akshat')
``` | 3 | |
312 | A | Whose sentence is it? | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100. | For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions. | [
"5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n"
] | [
"Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"
] | none | 500 | [
{
"input": "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .",
"output": "Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nLpAEKiHVJrzSZqBVSSyY\nYECGBlala.\nUZeGpeM.UCwiHmmA\nqt_,.b_.LSwJtJ.\nFAnXZtHlala.\nmiao.iape... | 1,370,589,631 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 93 | 0 | n = int(input())
a = []
for x in range(n) : a.append(str(input()))
for x in range(n):
if (a[x][-5:] == 'lala.' and a[x][:5] == 'miao.'): print('OMG>.< I don\'t know!')
elif (a[x][-5:] == 'lala.'): print('Freda\'s')
elif (a[x][:5] == 'miao.'): print('Rainbow\'s')
else: print('OMG>.< I don\'t know!') | Title: Whose sentence is it?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.
Output Specification:
For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.
Demo Input:
['5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n']
Demo Output:
["Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"]
Note:
none | ```python
n = int(input())
a = []
for x in range(n) : a.append(str(input()))
for x in range(n):
if (a[x][-5:] == 'lala.' and a[x][:5] == 'miao.'): print('OMG>.< I don\'t know!')
elif (a[x][-5:] == 'lala.'): print('Freda\'s')
elif (a[x][:5] == 'miao.'): print('Rainbow\'s')
else: print('OMG>.< I don\'t know!')
``` | 3 | |
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "... | 1,680,696,520 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 25 | 124 | 0 | n, m = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
a = list(sorted(a))
ans = 0
for i in range(n):
if a[i] < 0 and m != 0:
ans += a[i]
m -= 1
print(abs(ans)) | Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
n, m = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
a = list(sorted(a))
ans = 0
for i in range(n):
if a[i] < 0 and m != 0:
ans += a[i]
m -= 1
print(abs(ans))
``` | 3.969 |
60 | B | Serial Time! | PROGRAMMING | 1,400 | [
"dfs and similar",
"dsu"
] | B. Serial Time! | 2 | 256 | The Cereal Guy's friend Serial Guy likes to watch soap operas. An episode is about to start, and he hasn't washed his plate yet. But he decided to at least put in under the tap to be filled with water. The plate can be represented by a parallelepiped *k*<=×<=*n*<=×<=*m*, that is, it has *k* layers (the first layer is the upper one), each of which is a rectangle *n*<=×<=*m* with empty squares ('.') and obstacles ('#'). The water can only be present in the empty squares. The tap is positioned above the square (*x*,<=*y*) of the first layer, it is guaranteed that this square is empty. Every minute a cubical unit of water falls into the plate. Find out in how many minutes the Serial Guy should unglue himself from the soap opera and turn the water off for it not to overfill the plate. That is, you should find the moment of time when the plate is absolutely full and is going to be overfilled in the next moment.
Note: the water fills all the area within reach (see sample 4). Water flows in each of the 6 directions, through faces of 1<=×<=1<=×<=1 cubes. | The first line contains three numbers *k*, *n*, *m* (1<=≤<=*k*,<=*n*,<=*m*<=≤<=10) which are the sizes of the plate. Then follow *k* rectangles consisting of *n* lines each containing *m* characters '.' or '#', which represents the "layers" of the plate in the order from the top to the bottom. The rectangles are separated by empty lines (see the samples). The last line contains *x* and *y* (1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*y*<=≤<=*m*) which are the tap's coordinates. *x* is the number of the line and *y* is the number of the column. Lines of each layer are numbered from left to right by the integers from 1 to *n*, columns of each layer are numbered from top to bottom by the integers from 1 to *m*. | The answer should contain a single number, showing in how many minutes the plate will be filled. | [
"1 1 1\n\n.\n\n1 1\n",
"2 1 1\n\n.\n\n#\n\n1 1\n",
"2 2 2\n\n.#\n##\n\n..\n..\n\n1 1\n",
"3 2 2\n\n#.\n##\n\n#.\n.#\n\n..\n..\n\n1 2\n",
"3 3 3\n\n.#.\n###\n##.\n\n.##\n###\n##.\n\n...\n...\n...\n\n1 1\n"
] | [
"1\n",
"1\n",
"5\n",
"7\n",
"13\n"
] | none | 1,000 | [
{
"input": "1 1 1\n\n.\n\n1 1",
"output": "1"
},
{
"input": "2 1 1\n\n.\n\n#\n\n1 1",
"output": "1"
},
{
"input": "2 2 2\n\n.#\n##\n\n..\n..\n\n1 1",
"output": "5"
},
{
"input": "3 2 2\n\n#.\n##\n\n#.\n.#\n\n..\n..\n\n1 2",
"output": "7"
},
{
"input": "3 3 3\n\n.#... | 1,590,869,329 | 2,147,483,647 | PyPy 3 | OK | TESTS | 46 | 342 | 3,276,800 | (k, n, m) = map(int, input().split())
a = []
for i in range(k):
s = input()
a.append([])
for j in range(n):
a[-1].append(list(input()))
s = input()
(x, y) = map(int, input().split())
x -= 1
y -= 1
start = (0, x, y)
sosed = set()
sosed.add(start)
posetil = set()
posetil.add(start)
while len(sosed) >0:
sosed2 = set()
for i in sosed:
for dk in [-1,0,1]:
for dx in [-1,0,1]:
for dy in [-1,0,1]:
if abs(dk)+abs(dx)+abs(dy) == 1:
if (i[0]+dk <k) and (i[0]+dk>=0) and (i[1]+dx < n) and (i[1]+dx>=0) and (i[2]+dy < m) and (i[2]+dy >=0):
if a[i[0]+dk][i[1]+dx][i[2]+dy] == '.':
if (i[0]+dk, i[1]+dx, i[2]+dy) in posetil:
continue
else:
posetil.add((i[0]+dk,i[1]+dx,i[2]+dy))
sosed2.add((i[0]+dk,i[1]+dx,i[2]+dy))
sosed = sosed2.copy()
print(len(posetil))
| Title: Serial Time!
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Cereal Guy's friend Serial Guy likes to watch soap operas. An episode is about to start, and he hasn't washed his plate yet. But he decided to at least put in under the tap to be filled with water. The plate can be represented by a parallelepiped *k*<=×<=*n*<=×<=*m*, that is, it has *k* layers (the first layer is the upper one), each of which is a rectangle *n*<=×<=*m* with empty squares ('.') and obstacles ('#'). The water can only be present in the empty squares. The tap is positioned above the square (*x*,<=*y*) of the first layer, it is guaranteed that this square is empty. Every minute a cubical unit of water falls into the plate. Find out in how many minutes the Serial Guy should unglue himself from the soap opera and turn the water off for it not to overfill the plate. That is, you should find the moment of time when the plate is absolutely full and is going to be overfilled in the next moment.
Note: the water fills all the area within reach (see sample 4). Water flows in each of the 6 directions, through faces of 1<=×<=1<=×<=1 cubes.
Input Specification:
The first line contains three numbers *k*, *n*, *m* (1<=≤<=*k*,<=*n*,<=*m*<=≤<=10) which are the sizes of the plate. Then follow *k* rectangles consisting of *n* lines each containing *m* characters '.' or '#', which represents the "layers" of the plate in the order from the top to the bottom. The rectangles are separated by empty lines (see the samples). The last line contains *x* and *y* (1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*y*<=≤<=*m*) which are the tap's coordinates. *x* is the number of the line and *y* is the number of the column. Lines of each layer are numbered from left to right by the integers from 1 to *n*, columns of each layer are numbered from top to bottom by the integers from 1 to *m*.
Output Specification:
The answer should contain a single number, showing in how many minutes the plate will be filled.
Demo Input:
['1 1 1\n\n.\n\n1 1\n', '2 1 1\n\n.\n\n#\n\n1 1\n', '2 2 2\n\n.#\n##\n\n..\n..\n\n1 1\n', '3 2 2\n\n#.\n##\n\n#.\n.#\n\n..\n..\n\n1 2\n', '3 3 3\n\n.#.\n###\n##.\n\n.##\n###\n##.\n\n...\n...\n...\n\n1 1\n']
Demo Output:
['1\n', '1\n', '5\n', '7\n', '13\n']
Note:
none | ```python
(k, n, m) = map(int, input().split())
a = []
for i in range(k):
s = input()
a.append([])
for j in range(n):
a[-1].append(list(input()))
s = input()
(x, y) = map(int, input().split())
x -= 1
y -= 1
start = (0, x, y)
sosed = set()
sosed.add(start)
posetil = set()
posetil.add(start)
while len(sosed) >0:
sosed2 = set()
for i in sosed:
for dk in [-1,0,1]:
for dx in [-1,0,1]:
for dy in [-1,0,1]:
if abs(dk)+abs(dx)+abs(dy) == 1:
if (i[0]+dk <k) and (i[0]+dk>=0) and (i[1]+dx < n) and (i[1]+dx>=0) and (i[2]+dy < m) and (i[2]+dy >=0):
if a[i[0]+dk][i[1]+dx][i[2]+dy] == '.':
if (i[0]+dk, i[1]+dx, i[2]+dy) in posetil:
continue
else:
posetil.add((i[0]+dk,i[1]+dx,i[2]+dy))
sosed2.add((i[0]+dk,i[1]+dx,i[2]+dy))
sosed = sosed2.copy()
print(len(posetil))
``` | 3.908396 |
391 | A | Genetic Engineering | PROGRAMMING | 0 | [
"implementation",
"two pointers"
] | null | null | You will receive 3 points for solving this problem.
Manao is designing the genetic code for a new type of algae to efficiently produce fuel. Specifically, Manao is focusing on a stretch of DNA that encodes one protein. The stretch of DNA is represented by a string containing only the characters 'A', 'T', 'G' and 'C'.
Manao has determined that if the stretch of DNA contains a maximal sequence of consecutive identical nucleotides that is of even length, then the protein will be nonfunctional. For example, consider a protein described by DNA string "GTTAAAG". It contains four maximal sequences of consecutive identical nucleotides: "G", "TT", "AAA", and "G". The protein is nonfunctional because sequence "TT" has even length.
Manao is trying to obtain a functional protein from the protein he currently has. Manao can insert additional nucleotides into the DNA stretch. Each additional nucleotide is a character from the set {'A', 'T', 'G', 'C'}. Manao wants to determine the minimum number of insertions necessary to make the DNA encode a functional protein. | The input consists of a single line, containing a string *s* of length *n* (1<=≤<=*n*<=≤<=100). Each character of *s* will be from the set {'A', 'T', 'G', 'C'}.
This problem doesn't have subproblems. You will get 3 points for the correct submission. | The program should print on one line a single integer representing the minimum number of 'A', 'T', 'G', 'C' characters that are required to be inserted into the input string in order to make all runs of identical characters have odd length. | [
"GTTAAAG\n",
"AACCAACCAAAAC\n"
] | [
"1\n",
"5\n"
] | In the first example, it is sufficient to insert a single nucleotide of any type between the two 'T's in the sequence to restore the functionality of the protein. | 3 | [
{
"input": "GTTAAAG",
"output": "1"
},
{
"input": "AACCAACCAAAAC",
"output": "5"
},
{
"input": "GTGAATTTCC",
"output": "2"
},
{
"input": "CAGGGGGCCGCCCATGAAAAAAACCCGGCCCCTTGGGAAAACTTGGGTTA",
"output": "7"
},
{
"input": "CCCTTCACCCGGATCCAAATCCCTTAGAAATAATCCCCGACGGC... | 1,583,151,842 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 109 | 0 | S = input()
ans = 0
count = 1
for i in range(1, len(S)):
if(S[i] == S[i-1]):
count += 1
else:
if(not (count & 1)):
ans += 1
count = 1
print(ans)
| Title: Genetic Engineering
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You will receive 3 points for solving this problem.
Manao is designing the genetic code for a new type of algae to efficiently produce fuel. Specifically, Manao is focusing on a stretch of DNA that encodes one protein. The stretch of DNA is represented by a string containing only the characters 'A', 'T', 'G' and 'C'.
Manao has determined that if the stretch of DNA contains a maximal sequence of consecutive identical nucleotides that is of even length, then the protein will be nonfunctional. For example, consider a protein described by DNA string "GTTAAAG". It contains four maximal sequences of consecutive identical nucleotides: "G", "TT", "AAA", and "G". The protein is nonfunctional because sequence "TT" has even length.
Manao is trying to obtain a functional protein from the protein he currently has. Manao can insert additional nucleotides into the DNA stretch. Each additional nucleotide is a character from the set {'A', 'T', 'G', 'C'}. Manao wants to determine the minimum number of insertions necessary to make the DNA encode a functional protein.
Input Specification:
The input consists of a single line, containing a string *s* of length *n* (1<=≤<=*n*<=≤<=100). Each character of *s* will be from the set {'A', 'T', 'G', 'C'}.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
The program should print on one line a single integer representing the minimum number of 'A', 'T', 'G', 'C' characters that are required to be inserted into the input string in order to make all runs of identical characters have odd length.
Demo Input:
['GTTAAAG\n', 'AACCAACCAAAAC\n']
Demo Output:
['1\n', '5\n']
Note:
In the first example, it is sufficient to insert a single nucleotide of any type between the two 'T's in the sequence to restore the functionality of the protein. | ```python
S = input()
ans = 0
count = 1
for i in range(1, len(S)):
if(S[i] == S[i-1]):
count += 1
else:
if(not (count & 1)):
ans += 1
count = 1
print(ans)
``` | 0 | |
870 | C | Maximum splitting | PROGRAMMING | 1,300 | [
"dp",
"greedy",
"math",
"number theory"
] | null | null | You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself. | The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query. | For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings. | [
"1\n12\n",
"2\n6\n8\n",
"3\n1\n2\n3\n"
] | [
"3\n",
"1\n2\n",
"-1\n-1\n-1\n"
] | 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | 1,500 | [
{
"input": "1\n12",
"output": "3"
},
{
"input": "2\n6\n8",
"output": "1\n2"
},
{
"input": "3\n1\n2\n3",
"output": "-1\n-1\n-1"
},
{
"input": "6\n1\n2\n3\n5\n7\n11",
"output": "-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "3\n4\n6\n9",
"output": "1\n1\n1"
},
{
... | 1,510,498,025 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 6 | 62 | 0 | dp = [-1, -1, -1, 1, -1, 1, -1, 2, 1, 2, -1, 3, 2, 3, 2, 4, 3, 4, 3, 5, 4, 5, 4, 6, 5, 6, 5, 7, 6, 7, 6, 8, 7, 8, 7, 9, 8, 9, 8, 10, 9, 10, 9, 11, 10, 11, 10, 12, 11, 12, 11, 13, 12, 13, 12, 14, 13, 14, 13, 15, 14, 15, 14, 16, 15, 16, 15, 17, 16, 17, 16, 18, 17, 18, 17, 19, 18, 19, 18, 20, 19, 20, 19, 21, 20, 21, 20, 22, 21, 22, 21, 23, 22, 23, 22, 24, 23, 24, 23]
def main():
n = int(input())
if n < len(dp):
print(dp[n - 1])
return
res = (n - len(dp)) // 4
n -= ((n - len(dp)) // 4) * 4
while n > len(dp):
n -= 4
res += 1
res += dp[n]
print(res)
return
q = int(input())
for i in range(q):
main() | Title: Maximum splitting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
Input Specification:
The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query.
Output Specification:
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
Demo Input:
['1\n12\n', '2\n6\n8\n', '3\n1\n2\n3\n']
Demo Output:
['3\n', '1\n2\n', '-1\n-1\n-1\n']
Note:
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | ```python
dp = [-1, -1, -1, 1, -1, 1, -1, 2, 1, 2, -1, 3, 2, 3, 2, 4, 3, 4, 3, 5, 4, 5, 4, 6, 5, 6, 5, 7, 6, 7, 6, 8, 7, 8, 7, 9, 8, 9, 8, 10, 9, 10, 9, 11, 10, 11, 10, 12, 11, 12, 11, 13, 12, 13, 12, 14, 13, 14, 13, 15, 14, 15, 14, 16, 15, 16, 15, 17, 16, 17, 16, 18, 17, 18, 17, 19, 18, 19, 18, 20, 19, 20, 19, 21, 20, 21, 20, 22, 21, 22, 21, 23, 22, 23, 22, 24, 23, 24, 23]
def main():
n = int(input())
if n < len(dp):
print(dp[n - 1])
return
res = (n - len(dp)) // 4
n -= ((n - len(dp)) // 4) * 4
while n > len(dp):
n -= 4
res += 1
res += dp[n]
print(res)
return
q = int(input())
for i in range(q):
main()
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | A positive integer is called a 2-3-integer, if it is equal to 2*x*·3*y* for some non-negative integers *x* and *y*. In other words, these integers are such integers that only have 2 and 3 among their prime divisors. For example, integers 1, 6, 9, 16 and 108 — are 2-3 integers, while 5, 10, 21 and 120 are not.
Print the number of 2-3-integers on the given segment [*l*,<=*r*], i. e. the number of sich 2-3-integers *t* that *l*<=≤<=*t*<=≤<=*r*. | The only line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=2·109). | Print a single integer the number of 2-3-integers on the segment [*l*,<=*r*]. | [
"1 10\n",
"100 200\n",
"1 2000000000\n"
] | [
"7\n",
"5\n",
"326\n"
] | In the first example the 2-3-integers are 1, 2, 3, 4, 6, 8 and 9.
In the second example the 2-3-integers are 108, 128, 144, 162 and 192. | 0 | [
{
"input": "1 10",
"output": "7"
},
{
"input": "100 200",
"output": "5"
},
{
"input": "1 2000000000",
"output": "326"
},
{
"input": "1088391168 1934917632",
"output": "17"
},
{
"input": "1088391167 1934917632",
"output": "17"
},
{
"input": "1088391169 ... | 1,533,290,222 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 124 | 307,200 | l,r=map(int,input().split())
ans=0
for i in range(31):
for j in range(20):
if l <= 2**i * 3**j <= r:
ans+=1
print(ans)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A positive integer is called a 2-3-integer, if it is equal to 2*x*·3*y* for some non-negative integers *x* and *y*. In other words, these integers are such integers that only have 2 and 3 among their prime divisors. For example, integers 1, 6, 9, 16 and 108 — are 2-3 integers, while 5, 10, 21 and 120 are not.
Print the number of 2-3-integers on the given segment [*l*,<=*r*], i. e. the number of sich 2-3-integers *t* that *l*<=≤<=*t*<=≤<=*r*.
Input Specification:
The only line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=2·109).
Output Specification:
Print a single integer the number of 2-3-integers on the segment [*l*,<=*r*].
Demo Input:
['1 10\n', '100 200\n', '1 2000000000\n']
Demo Output:
['7\n', '5\n', '326\n']
Note:
In the first example the 2-3-integers are 1, 2, 3, 4, 6, 8 and 9.
In the second example the 2-3-integers are 108, 128, 144, 162 and 192. | ```python
l,r=map(int,input().split())
ans=0
for i in range(31):
for j in range(20):
if l <= 2**i * 3**j <= r:
ans+=1
print(ans)
``` | 3 | |
499 | A | Watching a movie | PROGRAMMING | 1,000 | [
"greedy",
"implementation"
] | null | null | You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments? | The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*. | Output a single number — the answer to the problem. | [
"2 3\n5 6\n10 12\n",
"1 1\n1 100000\n"
] | [
"6\n",
"100000\n"
] | In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie. | 500 | [
{
"input": "2 3\n5 6\n10 12",
"output": "6"
},
{
"input": "1 1\n1 100000",
"output": "100000"
},
{
"input": "10 1\n2156 3497\n4784 7775\n14575 31932\n33447 35902\n36426 47202\n48772 60522\n63982 68417\n78537 79445\n90081 90629\n94325 95728",
"output": "53974"
},
{
"input": "1... | 1,688,025,787 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 78 | 5,222,400 | L1= [int(x) for x in input().split(" ")]
n=L1[0]
suru=L1[1]
L3=[]
time=1
currtime=0
for i in range(0,n):
L2= [int(x) for x in input().split(" ")]
L3.append(L2)
usermax=L3[n-1][1]
L5=[]
ctr=0
string=""
for i in range(1,usermax+1):
L5.append(i)
#print(L5)
L6=[]
for i in L3:
skip=(i[0]-currtime)//suru
num=skip*suru
currtime=currtime+1
for j in range(0,num):
L6.append(time)
time=time+1
currtime=currtime+i[1]
time=currtime
#print(L6)
for i in L6:
if i in L5:
L5.remove(i)
if suru==1:
print(len(L5)+1)
else:
print(len(L5)) | Title: Watching a movie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments?
Input Specification:
The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*.
Output Specification:
Output a single number — the answer to the problem.
Demo Input:
['2 3\n5 6\n10 12\n', '1 1\n1 100000\n']
Demo Output:
['6\n', '100000\n']
Note:
In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie. | ```python
L1= [int(x) for x in input().split(" ")]
n=L1[0]
suru=L1[1]
L3=[]
time=1
currtime=0
for i in range(0,n):
L2= [int(x) for x in input().split(" ")]
L3.append(L2)
usermax=L3[n-1][1]
L5=[]
ctr=0
string=""
for i in range(1,usermax+1):
L5.append(i)
#print(L5)
L6=[]
for i in L3:
skip=(i[0]-currtime)//suru
num=skip*suru
currtime=currtime+1
for j in range(0,num):
L6.append(time)
time=time+1
currtime=currtime+i[1]
time=currtime
#print(L6)
for i in L6:
if i in L5:
L5.remove(i)
if suru==1:
print(len(L5)+1)
else:
print(len(L5))
``` | 0 | |
190 | A | Vasya and the Bus | PROGRAMMING | 1,100 | [
"greedy",
"math"
] | null | null | One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..."
The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride.
The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=><=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble.
We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups.
Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total. | The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly. | If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly.
Otherwise, print "Impossible" (without the quotes). | [
"1 2\n",
"0 5\n",
"2 2\n"
] | [
"2 2",
"Impossible",
"2 3"
] | In the first sample a grown-up rides with two children and pays two rubles.
In the second sample there are only children in the bus, so the situation is impossible.
In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total. | 500 | [
{
"input": "1 2",
"output": "2 2"
},
{
"input": "0 5",
"output": "Impossible"
},
{
"input": "2 2",
"output": "2 3"
},
{
"input": "2 7",
"output": "7 8"
},
{
"input": "4 10",
"output": "10 13"
},
{
"input": "6 0",
"output": "6 6"
},
{
"input... | 1,661,671,127 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | x = int(input('Enter a number: '))
if x%2 == 0
print('You have entered an even number.')
else:
print('You have entered an odd number.') | Title: Vasya and the Bus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..."
The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride.
The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=><=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble.
We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups.
Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total.
Input Specification:
The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly.
Output Specification:
If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly.
Otherwise, print "Impossible" (without the quotes).
Demo Input:
['1 2\n', '0 5\n', '2 2\n']
Demo Output:
['2 2', 'Impossible', '2 3']
Note:
In the first sample a grown-up rides with two children and pays two rubles.
In the second sample there are only children in the bus, so the situation is impossible.
In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total. | ```python
x = int(input('Enter a number: '))
if x%2 == 0
print('You have entered an even number.')
else:
print('You have entered an odd number.')
``` | -1 | |
522 | A | Reposts | PROGRAMMING | 1,200 | [
"*special",
"dfs and similar",
"dp",
"graphs",
"trees"
] | null | null | One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user. | Print a single integer — the maximum length of a repost chain. | [
"5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n",
"6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n",
"1\nSoMeStRaNgEgUe repos... | [
"6\n",
"2\n",
"2\n"
] | none | 500 | [
{
"input": "5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya",
"output": "6"
},
{
"input": "6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforc... | 1,425,753,420 | 13,020 | PyPy 3 | OK | TESTS | 36 | 109 | 307,200 | n = int(input())
a = {}
a['polycarp'] = 1
for i in range(n):
inp = input().split()
a[inp[0].lower()] = a[inp[2].lower()] + 1
mx = 0
for key in a:
mx = max(mx, a[key])
print(mx) | Title: Reposts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user.
Output Specification:
Print a single integer — the maximum length of a repost chain.
Demo Input:
['5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n', '6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n', '1\nSoMeStRaNgEgUe reposted PoLyCaRp\n']
Demo Output:
['6\n', '2\n', '2\n']
Note:
none | ```python
n = int(input())
a = {}
a['polycarp'] = 1
for i in range(n):
inp = input().split()
a[inp[0].lower()] = a[inp[2].lower()] + 1
mx = 0
for key in a:
mx = max(mx, a[key])
print(mx)
``` | 3 | |
935 | A | Fafa and his Company | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly. | The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company. | Print a single integer representing the answer to the problem. | [
"2\n",
"10\n"
] | [
"1\n",
"3\n"
] | In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | 500 | [
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
... | 1,655,791,689 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 46 | 0 | n=int(input())
a=""
for i in range(1,n,1):
if n%i==0 :
a=a+str(i)
print(len(a)) | Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | ```python
n=int(input())
a=""
for i in range(1,n,1):
if n%i==0 :
a=a+str(i)
print(len(a))
``` | 0 | |
433 | B | Kuriyama Mirai's Stones | PROGRAMMING | 1,200 | [
"dp",
"implementation",
"sortings"
] | null | null | Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one. | Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input. | [
"6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n",
"4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n"
] | [
"24\n9\n28\n",
"10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n"
] | Please note that the answers to the questions may overflow 32-bit integer type. | 1,500 | [
{
"input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6",
"output": "24\n9\n28"
},
{
"input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2",
"output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5"
},
{
"input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1... | 1,665,589,421 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 46 | 1,153 | 17,817,600 | n = int(input())
v = list(map(int,input().split()))
u = sorted(v)
dp = [0]
dps = [0]
for i in range(n):
dp.append(dp[-1]+v[i])
dps.append(dps[-1]+u[i])
m = int(input())
for _ in range(m):
type,l,r = map(int,input().split())
if type == 1:
print(dp[r]-dp[l-1])
else:
print(dps[r]-dps[l-1])
| Title: Kuriyama Mirai's Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output Specification:
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Demo Input:
['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n']
Demo Output:
['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n']
Note:
Please note that the answers to the questions may overflow 32-bit integer type. | ```python
n = int(input())
v = list(map(int,input().split()))
u = sorted(v)
dp = [0]
dps = [0]
for i in range(n):
dp.append(dp[-1]+v[i])
dps.append(dps[-1]+u[i])
m = int(input())
for _ in range(m):
type,l,r = map(int,input().split())
if type == 1:
print(dp[r]-dp[l-1])
else:
print(dps[r]-dps[l-1])
``` | 3 | |
779 | C | Dishonest Sellers | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items. | In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week). | Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now. | [
"3 1\n5 4 6\n3 1 5\n",
"5 3\n3 4 7 10 3\n4 5 5 12 5\n"
] | [
"10\n",
"25\n"
] | In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | 1,000 | [
{
"input": "3 1\n5 4 6\n3 1 5",
"output": "10"
},
{
"input": "5 3\n3 4 7 10 3\n4 5 5 12 5",
"output": "25"
},
{
"input": "1 0\n9\n8",
"output": "8"
},
{
"input": "2 0\n4 10\n1 2",
"output": "3"
},
{
"input": "4 2\n19 5 17 13\n3 18 8 10",
"output": "29"
},
... | 1,700,413,180 | 2,147,483,647 | Python 3 | OK | TESTS | 67 | 218 | 35,840,000 | n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
d = sorted([b_el - a_el for b_el, a_el in zip(b, a) if b_el - a_el < 0])
print(sum(a) + sum(d[:n - k]))
| Title: Dishonest Sellers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.
Input Specification:
In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).
Output Specification:
Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.
Demo Input:
['3 1\n5 4 6\n3 1 5\n', '5 3\n3 4 7 10 3\n4 5 5 12 5\n']
Demo Output:
['10\n', '25\n']
Note:
In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | ```python
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
d = sorted([b_el - a_el for b_el, a_el in zip(b, a) if b_el - a_el < 0])
print(sum(a) + sum(d[:n - k]))
``` | 3 | |
237 | A | Free Cash | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period. | Print a single integer — the minimum number of cashes, needed to serve all clients next day. | [
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] | [
"2\n",
"1\n"
] | In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash. | 500 | [
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "... | 1,680,777,726 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | def it(): return int(input())
def pr(a): return print(a)
def li(): return list(map(int, input().split()))
def ls(): return [str(i) for i in input()]
def no(): return print('NO')
def yes(): return print('YES')
kol = 1
m = []
for i in range(it()):
a, b = map(int, input().split())
m.append(a+b)
for i in range(len(m)-1):
if m[i] == m[i+1]:
kol += 1
print(kol)
'''??????????????????| -_- |???????????????????''' | Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash. | ```python
def it(): return int(input())
def pr(a): return print(a)
def li(): return list(map(int, input().split()))
def ls(): return [str(i) for i in input()]
def no(): return print('NO')
def yes(): return print('YES')
kol = 1
m = []
for i in range(it()):
a, b = map(int, input().split())
m.append(a+b)
for i in range(len(m)-1):
if m[i] == m[i+1]:
kol += 1
print(kol)
'''??????????????????| -_- |???????????????????'''
``` | 0 | |
129 | B | Students and Shoelaces | PROGRAMMING | 1,200 | [
"brute force",
"dfs and similar",
"graphs",
"implementation"
] | null | null | Anna and Maria are in charge of the math club for junior students. When the club gathers together, the students behave badly. They've brought lots of shoe laces to the club and got tied with each other. Specifically, each string ties together two students. Besides, if two students are tied, then the lace connects the first student with the second one as well as the second student with the first one.
To restore order, Anna and Maria do the following. First, for each student Anna finds out what other students he is tied to. If a student is tied to exactly one other student, Anna reprimands him. Then Maria gathers in a single group all the students who have been just reprimanded. She kicks them out from the club. This group of students immediately leaves the club. These students takes with them the laces that used to tie them. Then again for every student Anna finds out how many other students he is tied to and so on. And they do so until Anna can reprimand at least one student.
Determine how many groups of students will be kicked out of the club. | The first line contains two integers *n* and *m* — the initial number of students and laces (). The students are numbered from 1 to *n*, and the laces are numbered from 1 to *m*. Next *m* lines each contain two integers *a* and *b* — the numbers of students tied by the *i*-th lace (1<=≤<=*a*,<=*b*<=≤<=*n*,<=*a*<=≠<=*b*). It is guaranteed that no two students are tied with more than one lace. No lace ties a student to himself. | Print the single number — the number of groups of students that will be kicked out from the club. | [
"3 3\n1 2\n2 3\n3 1\n",
"6 3\n1 2\n2 3\n3 4\n",
"6 5\n1 4\n2 4\n3 4\n5 4\n6 4\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample Anna and Maria won't kick out any group of students — in the initial position every student is tied to two other students and Anna won't be able to reprimand anyone.
In the second sample four students are tied in a chain and two more are running by themselves. First Anna and Maria kick out the two students from both ends of the chain (1 and 4), then — two other students from the chain (2 and 3). At that the students who are running by themselves will stay in the club.
In the third sample Anna and Maria will momentarily kick out all students except for the fourth one and the process stops at that point. The correct answer is one. | 1,000 | [
{
"input": "3 3\n1 2\n2 3\n3 1",
"output": "0"
},
{
"input": "6 3\n1 2\n2 3\n3 4",
"output": "2"
},
{
"input": "6 5\n1 4\n2 4\n3 4\n5 4\n6 4",
"output": "1"
},
{
"input": "100 0",
"output": "0"
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "0"
},
... | 1,661,801,409 | 2,147,483,647 | Python 3 | OK | TESTS | 70 | 186 | 0 | sn, ln = map(int, input().split())
laces = [list(map(int, input().split())) for _ in range(ln)]
ans = 0
while True:
deleted_list = [i for i in range(1, sn + 1) if sum([lace.count(i) for lace in laces]) == 1]
if len(deleted_list) == 0:
break
laces = [lace for lace in laces if lace[0] not in deleted_list and lace[1] not in deleted_list]
ans += 1
print(ans)
| Title: Students and Shoelaces
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anna and Maria are in charge of the math club for junior students. When the club gathers together, the students behave badly. They've brought lots of shoe laces to the club and got tied with each other. Specifically, each string ties together two students. Besides, if two students are tied, then the lace connects the first student with the second one as well as the second student with the first one.
To restore order, Anna and Maria do the following. First, for each student Anna finds out what other students he is tied to. If a student is tied to exactly one other student, Anna reprimands him. Then Maria gathers in a single group all the students who have been just reprimanded. She kicks them out from the club. This group of students immediately leaves the club. These students takes with them the laces that used to tie them. Then again for every student Anna finds out how many other students he is tied to and so on. And they do so until Anna can reprimand at least one student.
Determine how many groups of students will be kicked out of the club.
Input Specification:
The first line contains two integers *n* and *m* — the initial number of students and laces (). The students are numbered from 1 to *n*, and the laces are numbered from 1 to *m*. Next *m* lines each contain two integers *a* and *b* — the numbers of students tied by the *i*-th lace (1<=≤<=*a*,<=*b*<=≤<=*n*,<=*a*<=≠<=*b*). It is guaranteed that no two students are tied with more than one lace. No lace ties a student to himself.
Output Specification:
Print the single number — the number of groups of students that will be kicked out from the club.
Demo Input:
['3 3\n1 2\n2 3\n3 1\n', '6 3\n1 2\n2 3\n3 4\n', '6 5\n1 4\n2 4\n3 4\n5 4\n6 4\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample Anna and Maria won't kick out any group of students — in the initial position every student is tied to two other students and Anna won't be able to reprimand anyone.
In the second sample four students are tied in a chain and two more are running by themselves. First Anna and Maria kick out the two students from both ends of the chain (1 and 4), then — two other students from the chain (2 and 3). At that the students who are running by themselves will stay in the club.
In the third sample Anna and Maria will momentarily kick out all students except for the fourth one and the process stops at that point. The correct answer is one. | ```python
sn, ln = map(int, input().split())
laces = [list(map(int, input().split())) for _ in range(ln)]
ans = 0
while True:
deleted_list = [i for i in range(1, sn + 1) if sum([lace.count(i) for lace in laces]) == 1]
if len(deleted_list) == 0:
break
laces = [lace for lace in laces if lace[0] not in deleted_list and lace[1] not in deleted_list]
ans += 1
print(ans)
``` | 3 | |
673 | A | Bear and Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game. | The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order. | Print the number of minutes Limak will watch the game. | [
"3\n7 20 88\n",
"9\n16 20 30 40 50 60 70 80 90\n",
"9\n15 20 30 40 50 60 70 80 90\n"
] | [
"35\n",
"15\n",
"90\n"
] | In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | 500 | [
{
"input": "3\n7 20 88",
"output": "35"
},
{
"input": "9\n16 20 30 40 50 60 70 80 90",
"output": "15"
},
{
"input": "9\n15 20 30 40 50 60 70 80 90",
"output": "90"
},
{
"input": "30\n6 11 12 15 22 24 30 31 32 33 34 35 40 42 44 45 47 50 53 54 57 58 63 67 75 77 79 81 83 88",
... | 1,629,176,403 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 6,758,400 | N = input()
T = list(map(int, input().split()))
number_of_minutes = 0
for t in T:
if t - number_of_minutes <= 15:
number_of_minutes = t
else:
break
if number_of_minutes < 90:
number_of_minutes += 15
print(number_of_minutes)
| Title: Bear and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game.
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order.
Output Specification:
Print the number of minutes Limak will watch the game.
Demo Input:
['3\n7 20 88\n', '9\n16 20 30 40 50 60 70 80 90\n', '9\n15 20 30 40 50 60 70 80 90\n']
Demo Output:
['35\n', '15\n', '90\n']
Note:
In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | ```python
N = input()
T = list(map(int, input().split()))
number_of_minutes = 0
for t in T:
if t - number_of_minutes <= 15:
number_of_minutes = t
else:
break
if number_of_minutes < 90:
number_of_minutes += 15
print(number_of_minutes)
``` | 0 | |
148 | A | Insomnia cure | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation",
"math"
] | null | null | «One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons? | Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105). | Output the number of damaged dragons. | [
"1\n2\n3\n4\n12\n",
"2\n3\n4\n5\n24\n"
] | [
"12\n",
"17\n"
] | In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed. | 1,000 | [
{
"input": "1\n2\n3\n4\n12",
"output": "12"
},
{
"input": "2\n3\n4\n5\n24",
"output": "17"
},
{
"input": "1\n1\n1\n1\n100000",
"output": "100000"
},
{
"input": "10\n9\n8\n7\n6",
"output": "0"
},
{
"input": "8\n4\n4\n3\n65437",
"output": "32718"
},
{
"i... | 1,697,099,720 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 186 | 2,764,800 | k = int(input())
l = int(input())
m = int(input())
n = int(input())
d = int(input())
lst = [0]*(d+1)
w,x,y,z = 1,1,1,1
while w*k<=d or x*l<=d or y*m<=d or z*n<=d:
if w*k<=d:
lst[w*k] = 1
w+=1
if x*l<=d:
lst[x*l] = 1
x+=1
if y*m<=d:
lst[y*m] = 1
y+=1
if z*n<=d:
lst[z*n] = 1
z+=1
cnt = 0
for i in range(1,d+1):
if lst[i] == 1:
cnt+=1
print(cnt) | Title: Insomnia cure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
Input Specification:
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
Output Specification:
Output the number of damaged dragons.
Demo Input:
['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n']
Demo Output:
['12\n', '17\n']
Note:
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed. | ```python
k = int(input())
l = int(input())
m = int(input())
n = int(input())
d = int(input())
lst = [0]*(d+1)
w,x,y,z = 1,1,1,1
while w*k<=d or x*l<=d or y*m<=d or z*n<=d:
if w*k<=d:
lst[w*k] = 1
w+=1
if x*l<=d:
lst[x*l] = 1
x+=1
if y*m<=d:
lst[y*m] = 1
y+=1
if z*n<=d:
lst[z*n] = 1
z+=1
cnt = 0
for i in range(1,d+1):
if lst[i] == 1:
cnt+=1
print(cnt)
``` | 3 | |
677 | A | Vanya and Fence | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? | The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. | Print a single integer — the minimum possible valid width of the road. | [
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] | [
"4\n",
"6\n",
"11\n"
] | In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | 500 | [
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481... | 1,692,211,626 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 46 | 0 | r,c=map(int,input().split())
lst=list(map(int,input().split()))
count=0
for ele in lst:
if ele> c:
count=count+2
else:
count=count+1
print(count) | Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | ```python
r,c=map(int,input().split())
lst=list(map(int,input().split()))
count=0
for ele in lst:
if ele> c:
count=count+2
else:
count=count+1
print(count)
``` | 3 | |
583 | A | Asphalting Roads | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct. | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | [
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] | [
"1 4 \n",
"1 \n"
] | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | 500 | [
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 ... | 1,461,495,779 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 92 | 5,120,000 | n = int(input())
G = []; V = []; res = []
for i in range(n*n):
h,v = map(int, input().split())
if G.count(h)==0 and V.count(v)==0:
G.append(h)
V.append(v)
res.append(i+1)
print(*res)
| Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | ```python
n = int(input())
G = []; V = []; res = []
for i in range(n*n):
h,v = map(int, input().split())
if G.count(h)==0 and V.count(v)==0:
G.append(h)
V.append(v)
res.append(i+1)
print(*res)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Gennady is one of the best child dentists in Berland. Today *n* children got an appointment with him, they lined up in front of his office.
All children love to cry loudly at the reception at the dentist. We enumerate the children with integers from 1 to *n* in the order they go in the line. Every child is associated with the value of his cofidence *p**i*. The children take turns one after another to come into the office; each time the child that is the first in the line goes to the doctor.
While Gennady treats the teeth of the *i*-th child, the child is crying with the volume of *v**i*. At that the confidence of the first child in the line is reduced by the amount of *v**i*, the second one — by value *v**i*<=-<=1, and so on. The children in the queue after the *v**i*-th child almost do not hear the crying, so their confidence remains unchanged.
If at any point in time the confidence of the *j*-th child is less than zero, he begins to cry with the volume of *d**j* and leaves the line, running towards the exit, without going to the doctor's office. At this the confidence of all the children after the *j*-th one in the line is reduced by the amount of *d**j*.
All these events occur immediately one after the other in some order. Some cries may lead to other cries, causing a chain reaction. Once in the hallway it is quiet, the child, who is first in the line, goes into the doctor's office.
Help Gennady the Dentist to determine the numbers of kids, whose teeth he will cure. Print their numbers in the chronological order. | The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=4000) — the number of kids in the line.
Next *n* lines contain three integers each *v**i*,<=*d**i*,<=*p**i* (1<=≤<=*v**i*,<=*d**i*,<=*p**i*<=≤<=106) — the volume of the cry in the doctor's office, the volume of the cry in the hall and the confidence of the *i*-th child. | In the first line print number *k* — the number of children whose teeth Gennady will cure.
In the second line print *k* integers — the numbers of the children who will make it to the end of the line in the increasing order. | [
"5\n4 2 2\n4 1 2\n5 2 4\n3 3 5\n5 1 2\n",
"5\n4 5 1\n5 3 9\n4 1 2\n2 1 8\n4 1 9\n"
] | [
"2\n1 3 ",
"4\n1 2 4 5 "
] | In the first example, Gennady first treats the teeth of the first child who will cry with volume 4. The confidences of the remaining children will get equal to - 2, 1, 3, 1, respectively. Thus, the second child also cries at the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 0, 2, 0. Then the third child will go to the office, and cry with volume 5. The other children won't bear this, and with a loud cry they will run to the exit.
In the second sample, first the first child goes into the office, he will cry with volume 4. The confidence of the remaining children will be equal to 5, - 1, 6, 8. Thus, the third child will cry with the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 5, 5, 7. After that, the second child goes to the office and cry with the volume of 5. The confidences of the remaining children will be equal to 0, 3. Then the fourth child will go into the office and cry with the volume of 2. Because of this the confidence of the fifth child will be 1, and he will go into the office last. | 0 | [
{
"input": "5\n4 2 2\n4 1 2\n5 2 4\n3 3 5\n5 1 2",
"output": "2\n1 3 "
},
{
"input": "5\n4 5 1\n5 3 9\n4 1 2\n2 1 8\n4 1 9",
"output": "4\n1 2 4 5 "
},
{
"input": "10\n10 7 10\n3 6 11\n8 4 10\n10 1 11\n7 3 13\n7 2 13\n7 6 14\n3 4 17\n9 4 20\n5 2 24",
"output": "3\n1 2 5 "
},
{
... | 1,460,524,639 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 61 | 0 | a=int(input())
b=[]*a
c=list()
c.append(1)
for i in range(a):
b.append(list(map(int,input().split())))
def cry(p,a,b):
for i in range(p+1,a):
b[i][2] -= b[p][0]-i+1+p
for i in range(p+1,a):
if b[i][2] <0:
for j in range(i+1,a):
b[j][2] -= b[i][1]
cry(0,a,b)
t=1
while t<a:
for i in range(t,a):
if b[i][2]>=0:
c.append(i+1)
t=i
break
cry(t,a,b)
t=t+1
print(len(c))
for i in range(len(c)):
print(c[i],end=' ') | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gennady is one of the best child dentists in Berland. Today *n* children got an appointment with him, they lined up in front of his office.
All children love to cry loudly at the reception at the dentist. We enumerate the children with integers from 1 to *n* in the order they go in the line. Every child is associated with the value of his cofidence *p**i*. The children take turns one after another to come into the office; each time the child that is the first in the line goes to the doctor.
While Gennady treats the teeth of the *i*-th child, the child is crying with the volume of *v**i*. At that the confidence of the first child in the line is reduced by the amount of *v**i*, the second one — by value *v**i*<=-<=1, and so on. The children in the queue after the *v**i*-th child almost do not hear the crying, so their confidence remains unchanged.
If at any point in time the confidence of the *j*-th child is less than zero, he begins to cry with the volume of *d**j* and leaves the line, running towards the exit, without going to the doctor's office. At this the confidence of all the children after the *j*-th one in the line is reduced by the amount of *d**j*.
All these events occur immediately one after the other in some order. Some cries may lead to other cries, causing a chain reaction. Once in the hallway it is quiet, the child, who is first in the line, goes into the doctor's office.
Help Gennady the Dentist to determine the numbers of kids, whose teeth he will cure. Print their numbers in the chronological order.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=4000) — the number of kids in the line.
Next *n* lines contain three integers each *v**i*,<=*d**i*,<=*p**i* (1<=≤<=*v**i*,<=*d**i*,<=*p**i*<=≤<=106) — the volume of the cry in the doctor's office, the volume of the cry in the hall and the confidence of the *i*-th child.
Output Specification:
In the first line print number *k* — the number of children whose teeth Gennady will cure.
In the second line print *k* integers — the numbers of the children who will make it to the end of the line in the increasing order.
Demo Input:
['5\n4 2 2\n4 1 2\n5 2 4\n3 3 5\n5 1 2\n', '5\n4 5 1\n5 3 9\n4 1 2\n2 1 8\n4 1 9\n']
Demo Output:
['2\n1 3 ', '4\n1 2 4 5 ']
Note:
In the first example, Gennady first treats the teeth of the first child who will cry with volume 4. The confidences of the remaining children will get equal to - 2, 1, 3, 1, respectively. Thus, the second child also cries at the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 0, 2, 0. Then the third child will go to the office, and cry with volume 5. The other children won't bear this, and with a loud cry they will run to the exit.
In the second sample, first the first child goes into the office, he will cry with volume 4. The confidence of the remaining children will be equal to 5, - 1, 6, 8. Thus, the third child will cry with the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 5, 5, 7. After that, the second child goes to the office and cry with the volume of 5. The confidences of the remaining children will be equal to 0, 3. Then the fourth child will go into the office and cry with the volume of 2. Because of this the confidence of the fifth child will be 1, and he will go into the office last. | ```python
a=int(input())
b=[]*a
c=list()
c.append(1)
for i in range(a):
b.append(list(map(int,input().split())))
def cry(p,a,b):
for i in range(p+1,a):
b[i][2] -= b[p][0]-i+1+p
for i in range(p+1,a):
if b[i][2] <0:
for j in range(i+1,a):
b[j][2] -= b[i][1]
cry(0,a,b)
t=1
while t<a:
for i in range(t,a):
if b[i][2]>=0:
c.append(i+1)
t=i
break
cry(t,a,b)
t=t+1
print(len(c))
for i in range(len(c)):
print(c[i],end=' ')
``` | 0 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,695,123,460 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 46 | 0 | import math
n,m,a=map(int,input().split())
leng=math.ceil(n/a)
bre=math.ceil(m/a)
total=leng*bre
print(total) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
import math
n,m,a=map(int,input().split())
leng=math.ceil(n/a)
bre=math.ceil(m/a)
total=leng*bre
print(total)
``` | 3.977 |
157 | B | Trace | PROGRAMMING | 1,000 | [
"geometry",
"sortings"
] | null | null | One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric. | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *r**i* (1<=≤<=*r**i*<=≤<=1000) — the circles' radii. It is guaranteed that all circles are different. | Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4. | [
"1\n1\n",
"3\n1 4 2\n"
] | [
"3.1415926536\n",
"40.8407044967\n"
] | In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 1<sup class="upper-index">2</sup> = π.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 4<sup class="upper-index">2</sup> - π × 2<sup class="upper-index">2</sup>) + π × 1<sup class="upper-index">2</sup> = π × 12 + π = 13π | 1,000 | [
{
"input": "1\n1",
"output": "3.1415926536"
},
{
"input": "3\n1 4 2",
"output": "40.8407044967"
},
{
"input": "4\n4 1 3 2",
"output": "31.4159265359"
},
{
"input": "4\n100 10 2 1",
"output": "31111.1920484997"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output"... | 1,615,526,530 | 2,147,483,647 | PyPy 3 | OK | TESTS | 44 | 248 | 512,000 | #!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from sys import stdin,stdout
from io import BytesIO, IOBase
from itertools import accumulate
from collections import deque
#sys.setrecursionlimit(10**5)
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-----------------------------------------------------------------
def regularbracket(t):
p=0
for i in t:
if i=="(":
p+=1
else:
p-=1
if p<0:
return False
else:
if p>0:
return False
else:
return True
#-------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid+1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
#------------------------------reverse string(pallindrome)
def reverse1(string):
pp=""
for i in string[::-1]:
pp+=i
if pp==string:
return True
return False
#--------------------------------reverse list(paindrome)
def reverse2(list1):
l=[]
for i in list1[::-1]:
l.append(i)
if l==list1:
return True
return False
# endregion
import math
def mex(list1):
#list1 = sorted(list1)
p = max(list1)+1
for i in range(len(list1)):
if list1[i]!=i:
p = i
break
return p
#----endregion
"""
def main():
n,m = map(int,input().split())
np = list(map(int,input().split()))
mp = list(map(int,input().split()))
np = sorted(np)
mp = sorted(mp)
cnt=0
for i in np:
if len(mp)==0:
cnt+=1
elif i in mp:
continue
elif i>max(mp):
cnt+=1
elif i<max(mp):
#cnt-=1
for j in mp:
if j>i:
mp.remove(j)
break
print(cnt)
"""
"""
def main():
n = int(input())
arr = list(map(int,input().split()))
max=0
if len(arr)==1:
print(1)
elif len(arr)==2:
print(2)
else:
cnt=2
for i in range(n-2):
if arr[i+2]==arr[i+1]+arr[i]:
cnt+=1
else:
if cnt>max:
max = cnt
cnt=2
if cnt>max:
max=cnt
print(max)
"""
def main():
n = int(input())
arr = list(map(int,input().split()))
arr = sorted(arr)
p=0
if len(arr)%2!=0:
c = (math.pi)*(pow(arr[0],2))
arr.pop(0)
i=0
p=0
while(i<len(arr)-1):
p+=(math.pi*(pow(arr[i+1],2))) - ((math.pi*(pow(arr[i],2))))
i+=2
print(p+c)
else:
i=0
while(i<len(arr)-1):
p += (math.pi * (pow(arr[i + 1], 2))) - ((math.pi * (pow(arr[i], 2))))
i+=2
print(p)
if __name__ == '__main__':
main()
| Title: Trace
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=100). The second line contains *n* space-separated integers *r**i* (1<=≤<=*r**i*<=≤<=1000) — the circles' radii. It is guaranteed that all circles are different.
Output Specification:
Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4.
Demo Input:
['1\n1\n', '3\n1 4 2\n']
Demo Output:
['3.1415926536\n', '40.8407044967\n']
Note:
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 1<sup class="upper-index">2</sup> = π.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 4<sup class="upper-index">2</sup> - π × 2<sup class="upper-index">2</sup>) + π × 1<sup class="upper-index">2</sup> = π × 12 + π = 13π | ```python
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from sys import stdin,stdout
from io import BytesIO, IOBase
from itertools import accumulate
from collections import deque
#sys.setrecursionlimit(10**5)
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-----------------------------------------------------------------
def regularbracket(t):
p=0
for i in t:
if i=="(":
p+=1
else:
p-=1
if p<0:
return False
else:
if p>0:
return False
else:
return True
#-------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid+1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
#------------------------------reverse string(pallindrome)
def reverse1(string):
pp=""
for i in string[::-1]:
pp+=i
if pp==string:
return True
return False
#--------------------------------reverse list(paindrome)
def reverse2(list1):
l=[]
for i in list1[::-1]:
l.append(i)
if l==list1:
return True
return False
# endregion
import math
def mex(list1):
#list1 = sorted(list1)
p = max(list1)+1
for i in range(len(list1)):
if list1[i]!=i:
p = i
break
return p
#----endregion
"""
def main():
n,m = map(int,input().split())
np = list(map(int,input().split()))
mp = list(map(int,input().split()))
np = sorted(np)
mp = sorted(mp)
cnt=0
for i in np:
if len(mp)==0:
cnt+=1
elif i in mp:
continue
elif i>max(mp):
cnt+=1
elif i<max(mp):
#cnt-=1
for j in mp:
if j>i:
mp.remove(j)
break
print(cnt)
"""
"""
def main():
n = int(input())
arr = list(map(int,input().split()))
max=0
if len(arr)==1:
print(1)
elif len(arr)==2:
print(2)
else:
cnt=2
for i in range(n-2):
if arr[i+2]==arr[i+1]+arr[i]:
cnt+=1
else:
if cnt>max:
max = cnt
cnt=2
if cnt>max:
max=cnt
print(max)
"""
def main():
n = int(input())
arr = list(map(int,input().split()))
arr = sorted(arr)
p=0
if len(arr)%2!=0:
c = (math.pi)*(pow(arr[0],2))
arr.pop(0)
i=0
p=0
while(i<len(arr)-1):
p+=(math.pi*(pow(arr[i+1],2))) - ((math.pi*(pow(arr[i],2))))
i+=2
print(p+c)
else:
i=0
while(i<len(arr)-1):
p += (math.pi * (pow(arr[i + 1], 2))) - ((math.pi * (pow(arr[i], 2))))
i+=2
print(p)
if __name__ == '__main__':
main()
``` | 3 | |
748 | A | Santa Claus and a Place in a Class | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right! | The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place. | Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right. | [
"4 3 9\n",
"4 3 24\n",
"2 4 4\n"
] | [
"2 2 L\n",
"4 3 R\n",
"1 2 R\n"
] | The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | 500 | [
{
"input": "4 3 9",
"output": "2 2 L"
},
{
"input": "4 3 24",
"output": "4 3 R"
},
{
"input": "2 4 4",
"output": "1 2 R"
},
{
"input": "3 10 24",
"output": "2 2 R"
},
{
"input": "10 3 59",
"output": "10 3 L"
},
{
"input": "10000 10000 160845880",
"... | 1,482,661,578 | 4,878 | Python 3 | OK | TESTS | 46 | 77 | 4,608,000 |
inp = str(input());
s = ''
for s in inp:
s = inp.split(' ')
n = int(s[0])
m = int(s[1])
k = int(s[2])
i = 0
r = 1
d = 1;
p = 0
while(k > 2 * m * r):
p = 2 * m * r
r += 1;
while(k > p):
p += 1
if (i == 2): i = 0; d += 1;
i += 1
if(k % 2 == 0):
print(r,d,'R')
else:print(r,d,'L')
| Title: Santa Claus and a Place in a Class
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input Specification:
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output Specification:
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Demo Input:
['4 3 9\n', '4 3 24\n', '2 4 4\n']
Demo Output:
['2 2 L\n', '4 3 R\n', '1 2 R\n']
Note:
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | ```python
inp = str(input());
s = ''
for s in inp:
s = inp.split(' ')
n = int(s[0])
m = int(s[1])
k = int(s[2])
i = 0
r = 1
d = 1;
p = 0
while(k > 2 * m * r):
p = 2 * m * r
r += 1;
while(k > p):
p += 1
if (i == 2): i = 0; d += 1;
i += 1
if(k % 2 == 0):
print(r,d,'R')
else:print(r,d,'L')
``` | 3 | |
611 | B | New Year and Old Property | PROGRAMMING | 1,300 | [
"bitmasks",
"brute force",
"implementation"
] | null | null | The year 2015 is almost over.
Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510<==<=111110111112. Note that he doesn't care about the number of zeros in the decimal representation.
Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?
Assume that all positive integers are always written without leading zeros. | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=1018) — the first year and the last year in Limak's interval respectively. | Print one integer – the number of years Limak will count in his chosen interval. | [
"5 10\n",
"2015 2015\n",
"100 105\n",
"72057594000000000 72057595000000000\n"
] | [
"2\n",
"1\n",
"0\n",
"26\n"
] | In the first sample Limak's interval contains numbers 5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>, 6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>, 7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>, 8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>, 9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub> and 10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>. Two of them (101<sub class="lower-index">2</sub> and 110<sub class="lower-index">2</sub>) have the described property. | 750 | [
{
"input": "5 10",
"output": "2"
},
{
"input": "2015 2015",
"output": "1"
},
{
"input": "100 105",
"output": "0"
},
{
"input": "72057594000000000 72057595000000000",
"output": "26"
},
{
"input": "1 100",
"output": "16"
},
{
"input": "100000000000000000... | 1,564,159,486 | 2,147,483,647 | Python 3 | OK | TESTS | 103 | 109 | 0 | n, m = map(int,input().split())
a = len(bin(n))-2
b = len(bin(m))-2
count = 0
for i in range(a, b+1):
for j in range(i-1):
if n <= 2**i-1-2**j <= m:
count+=1
print(count)
| Title: New Year and Old Property
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The year 2015 is almost over.
Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 201510<==<=111110111112. Note that he doesn't care about the number of zeros in the decimal representation.
Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?
Assume that all positive integers are always written without leading zeros.
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=1018) — the first year and the last year in Limak's interval respectively.
Output Specification:
Print one integer – the number of years Limak will count in his chosen interval.
Demo Input:
['5 10\n', '2015 2015\n', '100 105\n', '72057594000000000 72057595000000000\n']
Demo Output:
['2\n', '1\n', '0\n', '26\n']
Note:
In the first sample Limak's interval contains numbers 5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>, 6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>, 7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>, 8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>, 9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub> and 10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>. Two of them (101<sub class="lower-index">2</sub> and 110<sub class="lower-index">2</sub>) have the described property. | ```python
n, m = map(int,input().split())
a = len(bin(n))-2
b = len(bin(m))-2
count = 0
for i in range(a, b+1):
for j in range(i-1):
if n <= 2**i-1-2**j <= m:
count+=1
print(count)
``` | 3 | |
735 | A | Ostap and Grasshopper | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | On the way to Rio de Janeiro Ostap kills time playing with a grasshopper he took with him in a special box. Ostap builds a line of length *n* such that some cells of this line are empty and some contain obstacles. Then, he places his grasshopper to one of the empty cells and a small insect in another empty cell. The grasshopper wants to eat the insect.
Ostap knows that grasshopper is able to jump to any empty cell that is exactly *k* cells away from the current (to the left or to the right). Note that it doesn't matter whether intermediate cells are empty or not as the grasshopper makes a jump over them. For example, if *k*<==<=1 the grasshopper can jump to a neighboring cell only, and if *k*<==<=2 the grasshopper can jump over a single cell.
Your goal is to determine whether there is a sequence of jumps such that grasshopper will get from his initial position to the cell with an insect. | The first line of the input contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1) — the number of cells in the line and the length of one grasshopper's jump.
The second line contains a string of length *n* consisting of characters '.', '#', 'G' and 'T'. Character '.' means that the corresponding cell is empty, character '#' means that the corresponding cell contains an obstacle and grasshopper can't jump there. Character 'G' means that the grasshopper starts at this position and, finally, 'T' means that the target insect is located at this cell. It's guaranteed that characters 'G' and 'T' appear in this line exactly once. | If there exists a sequence of jumps (each jump of length *k*), such that the grasshopper can get from his initial position to the cell with the insect, print "YES" (without quotes) in the only line of the input. Otherwise, print "NO" (without quotes). | [
"5 2\n#G#T#\n",
"6 1\nT....G\n",
"7 3\nT..#..G\n",
"6 2\n..GT..\n"
] | [
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] | In the first sample, the grasshopper can make one jump to the right in order to get from cell 2 to cell 4.
In the second sample, the grasshopper is only able to jump to neighboring cells but the way to the insect is free — he can get there by jumping left 5 times.
In the third sample, the grasshopper can't make a single jump.
In the fourth sample, the grasshopper can only jump to the cells with odd indices, thus he won't be able to reach the insect. | 500 | [
{
"input": "5 2\n#G#T#",
"output": "YES"
},
{
"input": "6 1\nT....G",
"output": "YES"
},
{
"input": "7 3\nT..#..G",
"output": "NO"
},
{
"input": "6 2\n..GT..",
"output": "NO"
},
{
"input": "2 1\nGT",
"output": "YES"
},
{
"input": "100 5\nG####.####.###... | 1,662,415,351 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 83 | 62 | 0 | import sys
ONLINE_JUDGE = True
if not ONLINE_JUDGE:
sys.stdin = open("input.txt", "r")
############ ---- Input Functions ---- ############
def inp(): # integer
return(int(input()))
def inlt(): # lists
return(list(map(int,input().split())))
def insr(): # list of characters
s = input()
return(list(s))
def invr(): # space seperate integers
return(map(int,input().split()))
############ ---- Solution ---- ############
n, k = inlt()
line = insr()
start = line.index("G")
end = line.index("T")
if start > end:
k = -k
def valid_dir():
if k > 0:
return start <= end
else:
return start >= end
res = "NO"
while valid_dir() and (start < n and start >= 0):
if start == end:
res = "YES"
break
if line[start] == "#":
res = "NO"
break
start += k
sys.stdout.write(res)
| Title: Ostap and Grasshopper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the way to Rio de Janeiro Ostap kills time playing with a grasshopper he took with him in a special box. Ostap builds a line of length *n* such that some cells of this line are empty and some contain obstacles. Then, he places his grasshopper to one of the empty cells and a small insect in another empty cell. The grasshopper wants to eat the insect.
Ostap knows that grasshopper is able to jump to any empty cell that is exactly *k* cells away from the current (to the left or to the right). Note that it doesn't matter whether intermediate cells are empty or not as the grasshopper makes a jump over them. For example, if *k*<==<=1 the grasshopper can jump to a neighboring cell only, and if *k*<==<=2 the grasshopper can jump over a single cell.
Your goal is to determine whether there is a sequence of jumps such that grasshopper will get from his initial position to the cell with an insect.
Input Specification:
The first line of the input contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1) — the number of cells in the line and the length of one grasshopper's jump.
The second line contains a string of length *n* consisting of characters '.', '#', 'G' and 'T'. Character '.' means that the corresponding cell is empty, character '#' means that the corresponding cell contains an obstacle and grasshopper can't jump there. Character 'G' means that the grasshopper starts at this position and, finally, 'T' means that the target insect is located at this cell. It's guaranteed that characters 'G' and 'T' appear in this line exactly once.
Output Specification:
If there exists a sequence of jumps (each jump of length *k*), such that the grasshopper can get from his initial position to the cell with the insect, print "YES" (without quotes) in the only line of the input. Otherwise, print "NO" (without quotes).
Demo Input:
['5 2\n#G#T#\n', '6 1\nT....G\n', '7 3\nT..#..G\n', '6 2\n..GT..\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
In the first sample, the grasshopper can make one jump to the right in order to get from cell 2 to cell 4.
In the second sample, the grasshopper is only able to jump to neighboring cells but the way to the insect is free — he can get there by jumping left 5 times.
In the third sample, the grasshopper can't make a single jump.
In the fourth sample, the grasshopper can only jump to the cells with odd indices, thus he won't be able to reach the insect. | ```python
import sys
ONLINE_JUDGE = True
if not ONLINE_JUDGE:
sys.stdin = open("input.txt", "r")
############ ---- Input Functions ---- ############
def inp(): # integer
return(int(input()))
def inlt(): # lists
return(list(map(int,input().split())))
def insr(): # list of characters
s = input()
return(list(s))
def invr(): # space seperate integers
return(map(int,input().split()))
############ ---- Solution ---- ############
n, k = inlt()
line = insr()
start = line.index("G")
end = line.index("T")
if start > end:
k = -k
def valid_dir():
if k > 0:
return start <= end
else:
return start >= end
res = "NO"
while valid_dir() and (start < n and start >= 0):
if start == end:
res = "YES"
break
if line[start] == "#":
res = "NO"
break
start += k
sys.stdout.write(res)
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,694,179,870 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | import math
k=int(input())
y1 = 0
y = 0
y2 = 0
y3 = 0
y4= 0
y=k/1
y1=k/2
y2=k/3
y3=k/4
y4=k/5
l=[y,y1,y2,y3,y4]
print(math.ceil(min(l))) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
import math
k=int(input())
y1 = 0
y = 0
y2 = 0
y3 = 0
y4= 0
y=k/1
y1=k/2
y2=k/3
y3=k/4
y4=k/5
l=[y,y1,y2,y3,y4]
print(math.ceil(min(l)))
``` | -1 |
886 | A | ACM ICPC | PROGRAMMING | 1,000 | [
"brute force"
] | null | null | In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question. | The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants | Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | 500 | [
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
... | 1,520,658,460 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 62 | 5,632,000 | a=sorted(list(map(int,input().split())))
if (a[0]+a[3]+a[5]==a[1]+a[2]+a[4]) or (a[0]+a[2]+a[5]==a[1]+a[3]+a[4]) or (a[1]+a[2]+a[5]==a[0]+a[3]+a[4]):
print('YES')
else:
print('NO') | Title: ACM ICPC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | ```python
a=sorted(list(map(int,input().split())))
if (a[0]+a[3]+a[5]==a[1]+a[2]+a[4]) or (a[0]+a[2]+a[5]==a[1]+a[3]+a[4]) or (a[1]+a[2]+a[5]==a[0]+a[3]+a[4]):
print('YES')
else:
print('NO')
``` | 0 | |
264 | B | Good Sequences | PROGRAMMING | 1,500 | [
"dp",
"number theory"
] | null | null | Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good.
Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions:
- The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers.
Find the length of the longest good sequence. | The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1). | Print a single integer — the length of the longest good sequence. | [
"5\n2 3 4 6 9\n",
"9\n1 2 3 5 6 7 8 9 10\n"
] | [
"4\n",
"4\n"
] | In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4. | 1,000 | [
{
"input": "5\n2 3 4 6 9",
"output": "4"
},
{
"input": "9\n1 2 3 5 6 7 8 9 10",
"output": "4"
},
{
"input": "4\n1 2 4 6",
"output": "3"
},
{
"input": "7\n1 2 3 4 7 9 10",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "8\n3 4 5 6 7 8 9... | 1,595,221,400 | 2,147,483,647 | PyPy 3 | OK | TESTS | 86 | 684 | 31,027,200 | #iterate through numbers
#find prime factorization for each number
#compute max value among each prime factor
#increment each by 1
from collections import defaultdict
import math
def primeFactors(n):
prime = set()
while n % 2 == 0:
prime.add(2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
prime.add(i)
n = n//i
if n > 2:
prime.add(n)
return list(prime)
n = int(input())
a = list(map(int, input().split()))
l = defaultdict(int)
for i in range(n):
pf = primeFactors(a[i])
mx = 0
for j in pf:
mx = max(mx, l[j])
for j in pf:
l[j] = mx+1
mx = 1
for k,v in l.items():
mx = max(mx, v)
print(mx) | Title: Good Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*1,<=*a*2,<=...,<=*a**n* are good.
Now she is interested in good sequences. A sequence *x*1,<=*x*2,<=...,<=*x**k* is called good if it satisfies the following three conditions:
- The sequence is strictly increasing, i.e. *x**i*<=<<=*x**i*<=+<=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1). - No two adjacent elements are coprime, i.e. *gcd*(*x**i*,<=*x**i*<=+<=1)<=><=1 for each *i* (1<=≤<=*i*<=≤<=*k*<=-<=1) (where *gcd*(*p*,<=*q*) denotes the greatest common divisor of the integers *p* and *q*). - All elements of the sequence are good integers.
Find the length of the longest good sequence.
Input Specification:
The input consists of two lines. The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of good integers. The second line contains a single-space separated list of good integers *a*1,<=*a*2,<=...,<=*a**n* in strictly increasing order (1<=≤<=*a**i*<=≤<=105; *a**i*<=<<=*a**i*<=+<=1).
Output Specification:
Print a single integer — the length of the longest good sequence.
Demo Input:
['5\n2 3 4 6 9\n', '9\n1 2 3 5 6 7 8 9 10\n']
Demo Output:
['4\n', '4\n']
Note:
In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4. | ```python
#iterate through numbers
#find prime factorization for each number
#compute max value among each prime factor
#increment each by 1
from collections import defaultdict
import math
def primeFactors(n):
prime = set()
while n % 2 == 0:
prime.add(2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
prime.add(i)
n = n//i
if n > 2:
prime.add(n)
return list(prime)
n = int(input())
a = list(map(int, input().split()))
l = defaultdict(int)
for i in range(n):
pf = primeFactors(a[i])
mx = 0
for j in pf:
mx = max(mx, l[j])
for j in pf:
l[j] = mx+1
mx = 1
for k,v in l.items():
mx = max(mx, v)
print(mx)
``` | 3 | |
381 | A | Sereja and Dima | PROGRAMMING | 800 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. | On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. | [
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] | [
"12 5\n",
"16 12\n"
] | In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | 500 | [
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 ... | 1,678,009,922 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 93 | 0 | n = int(input())
c = [int(l) for l in input().split()]
c.sort(reverse=True)
s = 0
d = 0
for i in range(n):
if i % 2 == 0:
s += c[i]
else:
d += c[i]
print(s, d)
| Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | ```python
n = int(input())
c = [int(l) for l in input().split()]
c.sort(reverse=True)
s = 0
d = 0
for i in range(n):
if i % 2 == 0:
s += c[i]
else:
d += c[i]
print(s, d)
``` | 0 | |
1,004 | A | Sonya and Hotels | PROGRAMMING | 900 | [
"implementation"
] | null | null | Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$. | The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels. | Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$. | [
"4 3\n-3 2 9 16\n",
"5 2\n4 8 11 18 19\n"
] | [
"6\n",
"5\n"
] | In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$. | 500 | [
{
"input": "4 3\n-3 2 9 16",
"output": "6"
},
{
"input": "5 2\n4 8 11 18 19",
"output": "5"
},
{
"input": "10 10\n-67 -59 -49 -38 -8 20 41 59 74 83",
"output": "8"
},
{
"input": "10 10\n0 20 48 58 81 95 111 137 147 159",
"output": "9"
},
{
"input": "100 1\n0 1 2 3... | 1,532,144,754 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 124 | 0 | n,d=map(int,input().split())
a=list(map(int,input().split()))
ans=2
for i in range(n-1):
z=a[i+1]-a[i]
if z>2*d:
ans+=2
else:
if z==2*d:
ans+=1
print(ans)
| Title: Sonya and Hotels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$.
Input Specification:
The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$) — coordinates of Sonya's hotels.
Output Specification:
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$.
Demo Input:
['4 3\n-3 2 9 16\n', '5 2\n4 8 11 18 19\n']
Demo Output:
['6\n', '5\n']
Note:
In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$. | ```python
n,d=map(int,input().split())
a=list(map(int,input().split()))
ans=2
for i in range(n-1):
z=a[i+1]-a[i]
if z>2*d:
ans+=2
else:
if z==2*d:
ans+=1
print(ans)
``` | 3 | |
777 | A | Shell Game | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation",
"math"
] | null | null | Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball.
Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.).
Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball? | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator.
The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements. | Print one integer from 0 to 2 — the index of the shell where the ball was initially placed. | [
"4\n2\n",
"1\n1\n"
] | [
"1\n",
"0\n"
] | In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements.
1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell. | 500 | [
{
"input": "4\n2",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n2",
"output": "0"
},
{
"input": "3\n1",
"output": "1"
},
{
"input": "3\n2",
"output": "0"
},
{
"input": "3\n0",
"output": "2"
},
{
"input": "2000000000\n... | 1,613,320,862 | 4,262 | PyPy 3 | OK | TESTS | 68 | 109 | 0 | n = int(input())
x = int(input())
a = [[0, 1, 2], [1, 0, 2], [1, 2, 0], [2, 1, 0], [2, 0, 1], [0, 2, 1]]
print(a[n % 6][x])
| Title: Shell Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball.
Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.).
Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball?
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator.
The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements.
Output Specification:
Print one integer from 0 to 2 — the index of the shell where the ball was initially placed.
Demo Input:
['4\n2\n', '1\n1\n']
Demo Output:
['1\n', '0\n']
Note:
In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements.
1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell. | ```python
n = int(input())
x = int(input())
a = [[0, 1, 2], [1, 0, 2], [1, 2, 0], [2, 1, 0], [2, 0, 1], [0, 2, 1]]
print(a[n % 6][x])
``` | 3 | |
793 | A | Oleg and shares | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question? | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices. | Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. | [
"3 3\n12 9 15\n",
"2 2\n10 9\n",
"4 1\n1 1000000000 1000000000 1000000000\n"
] | [
"3",
"-1",
"2999999997"
] | Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3.
In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal.
In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time. | 500 | [
{
"input": "3 3\n12 9 15",
"output": "3"
},
{
"input": "2 2\n10 9",
"output": "-1"
},
{
"input": "4 1\n1 1000000000 1000000000 1000000000",
"output": "2999999997"
},
{
"input": "1 11\n123",
"output": "0"
},
{
"input": "20 6\n38 86 86 50 98 62 32 2 14 62 98 50 2 50... | 1,493,284,239 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 4,608,000 | import sys
[_, k, *nums] = (int(x) for x in sys.stdin.read().split())
smallest = min(nums)
steps = 0
for num in nums:
dist = num - smallest
if dist % k != 0:
steps = -1
break
steps += dist / k
print(str(steps), end='')
| Title: Oleg and shares
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial prices.
Output Specification:
Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible.
Demo Input:
['3 3\n12 9 15\n', '2 2\n10 9\n', '4 1\n1 1000000000 1000000000 1000000000\n']
Demo Output:
['3', '-1', '2999999997']
Note:
Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3.
In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal.
In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999 * 3 = 2999999997 seconds. We can note that this is the minimum possible time. | ```python
import sys
[_, k, *nums] = (int(x) for x in sys.stdin.read().split())
smallest = min(nums)
steps = 0
for num in nums:
dist = num - smallest
if dist % k != 0:
steps = -1
break
steps += dist / k
print(str(steps), end='')
``` | 0 | |
658 | A | Bear and Reverse Radewoosh | PROGRAMMING | 800 | [
"implementation"
] | null | null | Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems. | The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem. | Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points. | [
"3 2\n50 85 250\n10 15 25\n",
"3 6\n50 85 250\n10 15 25\n",
"8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n"
] | [
"Limak\n",
"Radewoosh\n",
"Tie\n"
] | In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4. | 500 | [
{
"input": "3 2\n50 85 250\n10 15 25",
"output": "Limak"
},
{
"input": "3 6\n50 85 250\n10 15 25",
"output": "Radewoosh"
},
{
"input": "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76",
"output": "Tie"
},
{
"input": "4 1\n3 5 6 9\n1 2 4 8",
"output": "Limak"
},
{... | 1,606,064,525 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 109 | 307,200 | # https://codeforces.com/problemset/problem/658/A
n, c = map(int, input().split())
p = tuple(map(int, input().split()))
t = list(map(int, input().split()))
limak_t = 0
radewoosh_t = 0
limak = 0
radewoosh = 0
for i in range(n):
limak_t += t[i]
radewoosh_t += t[n - i - 1]
limak += max(0, p[i] - c * limak_t)
radewoosh += max(0, p[n - i - 1] - c * radewoosh_t)
if limak > radewoosh:
print("Limak")
elif limak < radewoosh:
print("Radewoosh")
else:
print("Tie")
| Title: Bear and Reverse Radewoosh
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.
Output Specification:
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
Demo Input:
['3 2\n50 85 250\n10 15 25\n', '3 6\n50 85 250\n10 15 25\n', '8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n']
Demo Output:
['Limak\n', 'Radewoosh\n', 'Tie\n']
Note:
In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4. | ```python
# https://codeforces.com/problemset/problem/658/A
n, c = map(int, input().split())
p = tuple(map(int, input().split()))
t = list(map(int, input().split()))
limak_t = 0
radewoosh_t = 0
limak = 0
radewoosh = 0
for i in range(n):
limak_t += t[i]
radewoosh_t += t[n - i - 1]
limak += max(0, p[i] - c * limak_t)
radewoosh += max(0, p[n - i - 1] - c * radewoosh_t)
if limak > radewoosh:
print("Limak")
elif limak < radewoosh:
print("Radewoosh")
else:
print("Tie")
``` | 3 | |
884 | A | Book Reading | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day. | Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*. | [
"2 2\n86400 86398\n",
"2 86400\n0 86400\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "2 2\n86400 86398",
"output": "2"
},
{
"input": "2 86400\n0 86400",
"output": "1"
},
{
"input": "2 86400\n1 86399",
"output": "2"
},
{
"input": "100 1000000\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0... | 1,585,404,679 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 109 | 0 | i=lambda:map(int,input().split())
_,d=i();r=0
for x in i():
r+=1
d-=86400-x
if d<1:break
print(r)
| Title: Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Luba bought a very interesting book. She knows that it will take *t* seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of *n* next days. The number of seconds that Luba has to spend working during *i*-th day is *a**i*. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed *n*.
Remember that there are 86400 seconds in a day.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=100, 1<=≤<=*t*<=≤<=106) — the number of days and the time required to read the book.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=86400) — the time Luba has to spend on her work during *i*-th day.
Output Specification:
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed *n*.
Demo Input:
['2 2\n86400 86398\n', '2 86400\n0 86400\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
i=lambda:map(int,input().split())
_,d=i();r=0
for x in i():
r+=1
d-=86400-x
if d<1:break
print(r)
``` | 3 | |
270 | A | Fancy Fence | PROGRAMMING | 1,100 | [
"geometry",
"implementation",
"math"
] | null | null | Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? | The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. | For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. | [
"3\n30\n60\n90\n"
] | [
"NO\nYES\nYES\n"
] | In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | 500 | [
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,697,993,243 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | t = int(input())
for _ in range(t):
a = int(input())
found = False
for n in range(3, 181):
angle = (n - 2) * 180 / n
if abs(angle - a) < 1e-9: # Comparing angles with a small epsilon for floating-point precision
found = True
break
if found:
print("YES")
else:
print("NO") | Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | ```python
t = int(input())
for _ in range(t):
a = int(input())
found = False
for n in range(3, 181):
angle = (n - 2) * 180 / n
if abs(angle - a) < 1e-9: # Comparing angles with a small epsilon for floating-point precision
found = True
break
if found:
print("YES")
else:
print("NO")
``` | 0 | |
472 | A | Design Tutorial: Learn from Math | PROGRAMMING | 800 | [
"math",
"number theory"
] | null | null | One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers. | The only line contains an integer *n* (12<=≤<=*n*<=≤<=106). | Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them. | [
"12\n",
"15\n",
"23\n",
"1000000\n"
] | [
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] | In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | 500 | [
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
... | 1,681,253,734 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 33 | 62 | 1,331,200 | n = int(input())
l = []
def is_prime(n):
is_prime = True
for i in range(2,n):
if n%i == 0:
is_prime = False
return is_prime
for i in range(2,n-1):
if not is_prime(i):
if not is_prime(n-i):
l.append(i)
l.append(n-i)
break
print(*l, sep = ' ') | Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | ```python
n = int(input())
l = []
def is_prime(n):
is_prime = True
for i in range(2,n):
if n%i == 0:
is_prime = False
return is_prime
for i in range(2,n-1):
if not is_prime(i):
if not is_prime(n-i):
l.append(i)
l.append(n-i)
break
print(*l, sep = ' ')
``` | 3 | |
625 | A | Guest From the Past | PROGRAMMING | 1,700 | [
"implementation",
"math"
] | null | null | Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively. | Print the only integer — maximum number of liters of kefir, that Kolya can drink. | [
"10\n11\n9\n8\n",
"10\n5\n6\n1\n"
] | [
"2\n",
"2\n"
] | In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | 750 | [
{
"input": "10\n11\n9\n8",
"output": "2"
},
{
"input": "10\n5\n6\n1",
"output": "2"
},
{
"input": "2\n2\n2\n1",
"output": "1"
},
{
"input": "10\n3\n3\n1",
"output": "4"
},
{
"input": "10\n1\n2\n1",
"output": "10"
},
{
"input": "10\n2\n3\n1",
"outpu... | 1,696,692,798 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n = int(input())
a = int(input())
b = int(input())
c = int(input())
# Calculate the maximum number of glass bottles Kolya can buy with the initial money
max_glass_bottles = n // b
# Calculate the maximum number of plastic bottles Kolya can buy with the initial money
max_plastic_bottles = n // a
# Calculate the maximum number of glass bottles Kolya can buy with the refunded money
max_glass_with_refund = (n % b + c) // (b - c)
# Calculate the maximum number of kefir liters Kolya can drink using glass bottles
max_kefir_glass = max_glass_bottles * b
# Calculate the maximum number of kefir liters Kolya can drink using plastic bottles
max_kefir_plastic = max_plastic_bottles * a
# Calculate the maximum number of kefir liters Kolya can drink using glass bottles with refund
max_kefir_refund = max_glass_with_refund * b
# Find the maximum of the three possibilities
max_kefir_liters = max(max_kefir_glass, max_kefir_plastic, max_kefir_refund)
# Output the result
print(max_kefir_liters)
| Title: Guest From the Past
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output Specification:
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Demo Input:
['10\n11\n9\n8\n', '10\n5\n6\n1\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | ```python
n = int(input())
a = int(input())
b = int(input())
c = int(input())
# Calculate the maximum number of glass bottles Kolya can buy with the initial money
max_glass_bottles = n // b
# Calculate the maximum number of plastic bottles Kolya can buy with the initial money
max_plastic_bottles = n // a
# Calculate the maximum number of glass bottles Kolya can buy with the refunded money
max_glass_with_refund = (n % b + c) // (b - c)
# Calculate the maximum number of kefir liters Kolya can drink using glass bottles
max_kefir_glass = max_glass_bottles * b
# Calculate the maximum number of kefir liters Kolya can drink using plastic bottles
max_kefir_plastic = max_plastic_bottles * a
# Calculate the maximum number of kefir liters Kolya can drink using glass bottles with refund
max_kefir_refund = max_glass_with_refund * b
# Find the maximum of the three possibilities
max_kefir_liters = max(max_kefir_glass, max_kefir_plastic, max_kefir_refund)
# Output the result
print(max_kefir_liters)
``` | 0 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,563,721,458 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 202 | 0 | a=str(input())
b=str(input())
c=""
for i in range(0, len(a)):
if a[i]==b[i]:
c+="0"
else: c+="1"
for i in range(0, len(c)):
print(c[i], end="")
| Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
a=str(input())
b=str(input())
c=""
for i in range(0, len(a)):
if a[i]==b[i]:
c+="0"
else: c+="1"
for i in range(0, len(c)):
print(c[i], end="")
``` | 3.9495 |
20 | A | BerOS file system | PROGRAMMING | 1,700 | [
"implementation"
] | A. BerOS file system | 2 | 64 | The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form. | The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty. | The path in normalized form. | [
"//usr///local//nginx/sbin\n"
] | [
"/usr/local/nginx/sbin\n"
] | none | 500 | [
{
"input": "//usr///local//nginx/sbin",
"output": "/usr/local/nginx/sbin"
},
{
"input": "////a//b/////g",
"output": "/a/b/g"
},
{
"input": "/a/b/c",
"output": "/a/b/c"
},
{
"input": "/",
"output": "/"
},
{
"input": "////",
"output": "/"
},
{
"input": "... | 1,612,377,663 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 154 | 819,200 | #***************20A - BerOS file system***************#
#author - @Divyesh Chhabra
from math import *
import os
import random
import re
import sys
from itertools import *
path = input()
new = re.split("/",path)
new_path = ""
for i in new:
if i == "":
continue
new_path += (i+"/")
new_path = new_path[:-1]
print("/"+new_path) | Title: BerOS file system
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form.
Input Specification:
The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.
Output Specification:
The path in normalized form.
Demo Input:
['//usr///local//nginx/sbin\n']
Demo Output:
['/usr/local/nginx/sbin\n']
Note:
none | ```python
#***************20A - BerOS file system***************#
#author - @Divyesh Chhabra
from math import *
import os
import random
import re
import sys
from itertools import *
path = input()
new = re.split("/",path)
new_path = ""
for i in new:
if i == "":
continue
new_path += (i+"/")
new_path = new_path[:-1]
print("/"+new_path)
``` | 3.955396 |
313 | A | Ilya and Bank Account | PROGRAMMING | 900 | [
"implementation",
"number theory"
] | null | null | Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift. | The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account. | In a single line print an integer — the maximum state of the bank account that Ilya can get. | [
"2230\n",
"-10\n",
"-100003\n"
] | [
"2230\n",
"0\n",
"-10000\n"
] | In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0. | 500 | [
{
"input": "2230",
"output": "2230"
},
{
"input": "-10",
"output": "0"
},
{
"input": "-100003",
"output": "-10000"
},
{
"input": "544883178",
"output": "544883178"
},
{
"input": "-847251738",
"output": "-84725173"
},
{
"input": "423654797",
"output... | 1,697,883,343 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 30 | 0 | n = int(input())
if n >= 0:
print(n)
else:
num1 = []
num2 = []
num11 = ""
num22 = ""
for x in str(n):
num1.append(x)
num2.append(x)
num1.remove(num1[len(num1)-1])
num1.remove(num1[len(num1)-2])
print(num1)
print(num2)
| Title: Ilya and Bank Account
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
Input Specification:
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
Output Specification:
In a single line print an integer — the maximum state of the bank account that Ilya can get.
Demo Input:
['2230\n', '-10\n', '-100003\n']
Demo Output:
['2230\n', '0\n', '-10000\n']
Note:
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0. | ```python
n = int(input())
if n >= 0:
print(n)
else:
num1 = []
num2 = []
num11 = ""
num22 = ""
for x in str(n):
num1.append(x)
num2.append(x)
num1.remove(num1[len(num1)-1])
num1.remove(num1[len(num1)-2])
print(num1)
print(num2)
``` | -1 | |
365 | A | Good Number | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*). | The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109). | Print a single integer — the number of *k*-good numbers in *a*. | [
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] | [
"10\n",
"1\n"
] | none | 500 | [
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
... | 1,648,318,879 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 31 | 0 | n,k=map(int,input().split(" "))
con=0
l=[]
flag=False
for i in range(n):
x=list(map(int,input()))
z=max(x)
for j in range(len(x)):
if (x[j]<=k and min(x)==0):
flag=True
if flag==True:
con+=1
print(con) | Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none | ```python
n,k=map(int,input().split(" "))
con=0
l=[]
flag=False
for i in range(n):
x=list(map(int,input()))
z=max(x)
for j in range(len(x)):
if (x[j]<=k and min(x)==0):
flag=True
if flag==True:
con+=1
print(con)
``` | 0 | |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,645,038,430 | 2,147,483,647 | Python 3 | OK | TESTS | 88 | 62 | 0 | n = int(input())
a = list(map(int, input().split()))
lp, ls, lo = [0, 1], [0, 1], 0
for i in range(n):
if a[i] == 0:
lo = max(lo, lp[0]*lp[1], ls[0]*ls[1])
ls[1], lp[1] = 0, 0
elif a[i] == 1:
lo, lp = max(lo, lp[0] * lp[1], ls[0] * ls[1]), [1 + max(lo, ls[0] * ls[1]), 1]
ls[1], lp[1] = 0, 1
elif a[i] == 2:
lo, ls = max(lo, lp[0] * lp[1], ls[0] * ls[1]), [1 + max(lo, lp[0] * lp[1]), 1]
ls[1], lp[1] = 1, 0
else:
lo, ls, lp = max(lo, lp[0] * lp[1], ls[0] * ls[1]), [1 + max(lo, lp[0] * lp[1]), 1], [1 + max(lo, ls[0] * ls[1]), 1]
ls[1], lp[1] = 1, 1
print(n-max(lo, ls[0], lp[0])) | Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
n = int(input())
a = list(map(int, input().split()))
lp, ls, lo = [0, 1], [0, 1], 0
for i in range(n):
if a[i] == 0:
lo = max(lo, lp[0]*lp[1], ls[0]*ls[1])
ls[1], lp[1] = 0, 0
elif a[i] == 1:
lo, lp = max(lo, lp[0] * lp[1], ls[0] * ls[1]), [1 + max(lo, ls[0] * ls[1]), 1]
ls[1], lp[1] = 0, 1
elif a[i] == 2:
lo, ls = max(lo, lp[0] * lp[1], ls[0] * ls[1]), [1 + max(lo, lp[0] * lp[1]), 1]
ls[1], lp[1] = 1, 0
else:
lo, ls, lp = max(lo, lp[0] * lp[1], ls[0] * ls[1]), [1 + max(lo, lp[0] * lp[1]), 1], [1 + max(lo, ls[0] * ls[1]), 1]
ls[1], lp[1] = 1, 1
print(n-max(lo, ls[0], lp[0]))
``` | 3 | |
808 | C | Tea Party | PROGRAMMING | 1,400 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1. | The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1. | [
"2 10\n8 7\n",
"4 4\n1 1 1 1\n",
"3 10\n9 8 10\n"
] | [
"6 4 \n",
"1 1 1 1 \n",
"-1\n"
] | In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available. | 0 | [
{
"input": "2 10\n8 7",
"output": "6 4 "
},
{
"input": "4 4\n1 1 1 1",
"output": "1 1 1 1 "
},
{
"input": "3 10\n9 8 10",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "1 1\n2",
"output": "1 "
},
{
"input": "1 10\n20",
"output... | 1,502,909,307 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 62 | 0 |
cilv_sk, tejas_daudz=[int(x) for x in input().split()]
s=input().split()
tilpumi = [ [int(s[i]),i] for i in range(cilv_sk)]
aizpildijums = [0 for i in range(cilv_sk)]
for i in range(cilv_sk):
cur_t = tilpumi[i][0]//2 + tilpumi[i][0]%2
aizpildijums[i] = cur_t
tejas_daudz -= (cur_t)
if(tejas_daudz<0):
#print("Nevar!!!!")
print(-1)
exit(0)
tilpumi.sort(key=lambda a: a[0], reverse=True)
for i in range(cilv_sk):
starp=tilpumi[i][0]-aizpildijums[ tilpumi[i][1] ]
if(starp<tejas_daudz):
aizpildijums[ tilpumi[i][1] ]+=starp
tejas_daudz-=starp
else:
aizpildijums[ tilpumi[i][1] ]+=tejas_daudz
tejas_daudz=0
break
if(tejas_daudz!=0): print(-1)
else:
for i in aizpildijums: print(i, end=" ")
| Title: Tea Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input Specification:
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Demo Input:
['2 10\n8 7\n', '4 4\n1 1 1 1\n', '3 10\n9 8 10\n']
Demo Output:
['6 4 \n', '1 1 1 1 \n', '-1\n']
Note:
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available. | ```python
cilv_sk, tejas_daudz=[int(x) for x in input().split()]
s=input().split()
tilpumi = [ [int(s[i]),i] for i in range(cilv_sk)]
aizpildijums = [0 for i in range(cilv_sk)]
for i in range(cilv_sk):
cur_t = tilpumi[i][0]//2 + tilpumi[i][0]%2
aizpildijums[i] = cur_t
tejas_daudz -= (cur_t)
if(tejas_daudz<0):
#print("Nevar!!!!")
print(-1)
exit(0)
tilpumi.sort(key=lambda a: a[0], reverse=True)
for i in range(cilv_sk):
starp=tilpumi[i][0]-aizpildijums[ tilpumi[i][1] ]
if(starp<tejas_daudz):
aizpildijums[ tilpumi[i][1] ]+=starp
tejas_daudz-=starp
else:
aizpildijums[ tilpumi[i][1] ]+=tejas_daudz
tejas_daudz=0
break
if(tejas_daudz!=0): print(-1)
else:
for i in aizpildijums: print(i, end=" ")
``` | 3 | |
461 | A | Appleman and Toastman | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman. | Print a single integer — the largest possible score. | [
"3\n3 1 5\n",
"1\n10\n"
] | [
"26\n",
"10\n"
] | Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | 500 | [
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\... | 1,480,514,367 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 5,529,600 | num = eval(input()); data = [eval(i) for i in input().split()]
data = sorted(data)
score = 0
for i in range(num):
score += data[i] * (i + 2)
print(score - data[-1]) | Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | ```python
num = eval(input()); data = [eval(i) for i in input().split()]
data = sorted(data)
score = 0
for i in range(num):
score += data[i] * (i + 2)
print(score - data[-1])
``` | 0 | |
514 | B | Han Solo and Lazer Gun | PROGRAMMING | 1,400 | [
"brute force",
"data structures",
"geometry",
"implementation",
"math"
] | null | null | There are *n* Imperial stormtroopers on the field. The battle field is a plane with Cartesian coordinate system. Each stormtrooper is associated with his coordinates (*x*,<=*y*) on this plane.
Han Solo has the newest duplex lazer gun to fight these stormtroopers. It is situated at the point (*x*0,<=*y*0). In one shot it can can destroy all the stormtroopers, situated on some line that crosses point (*x*0,<=*y*0).
Your task is to determine what minimum number of shots Han Solo needs to defeat all the stormtroopers.
The gun is the newest invention, it shoots very quickly and even after a very large number of shots the stormtroopers don't have enough time to realize what's happening and change their location. | The first line contains three integers *n*, *x*0 и *y*0 (1<=≤<=*n*<=≤<=1000, <=-<=104<=≤<=*x*0,<=*y*0<=≤<=104) — the number of stormtroopers on the battle field and the coordinates of your gun.
Next *n* lines contain two integers each *x**i*, *y**i* (<=-<=104<=≤<=*x**i*,<=*y**i*<=≤<=104) — the coordinates of the stormtroopers on the battlefield. It is guaranteed that no stormtrooper stands at the same point with the gun. Multiple stormtroopers can stand at the same point. | Print a single integer — the minimum number of shots Han Solo needs to destroy all the stormtroopers. | [
"4 0 0\n1 1\n2 2\n2 0\n-1 -1\n",
"2 1 2\n1 1\n1 0\n"
] | [
"2\n",
"1\n"
] | Explanation to the first and second samples from the statement, respectively: | 1,000 | [
{
"input": "4 0 0\n1 1\n2 2\n2 0\n-1 -1",
"output": "2"
},
{
"input": "2 1 2\n1 1\n1 0",
"output": "1"
},
{
"input": "1 1 1\n0 0",
"output": "1"
},
{
"input": "2 0 0\n10000 -10000\n-10000 10000",
"output": "1"
},
{
"input": "2 0 0\n10000 -10000\n10000 10000",
... | 1,680,189,359 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 46 | 102,400 | from collections import defaultdict
n, x, y = map(int, input().split())
dic = defaultdict(int)
cnt = ok = 0
while n:
n -= 1
a, b = map(int, input().split())
if x == a: ok = 1
else: dic[(b - y) / (a - x)] += 1
for i in dic:
cnt += dic[i] != 0
print(cnt + ok)
| Title: Han Solo and Lazer Gun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* Imperial stormtroopers on the field. The battle field is a plane with Cartesian coordinate system. Each stormtrooper is associated with his coordinates (*x*,<=*y*) on this plane.
Han Solo has the newest duplex lazer gun to fight these stormtroopers. It is situated at the point (*x*0,<=*y*0). In one shot it can can destroy all the stormtroopers, situated on some line that crosses point (*x*0,<=*y*0).
Your task is to determine what minimum number of shots Han Solo needs to defeat all the stormtroopers.
The gun is the newest invention, it shoots very quickly and even after a very large number of shots the stormtroopers don't have enough time to realize what's happening and change their location.
Input Specification:
The first line contains three integers *n*, *x*0 и *y*0 (1<=≤<=*n*<=≤<=1000, <=-<=104<=≤<=*x*0,<=*y*0<=≤<=104) — the number of stormtroopers on the battle field and the coordinates of your gun.
Next *n* lines contain two integers each *x**i*, *y**i* (<=-<=104<=≤<=*x**i*,<=*y**i*<=≤<=104) — the coordinates of the stormtroopers on the battlefield. It is guaranteed that no stormtrooper stands at the same point with the gun. Multiple stormtroopers can stand at the same point.
Output Specification:
Print a single integer — the minimum number of shots Han Solo needs to destroy all the stormtroopers.
Demo Input:
['4 0 0\n1 1\n2 2\n2 0\n-1 -1\n', '2 1 2\n1 1\n1 0\n']
Demo Output:
['2\n', '1\n']
Note:
Explanation to the first and second samples from the statement, respectively: | ```python
from collections import defaultdict
n, x, y = map(int, input().split())
dic = defaultdict(int)
cnt = ok = 0
while n:
n -= 1
a, b = map(int, input().split())
if x == a: ok = 1
else: dic[(b - y) / (a - x)] += 1
for i in dic:
cnt += dic[i] != 0
print(cnt + ok)
``` | 3 | |
515 | A | Drazil and Date | PROGRAMMING | 1,000 | [
"math"
] | null | null | Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0,<=0) and Varda's home is located in point (*a*,<=*b*). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (*x*,<=*y*) he can go to positions (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1) or (*x*,<=*y*<=-<=1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (*a*,<=*b*) and continue travelling.
Luckily, Drazil arrived to the position (*a*,<=*b*) successfully. Drazil said to Varda: "It took me exactly *s* steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0,<=0) to (*a*,<=*b*) in exactly *s* steps. Can you find out if it is possible for Varda? | You are given three integers *a*, *b*, and *s* (<=-<=109<=≤<=*a*,<=*b*<=≤<=109, 1<=≤<=*s*<=≤<=2·109) in a single line. | If you think Drazil made a mistake and it is impossible to take exactly *s* steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes". | [
"5 5 11\n",
"10 15 25\n",
"0 5 1\n",
"0 0 2\n"
] | [
"No\n",
"Yes\n",
"No\n",
"Yes\n"
] | In fourth sample case one possible route is: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d30660ddf6eb6c64ffd071055a4e8ddd016cde5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 500 | [
{
"input": "5 5 11",
"output": "No"
},
{
"input": "10 15 25",
"output": "Yes"
},
{
"input": "0 5 1",
"output": "No"
},
{
"input": "0 0 2",
"output": "Yes"
},
{
"input": "999999999 999999999 2000000000",
"output": "Yes"
},
{
"input": "-606037695 9983201... | 1,658,838,105 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | a, b, c = map(int, input().split())
if not (a + b) or a+b == c:
print('YES')
else:
print('NO') | Title: Drazil and Date
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0,<=0) and Varda's home is located in point (*a*,<=*b*). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (*x*,<=*y*) he can go to positions (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1) or (*x*,<=*y*<=-<=1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (*a*,<=*b*) and continue travelling.
Luckily, Drazil arrived to the position (*a*,<=*b*) successfully. Drazil said to Varda: "It took me exactly *s* steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0,<=0) to (*a*,<=*b*) in exactly *s* steps. Can you find out if it is possible for Varda?
Input Specification:
You are given three integers *a*, *b*, and *s* (<=-<=109<=≤<=*a*,<=*b*<=≤<=109, 1<=≤<=*s*<=≤<=2·109) in a single line.
Output Specification:
If you think Drazil made a mistake and it is impossible to take exactly *s* steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Demo Input:
['5 5 11\n', '10 15 25\n', '0 5 1\n', '0 0 2\n']
Demo Output:
['No\n', 'Yes\n', 'No\n', 'Yes\n']
Note:
In fourth sample case one possible route is: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d30660ddf6eb6c64ffd071055a4e8ddd016cde5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
a, b, c = map(int, input().split())
if not (a + b) or a+b == c:
print('YES')
else:
print('NO')
``` | 0 | |
581 | C | Developing Skills | PROGRAMMING | 1,400 | [
"implementation",
"math",
"sortings"
] | null | null | Petya loves computer games. Finally a game that he's been waiting for so long came out!
The main character of this game has *n* different skills, each of which is characterized by an integer *a**i* from 0 to 100. The higher the number *a**i* is, the higher is the *i*-th skill of the character. The total rating of the character is calculated as the sum of the values of for all *i* from 1 to *n*. The expression ⌊ *x*⌋ denotes the result of rounding the number *x* down to the nearest integer.
At the beginning of the game Petya got *k* improvement units as a bonus that he can use to increase the skills of his character and his total rating. One improvement unit can increase any skill of Petya's character by exactly one. For example, if *a*4<==<=46, after using one imporvement unit to this skill, it becomes equal to 47. A hero's skill cannot rise higher more than 100. Thus, it is permissible that some of the units will remain unused.
Your task is to determine the optimal way of using the improvement units so as to maximize the overall rating of the character. It is not necessary to use all the improvement units. | The first line of the input contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=107) — the number of skills of the character and the number of units of improvements at Petya's disposal.
The second line of the input contains a sequence of *n* integers *a**i* (0<=≤<=*a**i*<=≤<=100), where *a**i* characterizes the level of the *i*-th skill of the character. | The first line of the output should contain a single non-negative integer — the maximum total rating of the character that Petya can get using *k* or less improvement units. | [
"2 4\n7 9\n",
"3 8\n17 15 19\n",
"2 2\n99 100\n"
] | [
"2\n",
"5\n",
"20\n"
] | In the first test case the optimal strategy is as follows. Petya has to improve the first skill to 10 by spending 3 improvement units, and the second skill to 10, by spending one improvement unit. Thus, Petya spends all his improvement units and the total rating of the character becomes equal to *lfloor* *frac*{100}{10} *rfloor* + *lfloor* *frac*{100}{10} *rfloor* = 10 + 10 = 20.
In the second test the optimal strategy for Petya is to improve the first skill to 20 (by spending 3 improvement units) and to improve the third skill to 20 (in this case by spending 1 improvement units). Thus, Petya is left with 4 improvement units and he will be able to increase the second skill to 19 (which does not change the overall rating, so Petya does not necessarily have to do it). Therefore, the highest possible total rating in this example is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ccaa4e1e435ea3a339c322e03a32de69d214a257.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third test case the optimal strategy for Petya is to increase the first skill to 100 by spending 1 improvement unit. Thereafter, both skills of the character will be equal to 100, so Petya will not be able to spend the remaining improvement unit. So the answer is equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b246630ca7d1b95b91970759bd8455cb3e930bf9.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,500 | [
{
"input": "2 4\n7 9",
"output": "2"
},
{
"input": "3 8\n17 15 19",
"output": "5"
},
{
"input": "2 2\n99 100",
"output": "20"
},
{
"input": "100 10000\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ... | 1,699,928,418 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 31 | 0 | arr = input().split(" ")
points = int(arr[1])
shops = input().split(" ")
buckets = [int(i) for i in shops]
mods = [10 - i%10 for i in buckets]
startPoints = sum([i//10 for i in buckets])
mods = sorted(mods)
ctr = 0
while points != 0:
if points < mods[ctr]:
break
else:
points -= mods[ctr]
ctr += 1
print(ctr + startPoints)
| Title: Developing Skills
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves computer games. Finally a game that he's been waiting for so long came out!
The main character of this game has *n* different skills, each of which is characterized by an integer *a**i* from 0 to 100. The higher the number *a**i* is, the higher is the *i*-th skill of the character. The total rating of the character is calculated as the sum of the values of for all *i* from 1 to *n*. The expression ⌊ *x*⌋ denotes the result of rounding the number *x* down to the nearest integer.
At the beginning of the game Petya got *k* improvement units as a bonus that he can use to increase the skills of his character and his total rating. One improvement unit can increase any skill of Petya's character by exactly one. For example, if *a*4<==<=46, after using one imporvement unit to this skill, it becomes equal to 47. A hero's skill cannot rise higher more than 100. Thus, it is permissible that some of the units will remain unused.
Your task is to determine the optimal way of using the improvement units so as to maximize the overall rating of the character. It is not necessary to use all the improvement units.
Input Specification:
The first line of the input contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=107) — the number of skills of the character and the number of units of improvements at Petya's disposal.
The second line of the input contains a sequence of *n* integers *a**i* (0<=≤<=*a**i*<=≤<=100), where *a**i* characterizes the level of the *i*-th skill of the character.
Output Specification:
The first line of the output should contain a single non-negative integer — the maximum total rating of the character that Petya can get using *k* or less improvement units.
Demo Input:
['2 4\n7 9\n', '3 8\n17 15 19\n', '2 2\n99 100\n']
Demo Output:
['2\n', '5\n', '20\n']
Note:
In the first test case the optimal strategy is as follows. Petya has to improve the first skill to 10 by spending 3 improvement units, and the second skill to 10, by spending one improvement unit. Thus, Petya spends all his improvement units and the total rating of the character becomes equal to *lfloor* *frac*{100}{10} *rfloor* + *lfloor* *frac*{100}{10} *rfloor* = 10 + 10 = 20.
In the second test the optimal strategy for Petya is to improve the first skill to 20 (by spending 3 improvement units) and to improve the third skill to 20 (in this case by spending 1 improvement units). Thus, Petya is left with 4 improvement units and he will be able to increase the second skill to 19 (which does not change the overall rating, so Petya does not necessarily have to do it). Therefore, the highest possible total rating in this example is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ccaa4e1e435ea3a339c322e03a32de69d214a257.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third test case the optimal strategy for Petya is to increase the first skill to 100 by spending 1 improvement unit. Thereafter, both skills of the character will be equal to 100, so Petya will not be able to spend the remaining improvement unit. So the answer is equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b246630ca7d1b95b91970759bd8455cb3e930bf9.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
arr = input().split(" ")
points = int(arr[1])
shops = input().split(" ")
buckets = [int(i) for i in shops]
mods = [10 - i%10 for i in buckets]
startPoints = sum([i//10 for i in buckets])
mods = sorted(mods)
ctr = 0
while points != 0:
if points < mods[ctr]:
break
else:
points -= mods[ctr]
ctr += 1
print(ctr + startPoints)
``` | -1 | |
288 | A | Polo the Penguin and Strings | PROGRAMMING | 1,300 | [
"greedy"
] | null | null | Little penguin Polo adores strings. But most of all he adores strings of length *n*.
One day he wanted to find a string that meets the following conditions:
1. The string consists of *n* lowercase English letters (that is, the string's length equals *n*), exactly *k* of these letters are distinct. 1. No two neighbouring letters of a string coincide; that is, if we represent a string as *s*<==<=*s*1*s*2... *s**n*, then the following inequality holds, *s**i*<=≠<=*s**i*<=+<=1(1<=≤<=*i*<=<<=*n*). 1. Among all strings that meet points 1 and 2, the required string is lexicographically smallest.
Help him find such string or state that such string doesn't exist.
String *x*<==<=*x*1*x*2... *x**p* is lexicographically less than string *y*<==<=*y*1*y*2... *y**q*, if either *p*<=<<=*q* and *x*1<==<=*y*1,<=*x*2<==<=*y*2,<=... ,<=*x**p*<==<=*y**p*, or there is such number *r* (*r*<=<<=*p*,<=*r*<=<<=*q*), that *x*1<==<=*y*1,<=*x*2<==<=*y*2,<=... ,<=*x**r*<==<=*y**r* and *x**r*<=+<=1<=<<=*y**r*<=+<=1. The characters of the strings are compared by their ASCII codes. | A single line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=106,<=1<=≤<=*k*<=≤<=26) — the string's length and the number of distinct letters. | In a single line print the required string. If there isn't such string, print "-1" (without the quotes). | [
"7 4\n",
"4 7\n"
] | [
"ababacd\n",
"-1\n"
] | none | 500 | [
{
"input": "7 4",
"output": "ababacd"
},
{
"input": "4 7",
"output": "-1"
},
{
"input": "10 5",
"output": "abababacde"
},
{
"input": "47 2",
"output": "abababababababababababababababababababababababa"
},
{
"input": "10 7",
"output": "ababacdefg"
},
{
"... | 1,622,334,528 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 7 | 218 | 2,048,000 | import collections
import functools
import math
import sys
import bisect
def In():
return map(int, sys.stdin.readline().split())
input = sys.stdin.readline
def pologen():
n, k = In()
if n < k:
return -1
ans = []
for i in range(n - k + 2):
ans.append(chr(97 + i % 2))
for i in range(2,k):
ans.append(chr(97 + i))
return ''.join(ans)
print(pologen())
| Title: Polo the Penguin and Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little penguin Polo adores strings. But most of all he adores strings of length *n*.
One day he wanted to find a string that meets the following conditions:
1. The string consists of *n* lowercase English letters (that is, the string's length equals *n*), exactly *k* of these letters are distinct. 1. No two neighbouring letters of a string coincide; that is, if we represent a string as *s*<==<=*s*1*s*2... *s**n*, then the following inequality holds, *s**i*<=≠<=*s**i*<=+<=1(1<=≤<=*i*<=<<=*n*). 1. Among all strings that meet points 1 and 2, the required string is lexicographically smallest.
Help him find such string or state that such string doesn't exist.
String *x*<==<=*x*1*x*2... *x**p* is lexicographically less than string *y*<==<=*y*1*y*2... *y**q*, if either *p*<=<<=*q* and *x*1<==<=*y*1,<=*x*2<==<=*y*2,<=... ,<=*x**p*<==<=*y**p*, or there is such number *r* (*r*<=<<=*p*,<=*r*<=<<=*q*), that *x*1<==<=*y*1,<=*x*2<==<=*y*2,<=... ,<=*x**r*<==<=*y**r* and *x**r*<=+<=1<=<<=*y**r*<=+<=1. The characters of the strings are compared by their ASCII codes.
Input Specification:
A single line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=106,<=1<=≤<=*k*<=≤<=26) — the string's length and the number of distinct letters.
Output Specification:
In a single line print the required string. If there isn't such string, print "-1" (without the quotes).
Demo Input:
['7 4\n', '4 7\n']
Demo Output:
['ababacd\n', '-1\n']
Note:
none | ```python
import collections
import functools
import math
import sys
import bisect
def In():
return map(int, sys.stdin.readline().split())
input = sys.stdin.readline
def pologen():
n, k = In()
if n < k:
return -1
ans = []
for i in range(n - k + 2):
ans.append(chr(97 + i % 2))
for i in range(2,k):
ans.append(chr(97 + i))
return ''.join(ans)
print(pologen())
``` | 0 | |
877 | A | Alex and broken contest | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive. | The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem. | Print "YES", if problem is from this contest, and "NO" otherwise. | [
"Alex_and_broken_contest\n",
"NikitaAndString\n",
"Danil_and_Olya\n"
] | [
"NO",
"YES",
"NO"
] | none | 500 | [
{
"input": "Alex_and_broken_contest",
"output": "NO"
},
{
"input": "NikitaAndString",
"output": "YES"
},
{
"input": "Danil_and_Olya",
"output": "NO"
},
{
"input": "Slava____and_the_game",
"output": "YES"
},
{
"input": "Olya_and_energy_drinks",
"output": "YES"
... | 1,509,551,273 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 62 | 0 | name = ['Danil','Olya','Slava','Ann','Nikita']
real = []
word = input()
for i in name:
if word.find(i) != -1:
real.append(i)
if len(real) == 1:
if word.find(real[0],word.find(real[0])+len(real[0])) == -1:
print("YES")
else:
print("NO")
else:
print("NO")
| Title: Alex and broken contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input Specification:
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.
Output Specification:
Print "YES", if problem is from this contest, and "NO" otherwise.
Demo Input:
['Alex_and_broken_contest\n', 'NikitaAndString\n', 'Danil_and_Olya\n']
Demo Output:
['NO', 'YES', 'NO']
Note:
none | ```python
name = ['Danil','Olya','Slava','Ann','Nikita']
real = []
word = input()
for i in name:
if word.find(i) != -1:
real.append(i)
if len(real) == 1:
if word.find(real[0],word.find(real[0])+len(real[0])) == -1:
print("YES")
else:
print("NO")
else:
print("NO")
``` | 3 | |
900 | B | Position in Fraction | PROGRAMMING | 1,300 | [
"math",
"number theory"
] | null | null | You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point. | The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9). | Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1. | [
"1 2 0\n",
"2 3 7\n"
] | [
"2",
"-1"
] | The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction. | 1,000 | [
{
"input": "1 2 0",
"output": "2"
},
{
"input": "2 3 7",
"output": "-1"
},
{
"input": "1 100000 1",
"output": "5"
},
{
"input": "1 7 7",
"output": "6"
},
{
"input": "99999 100000 8",
"output": "-1"
},
{
"input": "44102 73848 2",
"output": "132"
}... | 1,513,387,876 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 17 | 77 | 5,529,600 |
def Decimals(a, b):
denom = list()
while a != 0 and a not in denom:
denom.append(a)
a *= 10
yield a // b
a = a % b
def main():
a, b, c = map(int, input().strip().split())
for (i, n) in enumerate(Decimals(a, b)):
if n == c:
print(i + 1)
return
if c == 0:
print(i + 2)
else:
print(-1)
if __name__ == "__main__":
main()
| Title: Position in Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
Input Specification:
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
Output Specification:
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Demo Input:
['1 2 0\n', '2 3 7\n']
Demo Output:
['2', '-1']
Note:
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction. | ```python
def Decimals(a, b):
denom = list()
while a != 0 and a not in denom:
denom.append(a)
a *= 10
yield a // b
a = a % b
def main():
a, b, c = map(int, input().strip().split())
for (i, n) in enumerate(Decimals(a, b)):
if n == c:
print(i + 1)
return
if c == 0:
print(i + 2)
else:
print(-1)
if __name__ == "__main__":
main()
``` | 0 | |
888 | B | Buggy Robot | PROGRAMMING | 1,000 | [
"greedy"
] | null | null | Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform:
- U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*).
Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations! | The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100).
The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R. | Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell. | [
"4\nLDUR\n",
"5\nRRRUU\n",
"6\nLLRRRR\n"
] | [
"4\n",
"0\n",
"4\n"
] | none | 0 | [
{
"input": "4\nLDUR",
"output": "4"
},
{
"input": "5\nRRRUU",
"output": "0"
},
{
"input": "6\nLLRRRR",
"output": "4"
},
{
"input": "88\nLLUUULRDRRURDDLURRLRDRLLRULRUUDDLLLLRRDDURDURRLDURRLDRRRUULDDLRRRDDRRLUULLURDURUDDDDDLDR",
"output": "76"
},
{
"input": "89\nLDL... | 1,690,397,888 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 31 | 0 | a=int(input())
s=input()
dict1={'U':1,'D':-1}
dict2={'L':-1,'R':1}
c1=0
s1=0
c2=0
s2=0
l=[]
for k in s:
if k in dict1:
s1=s1+dict1[k]
c1+=1
else:
s2=s2+dict2[k]
c2+=1
if s1+s2==0:
l.append(c1+c2)
if len(l)>0:
print(max(l))
else:
print(0) | Title: Buggy Robot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0,<=0). The robot can process commands. There are four types of commands it can perform:
- U — move from the cell (*x*,<=*y*) to (*x*,<=*y*<=+<=1); - D — move from (*x*,<=*y*) to (*x*,<=*y*<=-<=1); - L — move from (*x*,<=*y*) to (*x*<=-<=1,<=*y*); - R — move from (*x*,<=*y*) to (*x*<=+<=1,<=*y*).
Ivan entered a sequence of *n* commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0,<=0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!
Input Specification:
The first line contains one number *n* — the length of sequence of commands entered by Ivan (1<=≤<=*n*<=≤<=100).
The second line contains the sequence itself — a string consisting of *n* characters. Each character can be U, D, L or R.
Output Specification:
Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.
Demo Input:
['4\nLDUR\n', '5\nRRRUU\n', '6\nLLRRRR\n']
Demo Output:
['4\n', '0\n', '4\n']
Note:
none | ```python
a=int(input())
s=input()
dict1={'U':1,'D':-1}
dict2={'L':-1,'R':1}
c1=0
s1=0
c2=0
s2=0
l=[]
for k in s:
if k in dict1:
s1=s1+dict1[k]
c1+=1
else:
s2=s2+dict2[k]
c2+=1
if s1+s2==0:
l.append(c1+c2)
if len(l)>0:
print(max(l))
else:
print(0)
``` | 0 | |
675 | A | Infinite Sequence | PROGRAMMING | 1,100 | [
"math"
] | null | null | Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help. | The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively. | If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes). | [
"1 7 3\n",
"10 10 0\n",
"1 -4 5\n",
"0 60 50\n"
] | [
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] | In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer. | 500 | [
{
"input": "1 7 3",
"output": "YES"
},
{
"input": "10 10 0",
"output": "YES"
},
{
"input": "1 -4 5",
"output": "NO"
},
{
"input": "0 60 50",
"output": "NO"
},
{
"input": "1 -4 -5",
"output": "YES"
},
{
"input": "0 1 0",
"output": "NO"
},
{
... | 1,593,742,166 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 1,000 | 0 | a,b,c=[int(x) for x in input().split()]
if c>0:
while a<b:
a+=c
else:
while b<a:
a+=c
if a==b:
print('YES')
else:
print('NO')
| Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help.
Input Specification:
The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
Output Specification:
If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes).
Demo Input:
['1 7 3\n', '10 10 0\n', '1 -4 5\n', '0 60 50\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer. | ```python
a,b,c=[int(x) for x in input().split()]
if c>0:
while a<b:
a+=c
else:
while b<a:
a+=c
if a==b:
print('YES')
else:
print('NO')
``` | 0 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,602,645,903 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 109 | 0 | sample=list("hello")
counter=0
word=input()
for i in range(len(word)):
if word[i]==sample[counter]:
counter+=1
if counter==5:
print ("YES")
break
if counter<5:
print ("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
sample=list("hello")
counter=0
word=input()
for i in range(len(word)):
if word[i]==sample[counter]:
counter+=1
if counter==5:
print ("YES")
break
if counter<5:
print ("NO")
``` | 3.9455 |
789 | A | Anastasia and pebbles | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | null | null | Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park.
She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day.
Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket.
The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type. | The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles. | [
"3 2\n2 3 4\n",
"5 4\n3 1 8 9 7\n"
] | [
"3\n",
"5\n"
] | In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day.
Optimal sequence of actions in the second sample case:
- In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type. | 500 | [
{
"input": "3 2\n2 3 4",
"output": "3"
},
{
"input": "5 4\n3 1 8 9 7",
"output": "5"
},
{
"input": "1 22\n1",
"output": "1"
},
{
"input": "3 57\n78 165 54",
"output": "3"
},
{
"input": "5 72\n74 10 146 189 184",
"output": "6"
},
{
"input": "9 13\n132 8... | 1,490,805,170 | 1,670 | Python 3 | OK | TESTS | 31 | 124 | 12,902,400 | """def main():
n, k = map(int, input().split())
lst = [int(i) for i in input().split()]
counter = 0
while sum(lst) > 0:
max_c = max(lst)
if max_c >= 2*k:
lst[lst.index(max_c)] -= 2 * k
elif k<=max_c <= 2*k:
lst[lst.index(max_c)] = 0
else:
lst[lst.index(max_c)] = 0
counter += 1
print(main())
"""
n, k = map(int, input().split())
lst = [int(i) for i in input().split()]
count = 0
for i in range(n):
count += (lst[i] + k -1)//k
count = (count + 1)//2
print(count) | Title: Anastasia and pebbles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park.
She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day.
Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket.
The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type.
Output Specification:
The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.
Demo Input:
['3 2\n2 3 4\n', '5 4\n3 1 8 9 7\n']
Demo Output:
['3\n', '5\n']
Note:
In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day.
Optimal sequence of actions in the second sample case:
- In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type. | ```python
"""def main():
n, k = map(int, input().split())
lst = [int(i) for i in input().split()]
counter = 0
while sum(lst) > 0:
max_c = max(lst)
if max_c >= 2*k:
lst[lst.index(max_c)] -= 2 * k
elif k<=max_c <= 2*k:
lst[lst.index(max_c)] = 0
else:
lst[lst.index(max_c)] = 0
counter += 1
print(main())
"""
n, k = map(int, input().split())
lst = [int(i) for i in input().split()]
count = 0
for i in range(n):
count += (lst[i] + k -1)//k
count = (count + 1)//2
print(count)
``` | 3 | |
830 | B | Cards Sorting | PROGRAMMING | 1,600 | [
"data structures",
"implementation",
"sortings"
] | null | null | Vasily has a deck of cards consisting of *n* cards. There is an integer on each of the cards, this integer is between 1 and 100<=000, inclusive. It is possible that some cards have the same integers on them.
Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.
You are to determine the total number of times Vasily takes the top card from the deck. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cards in the deck.
The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), where *a**i* is the number written on the *i*-th from top card in the deck. | Print the total number of times Vasily takes the top card from the deck. | [
"4\n6 3 1 2\n",
"1\n1000\n",
"7\n3 3 3 3 3 3 3\n"
] | [
"7\n",
"1\n",
"7\n"
] | In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards. | 1,000 | [
{
"input": "4\n6 3 1 2",
"output": "7"
},
{
"input": "1\n1000",
"output": "1"
},
{
"input": "7\n3 3 3 3 3 3 3",
"output": "7"
},
{
"input": "64\n826 142 89 337 897 891 1004 704 281 644 910 852 147 193 289 384 625 695 416 944 162 939 164 1047 359 114 499 99 713 300 268 316 256... | 1,503,470,619 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 1,000 | 7,270,400 | n = int(input())
l = list(map(int, input().split()))
z = sorted(list(map(int, set(l))))
q = []
for i in z:
q.append(l.count(i))
k = 0
g = 0
while len(l) != 0:
g += 1
if q[k] == 0:
k += 1
if l[0] == z[k]:
q[k] -= 1
l = l[1:]
else:
l = l[1:] + l[:1]
print(g)
| Title: Cards Sorting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily has a deck of cards consisting of *n* cards. There is an integer on each of the cards, this integer is between 1 and 100<=000, inclusive. It is possible that some cards have the same integers on them.
Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.
You are to determine the total number of times Vasily takes the top card from the deck.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cards in the deck.
The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), where *a**i* is the number written on the *i*-th from top card in the deck.
Output Specification:
Print the total number of times Vasily takes the top card from the deck.
Demo Input:
['4\n6 3 1 2\n', '1\n1000\n', '7\n3 3 3 3 3 3 3\n']
Demo Output:
['7\n', '1\n', '7\n']
Note:
In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards. | ```python
n = int(input())
l = list(map(int, input().split()))
z = sorted(list(map(int, set(l))))
q = []
for i in z:
q.append(l.count(i))
k = 0
g = 0
while len(l) != 0:
g += 1
if q[k] == 0:
k += 1
if l[0] == z[k]:
q[k] -= 1
l = l[1:]
else:
l = l[1:] + l[:1]
print(g)
``` | 0 | |
814 | C | An impassioned circulation of affection | PROGRAMMING | 1,600 | [
"brute force",
"dp",
"strings",
"two pointers"
] | null | null | Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!
Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has *n* pieces numbered from 1 to *n* from left to right, and the *i*-th piece has a colour *s**i*, denoted by a lowercase English letter. Nadeko will repaint at most *m* of the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour *c* — Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.
For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o". Among all subsegments containing pieces of "o" only, "ooo" is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.
But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has *q* plans on this, each of which can be expressed as a pair of an integer *m**i* and a lowercase letter *c**i*, meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan. | The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=500) — the length of the garland.
The second line contains *n* lowercase English letters *s*1*s*2... *s**n* as a string — the initial colours of paper pieces on the garland.
The third line contains a positive integer *q* (1<=≤<=*q*<=≤<=200<=000) — the number of plans Nadeko has.
The next *q* lines describe one plan each: the *i*-th among them contains an integer *m**i* (1<=≤<=*m**i*<=≤<=*n*) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter *c**i* — Koyomi's possible favourite colour. | Output *q* lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it. | [
"6\nkoyomi\n3\n1 o\n4 o\n4 m\n",
"15\nyamatonadeshiko\n10\n1 a\n2 a\n3 a\n4 a\n5 a\n1 b\n2 b\n3 b\n4 b\n5 b\n",
"10\naaaaaaaaaa\n2\n10 b\n10 z\n"
] | [
"3\n6\n5\n",
"3\n4\n5\n7\n8\n1\n2\n3\n4\n5\n",
"10\n10\n"
] | In the first sample, there are three plans:
- In the first plan, at most 1 piece can be repainted. Repainting the "y" piece to become "o" results in "kooomi", whose Koyomity of 3 is the best achievable; - In the second plan, at most 4 pieces can be repainted, and "oooooo" results in a Koyomity of 6; - In the third plan, at most 4 pieces can be repainted, and "mmmmmi" and "kmmmmm" both result in a Koyomity of 5. | 1,750 | [
{
"input": "6\nkoyomi\n3\n1 o\n4 o\n4 m",
"output": "3\n6\n5"
},
{
"input": "15\nyamatonadeshiko\n10\n1 a\n2 a\n3 a\n4 a\n5 a\n1 b\n2 b\n3 b\n4 b\n5 b",
"output": "3\n4\n5\n7\n8\n1\n2\n3\n4\n5"
},
{
"input": "10\naaaaaaaaaa\n2\n10 b\n10 z",
"output": "10\n10"
},
{
"input": "1... | 1,499,701,361 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 5 | 46 | 5,529,600 | def im(i=None):
if not i:
i = input()
if i.isdigit():
return int(i)
else:
return map(int, i.split(' '))
n = int(input())
s = input()
q = int(input())
smb = {}
for c in set(s):
smb[c] = [0]*n
i = 0
while i < n:
repl_cnt = 0
j = i
while j < n:
if s[j] != c:
repl_cnt += 1
smb[c][repl_cnt] = max(smb[c][repl_cnt], j-i+1)
j += 1
i += 1
for i in range(1, n):
smb[c][i] = max(smb[c][i], smb[c][i-1])
for _ in range(q):
buf = input().split(' ')
m = int(buf[0])
c = buf[1]
if c not in smb:
print(m)
else:
print(smb[c][m])
| Title: An impassioned circulation of affection
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!
Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has *n* pieces numbered from 1 to *n* from left to right, and the *i*-th piece has a colour *s**i*, denoted by a lowercase English letter. Nadeko will repaint at most *m* of the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour *c* — Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.
For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o". Among all subsegments containing pieces of "o" only, "ooo" is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.
But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has *q* plans on this, each of which can be expressed as a pair of an integer *m**i* and a lowercase letter *c**i*, meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.
Input Specification:
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=500) — the length of the garland.
The second line contains *n* lowercase English letters *s*1*s*2... *s**n* as a string — the initial colours of paper pieces on the garland.
The third line contains a positive integer *q* (1<=≤<=*q*<=≤<=200<=000) — the number of plans Nadeko has.
The next *q* lines describe one plan each: the *i*-th among them contains an integer *m**i* (1<=≤<=*m**i*<=≤<=*n*) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter *c**i* — Koyomi's possible favourite colour.
Output Specification:
Output *q* lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it.
Demo Input:
['6\nkoyomi\n3\n1 o\n4 o\n4 m\n', '15\nyamatonadeshiko\n10\n1 a\n2 a\n3 a\n4 a\n5 a\n1 b\n2 b\n3 b\n4 b\n5 b\n', '10\naaaaaaaaaa\n2\n10 b\n10 z\n']
Demo Output:
['3\n6\n5\n', '3\n4\n5\n7\n8\n1\n2\n3\n4\n5\n', '10\n10\n']
Note:
In the first sample, there are three plans:
- In the first plan, at most 1 piece can be repainted. Repainting the "y" piece to become "o" results in "kooomi", whose Koyomity of 3 is the best achievable; - In the second plan, at most 4 pieces can be repainted, and "oooooo" results in a Koyomity of 6; - In the third plan, at most 4 pieces can be repainted, and "mmmmmi" and "kmmmmm" both result in a Koyomity of 5. | ```python
def im(i=None):
if not i:
i = input()
if i.isdigit():
return int(i)
else:
return map(int, i.split(' '))
n = int(input())
s = input()
q = int(input())
smb = {}
for c in set(s):
smb[c] = [0]*n
i = 0
while i < n:
repl_cnt = 0
j = i
while j < n:
if s[j] != c:
repl_cnt += 1
smb[c][repl_cnt] = max(smb[c][repl_cnt], j-i+1)
j += 1
i += 1
for i in range(1, n):
smb[c][i] = max(smb[c][i], smb[c][i-1])
for _ in range(q):
buf = input().split(' ')
m = int(buf[0])
c = buf[1]
if c not in smb:
print(m)
else:
print(smb[c][m])
``` | -1 | |
519 | A | A and B and Chess | PROGRAMMING | 900 | [
"implementation"
] | null | null | A and B are preparing themselves for programming contests.
To train their logical thinking and solve problems better, A and B decided to play chess. During the game A wondered whose position is now stronger.
For each chess piece we know its weight:
- the queen's weight is 9, - the rook's weight is 5, - the bishop's weight is 3, - the knight's weight is 3, - the pawn's weight is 1, - the king's weight isn't considered in evaluating position.
The player's weight equals to the sum of weights of all his pieces on the board.
As A doesn't like counting, he asked you to help him determine which player has the larger position weight. | The input contains eight lines, eight characters each — the board's description.
The white pieces on the board are marked with uppercase letters, the black pieces are marked with lowercase letters.
The white pieces are denoted as follows: the queen is represented is 'Q', the rook — as 'R', the bishop — as'B', the knight — as 'N', the pawn — as 'P', the king — as 'K'.
The black pieces are denoted as 'q', 'r', 'b', 'n', 'p', 'k', respectively.
An empty square of the board is marked as '.' (a dot).
It is not guaranteed that the given chess position can be achieved in a real game. Specifically, there can be an arbitrary (possibly zero) number pieces of each type, the king may be under attack and so on. | Print "White" (without quotes) if the weight of the position of the white pieces is more than the weight of the position of the black pieces, print "Black" if the weight of the black pieces is more than the weight of the white pieces and print "Draw" if the weights of the white and black pieces are equal. | [
"...QK...\n........\n........\n........\n........\n........\n........\n...rk...\n",
"rnbqkbnr\npppppppp\n........\n........\n........\n........\nPPPPPPPP\nRNBQKBNR\n",
"rppppppr\n...k....\n........\n........\n........\n........\nK...Q...\n........\n"
] | [
"White\n",
"Draw\n",
"Black\n"
] | In the first test sample the weight of the position of the white pieces equals to 9, the weight of the position of the black pieces equals 5.
In the second test sample the weights of the positions of the black and the white pieces are equal to 39.
In the third test sample the weight of the position of the white pieces equals to 9, the weight of the position of the black pieces equals to 16. | 500 | [
{
"input": "rnbqkbnr\npppppppp\n........\n........\n........\n........\nPPPPPPPP\nRNBQKBNR",
"output": "Draw"
},
{
"input": "....bQ.K\n.B......\n.....P..\n........\n........\n........\n...N.P..\n.....R..",
"output": "White"
},
{
"input": "b....p..\nR.......\n.pP...b.\npp......\nq.PPNpPR\... | 1,668,531,253 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 65 | 61 | 0 | w=b=0
for i in range(8):
s=input()
for j in s:
if j!=".":
if j=="p":
b+=1
elif j=="n":
b+=3
elif j=="b":
b+=3
elif j=="r":
b+=5
elif j=="q":
b+=9
elif j=="P":
w+=1
elif j=="N":
w+=3
elif j=="B":
w+=3
elif j=="R":
w+=5
elif j=="Q":
w+=9
if w<b:
print("Black")
elif w==b:
print("Draw")
else :
print("White") | Title: A and B and Chess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
To train their logical thinking and solve problems better, A and B decided to play chess. During the game A wondered whose position is now stronger.
For each chess piece we know its weight:
- the queen's weight is 9, - the rook's weight is 5, - the bishop's weight is 3, - the knight's weight is 3, - the pawn's weight is 1, - the king's weight isn't considered in evaluating position.
The player's weight equals to the sum of weights of all his pieces on the board.
As A doesn't like counting, he asked you to help him determine which player has the larger position weight.
Input Specification:
The input contains eight lines, eight characters each — the board's description.
The white pieces on the board are marked with uppercase letters, the black pieces are marked with lowercase letters.
The white pieces are denoted as follows: the queen is represented is 'Q', the rook — as 'R', the bishop — as'B', the knight — as 'N', the pawn — as 'P', the king — as 'K'.
The black pieces are denoted as 'q', 'r', 'b', 'n', 'p', 'k', respectively.
An empty square of the board is marked as '.' (a dot).
It is not guaranteed that the given chess position can be achieved in a real game. Specifically, there can be an arbitrary (possibly zero) number pieces of each type, the king may be under attack and so on.
Output Specification:
Print "White" (without quotes) if the weight of the position of the white pieces is more than the weight of the position of the black pieces, print "Black" if the weight of the black pieces is more than the weight of the white pieces and print "Draw" if the weights of the white and black pieces are equal.
Demo Input:
['...QK...\n........\n........\n........\n........\n........\n........\n...rk...\n', 'rnbqkbnr\npppppppp\n........\n........\n........\n........\nPPPPPPPP\nRNBQKBNR\n', 'rppppppr\n...k....\n........\n........\n........\n........\nK...Q...\n........\n']
Demo Output:
['White\n', 'Draw\n', 'Black\n']
Note:
In the first test sample the weight of the position of the white pieces equals to 9, the weight of the position of the black pieces equals 5.
In the second test sample the weights of the positions of the black and the white pieces are equal to 39.
In the third test sample the weight of the position of the white pieces equals to 9, the weight of the position of the black pieces equals to 16. | ```python
w=b=0
for i in range(8):
s=input()
for j in s:
if j!=".":
if j=="p":
b+=1
elif j=="n":
b+=3
elif j=="b":
b+=3
elif j=="r":
b+=5
elif j=="q":
b+=9
elif j=="P":
w+=1
elif j=="N":
w+=3
elif j=="B":
w+=3
elif j=="R":
w+=5
elif j=="Q":
w+=9
if w<b:
print("Black")
elif w==b:
print("Draw")
else :
print("White")
``` | 3 | |
414 | B | Mashmokh and ACM | PROGRAMMING | 1,400 | [
"combinatorics",
"dp",
"number theory"
] | null | null | Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of *l* integers *b*1,<=*b*2,<=...,<=*b**l* (1<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**l*<=≤<=*n*) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all *i* (1<=≤<=*i*<=≤<=*l*<=-<=1).
Given *n* and *k* find the number of good sequences of length *k*. As the answer can be rather large print it modulo 1000000007 (109<=+<=7). | The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*,<=*k*<=≤<=2000). | Output a single integer — the number of good sequences of length *k* modulo 1000000007 (109<=+<=7). | [
"3 2\n",
"6 4\n",
"2 1\n"
] | [
"5\n",
"39\n",
"2\n"
] | In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | 1,000 | [
{
"input": "3 2",
"output": "5"
},
{
"input": "6 4",
"output": "39"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "1478 194",
"output": "312087753"
},
{
"input": "1415 562",
"output": "953558593"
},
{
"input": "1266 844",
"output": "735042656"
... | 1,655,192,910 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 |
# import sys
import pyrival.misc
@pyrival.misc.bootstrap
a,b=list(map(int,input().split()))
g={}
for i in range(1,a+1):
g[i]=[]
c=1
while i*c<a+1:
g[i].append(i*c)
c=c+1
#print(g)
# sys.setrecursionlimit(10**7)
h={}
def recr(val,ind):
if (val,ind) in h:
yield h[(val,ind)]
if ind==b:
yield 1
ans=0
for i in g[val]:
if i%val==0:
ans=ans+recr(i,ind+1)
h[(val,ind)]=ans
yield ans
print(recr(1,0)%(10**9+7)) | Title: Mashmokh and ACM
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of *l* integers *b*1,<=*b*2,<=...,<=*b**l* (1<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**l*<=≤<=*n*) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all *i* (1<=≤<=*i*<=≤<=*l*<=-<=1).
Given *n* and *k* find the number of good sequences of length *k*. As the answer can be rather large print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*,<=*k*<=≤<=2000).
Output Specification:
Output a single integer — the number of good sequences of length *k* modulo 1000000007 (109<=+<=7).
Demo Input:
['3 2\n', '6 4\n', '2 1\n']
Demo Output:
['5\n', '39\n', '2\n']
Note:
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | ```python
# import sys
import pyrival.misc
@pyrival.misc.bootstrap
a,b=list(map(int,input().split()))
g={}
for i in range(1,a+1):
g[i]=[]
c=1
while i*c<a+1:
g[i].append(i*c)
c=c+1
#print(g)
# sys.setrecursionlimit(10**7)
h={}
def recr(val,ind):
if (val,ind) in h:
yield h[(val,ind)]
if ind==b:
yield 1
ans=0
for i in g[val]:
if i%val==0:
ans=ans+recr(i,ind+1)
h[(val,ind)]=ans
yield ans
print(recr(1,0)%(10**9+7))
``` | -1 | |
372 | A | Counting Kangaroos is Fun | PROGRAMMING | 1,600 | [
"binary search",
"greedy",
"sortings",
"two pointers"
] | null | null | There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible. | The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105). | Output a single integer — the optimal number of visible kangaroos. | [
"8\n2\n5\n7\n6\n9\n8\n4\n2\n",
"8\n9\n1\n6\n2\n6\n5\n8\n3\n"
] | [
"5\n",
"5\n"
] | none | 500 | [
{
"input": "8\n2\n5\n7\n6\n9\n8\n4\n2",
"output": "5"
},
{
"input": "8\n9\n1\n6\n2\n6\n5\n8\n3",
"output": "5"
},
{
"input": "12\n3\n99\n24\n46\n75\n63\n57\n55\n10\n62\n34\n52",
"output": "7"
},
{
"input": "12\n55\n75\n1\n98\n63\n64\n9\n39\n82\n18\n47\n9",
"output": "6"
... | 1,670,564,465 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 623 | 12,288,000 | import sys, math
n=int(input())
a=sorted(int(x) for x in sys.stdin)
i=(n//2)-1
j=n-1
k=0
while j>((n//2)-1) and i>=0:
if 2*a[i]<=a[j]:
j-=1
k+=1
i-=1
#print(a)
print(n-k)
| Title: Counting Kangaroos is Fun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input Specification:
The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105).
Output Specification:
Output a single integer — the optimal number of visible kangaroos.
Demo Input:
['8\n2\n5\n7\n6\n9\n8\n4\n2\n', '8\n9\n1\n6\n2\n6\n5\n8\n3\n']
Demo Output:
['5\n', '5\n']
Note:
none | ```python
import sys, math
n=int(input())
a=sorted(int(x) for x in sys.stdin)
i=(n//2)-1
j=n-1
k=0
while j>((n//2)-1) and i>=0:
if 2*a[i]<=a[j]:
j-=1
k+=1
i-=1
#print(a)
print(n-k)
``` | 3 | |
461 | A | Appleman and Toastman | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman. | Print a single integer — the largest possible score. | [
"3\n3 1 5\n",
"1\n10\n"
] | [
"26\n",
"10\n"
] | Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | 500 | [
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\... | 1,512,550,052 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 11,059,200 | n=int(input())
number=list(map(int,input().split( )))
number.sort()
toastman=sum(number)
for i in range(len(number)-1):
toastman+=sum(number[i:len(number)])
print(toastman)
| Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | ```python
n=int(input())
number=list(map(int,input().split( )))
number.sort()
toastman=sum(number)
for i in range(len(number)-1):
toastman+=sum(number[i:len(number)])
print(toastman)
``` | 0 | |
556 | A | Case of the Zeros and Ones | PROGRAMMING | 900 | [
"greedy"
] | null | null | Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones. | Output the minimum length of the string that may remain after applying the described operations several times. | [
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] | [
"0\n",
"1\n",
"6\n"
] | In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 250 | [
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input"... | 1,632,950,373 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | def main():
n = int(input())
s = list(input())
i = 0
c = 0
while c == 0 and i < len(s)-1:
if len(s)==1 or s.count(0)==0 or s.count(1==0):
break
if s[i] == "1":
if s[i + 1] == "0":
s = s[:i] + s[i + 1 + 1:]
i = 0
else:
i = i + 1
elif s[i] == "0":
if s[i + 1] == "1":
s = s[:i] + s[i + 1 + 1:]
i = 0
else:
i = i + 1
print(len(s))
if __name__ == '__main__':
main() | Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
def main():
n = int(input())
s = list(input())
i = 0
c = 0
while c == 0 and i < len(s)-1:
if len(s)==1 or s.count(0)==0 or s.count(1==0):
break
if s[i] == "1":
if s[i + 1] == "0":
s = s[:i] + s[i + 1 + 1:]
i = 0
else:
i = i + 1
elif s[i] == "0":
if s[i + 1] == "1":
s = s[:i] + s[i + 1 + 1:]
i = 0
else:
i = i + 1
print(len(s))
if __name__ == '__main__':
main()
``` | 0 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,660,629,991 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | s = int(input())
t = int(input())
print(s^t) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
s = int(input())
t = int(input())
print(s^t)
``` | 0 |
17 | D | Notepad | PROGRAMMING | 2,400 | [
"number theory"
] | D. Notepad | 2 | 64 | Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base *b* caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length *n* without leading zeros in this number system. Each page in Nick's notepad has enough space for *c* numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page. | The only input line contains three space-separated integers *b*, *n* and *c* (2<=≤<=*b*<=<<=10106, 1<=≤<=*n*<=<<=10106, 1<=≤<=*c*<=≤<=109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros. | In the only line output the amount of numbers written on the same page as the last number. | [
"2 3 3\n",
"2 3 4\n"
] | [
"1",
"4"
] | In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page. | 0 | [
{
"input": "2 3 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "4"
},
{
"input": "9 1 79",
"output": "8"
},
{
"input": "9 1 345",
"output": "8"
},
{
"input": "9 9 999982045",
"output": "344373768"
},
{
"input": "4 42 44",
"output": "12"
},
... | 1,548,888,694 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 186 | 0 | def gcd(a, b):
if (a == 0):
return b
return gcd(b % a, a)
def phi(n):
result = 1
for i in range(2, n):
if (gcd(i, n) == 1):
result+=1
return result
[b,n,c] = [int(k) for k in input().split(' ')]
result = (pow(b%c,(n-1) % phi(c),c) * ((b-1) % c)) % c
if result == 0:
print(c)
else:
print(result)
| Title: Notepad
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base *b* caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length *n* without leading zeros in this number system. Each page in Nick's notepad has enough space for *c* numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.
Would you help Nick find out how many numbers will be written on the last page.
Input Specification:
The only input line contains three space-separated integers *b*, *n* and *c* (2<=≤<=*b*<=<<=10106, 1<=≤<=*n*<=<<=10106, 1<=≤<=*c*<=≤<=109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros.
Output Specification:
In the only line output the amount of numbers written on the same page as the last number.
Demo Input:
['2 3 3\n', '2 3 4\n']
Demo Output:
['1', '4']
Note:
In both samples there are exactly 4 numbers of length 3 in binary number system. In the first sample Nick writes 3 numbers on the first page and 1 on the second page. In the second sample all the 4 numbers can be written on the first page. | ```python
def gcd(a, b):
if (a == 0):
return b
return gcd(b % a, a)
def phi(n):
result = 1
for i in range(2, n):
if (gcd(i, n) == 1):
result+=1
return result
[b,n,c] = [int(k) for k in input().split(' ')]
result = (pow(b%c,(n-1) % phi(c),c) * ((b-1) % c)) % c
if result == 0:
print(c)
else:
print(result)
``` | 0 |
43 | A | Football | PROGRAMMING | 1,000 | [
"strings"
] | A. Football | 2 | 256 | One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. | Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. | [
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] | [
"ABC\n",
"A\n"
] | none | 500 | [
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
... | 1,609,697,388 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 216 | 0 | n = int(input())
l = [input() for _ in range(n)]
a = l[:]
a = list(set(a))
save = []
for items in a:
save.append(l.count(items))
for c in l:
if max(save)==l.count(c):
print(c)
break
| Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none | ```python
n = int(input())
l = [input() for _ in range(n)]
a = l[:]
a = list(set(a))
save = []
for items in a:
save.append(l.count(items))
for c in l:
if max(save)==l.count(c):
print(c)
break
``` | 3.946 |
682 | A | Alyona and Numbers | PROGRAMMING | 1,100 | [
"constructive algorithms",
"math",
"number theory"
] | null | null | After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help. | The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000). | Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5. | [
"6 12\n",
"11 14\n",
"1 5\n",
"3 8\n",
"5 7\n",
"21 21\n"
] | [
"14\n",
"31\n",
"1\n",
"5\n",
"7\n",
"88\n"
] | Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | 500 | [
{
"input": "6 12",
"output": "14"
},
{
"input": "11 14",
"output": "31"
},
{
"input": "1 5",
"output": "1"
},
{
"input": "3 8",
"output": "5"
},
{
"input": "5 7",
"output": "7"
},
{
"input": "21 21",
"output": "88"
},
{
"input": "10 15",
... | 1,573,251,340 | 2,147,483,647 | Python 3 | OK | TESTS | 128 | 608 | 0 | def inp():
return map(int, input().split())
def fill_arr(n):
return [0 for x in range(n)]
x, y = inp()
a, b = fill_arr(5), fill_arr(5)
for i in range(1, x + 1):
a[i % 5] += 1
for i in range(1, y + 1):
b[i % 5] += 1
sum = a[0] * b[0]
for i in range(1,5):
sum += a[i] * b[5 - i]
print(sum)
| Title: Alyona and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).
Output Specification:
Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.
Demo Input:
['6 12\n', '11 14\n', '1 5\n', '3 8\n', '5 7\n', '21 21\n']
Demo Output:
['14\n', '31\n', '1\n', '5\n', '7\n', '88\n']
Note:
Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | ```python
def inp():
return map(int, input().split())
def fill_arr(n):
return [0 for x in range(n)]
x, y = inp()
a, b = fill_arr(5), fill_arr(5)
for i in range(1, x + 1):
a[i % 5] += 1
for i in range(1, y + 1):
b[i % 5] += 1
sum = a[0] * b[0]
for i in range(1,5):
sum += a[i] * b[5 - i]
print(sum)
``` | 3 | |
437 | A | The Child and Homework | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
- If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose? | The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_". | Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes). | [
"A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n",
"A.ab\nB.abcde\nC.ab\nD.abc\n",
"A.c\nB.cc\nC.c\nD.c\n"
] | [
"D\n",
"C\n",
"B\n"
] | In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | 500 | [
{
"input": "A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute",
"output": "D"
},
{
"input": "A.ab\nB.abcde\nC.ab\nD.abc",
"output": "C"
},
{
"input": "A.c\nB.cc\nC.c\nD.c",
"output": "B"
},
... | 1,568,778,401 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
struct ppp{
string h;int len;
}ln[5];
bool comp(ppp a,ppp b)
{
return a.len<b.len;
}
char ttt='9',yyy='9';
int main()
{
for(int i=1;i<=4;i++)
{
cin>>ln[i].h;
ln[i].len=ln[i].h.size()-2;
}
sort(ln+1,ln+5,comp);
if(ln[1].len<=ln[2].len/2)ttt=ln[1].h[0];
if(ln[4].len>=ln[3].len*2)yyy=ln[4].h[0];
if(ttt==yyy||ttt>='A'&&ttt<='D'&&yyy>='A'&&yyy<='D')printf("C");
else if(ttt=='9')putchar(yyy);
else putchar(ttt);
return 0;
} | Title: The Child and Homework
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
- If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input Specification:
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output Specification:
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Demo Input:
['A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n', 'A.ab\nB.abcde\nC.ab\nD.abc\n', 'A.c\nB.cc\nC.c\nD.c\n']
Demo Output:
['D\n', 'C\n', 'B\n']
Note:
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | ```python
#include<bits/stdc++.h>
using namespace std;
struct ppp{
string h;int len;
}ln[5];
bool comp(ppp a,ppp b)
{
return a.len<b.len;
}
char ttt='9',yyy='9';
int main()
{
for(int i=1;i<=4;i++)
{
cin>>ln[i].h;
ln[i].len=ln[i].h.size()-2;
}
sort(ln+1,ln+5,comp);
if(ln[1].len<=ln[2].len/2)ttt=ln[1].h[0];
if(ln[4].len>=ln[3].len*2)yyy=ln[4].h[0];
if(ttt==yyy||ttt>='A'&&ttt<='D'&&yyy>='A'&&yyy<='D')printf("C");
else if(ttt=='9')putchar(yyy);
else putchar(ttt);
return 0;
}
``` | -1 | |
818 | D | Multicolored Cars | PROGRAMMING | 1,700 | [
"data structures",
"implementation"
] | null | null | Alice and Bob got very bored during a long car trip so they decided to play a game. From the window they can see cars of different colors running past them. Cars are going one after another.
The game rules are like this. Firstly Alice chooses some color *A*, then Bob chooses some color *B* (*A*<=≠<=*B*). After each car they update the number of cars of their chosen color that have run past them. Let's define this numbers after *i*-th car *cnt**A*(*i*) and *cnt**B*(*i*).
- If *cnt**A*(*i*)<=><=*cnt**B*(*i*) for every *i* then the winner is Alice. - If *cnt**B*(*i*)<=≥<=*cnt**A*(*i*) for every *i* then the winner is Bob. - Otherwise it's a draw.
Bob knows all the colors of cars that they will encounter and order of their appearance. Alice have already chosen her color *A* and Bob now wants to choose such color *B* that he will win the game (draw is not a win). Help him find this color.
If there are multiple solutions, print any of them. If there is no such color then print -1. | The first line contains two integer numbers *n* and *A* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*A*<=≤<=106) – number of cars and the color chosen by Alice.
The second line contains *n* integer numbers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — colors of the cars that Alice and Bob will encounter in the order of their appearance. | Output such color *B* (1<=≤<=*B*<=≤<=106) that if Bob chooses it then he will win the game. If there are multiple solutions, print any of them. If there is no such color then print -1.
It is guaranteed that if there exists any solution then there exists solution with (1<=≤<=*B*<=≤<=106). | [
"4 1\n2 1 4 2\n",
"5 2\n2 2 4 5 3\n",
"3 10\n1 2 3\n"
] | [
"2\n",
"-1\n",
"4\n"
] | Let's consider availability of colors in the first example:
- *cnt*<sub class="lower-index">2</sub>(*i*) ≥ *cnt*<sub class="lower-index">1</sub>(*i*) for every *i*, and color 2 can be the answer. - *cnt*<sub class="lower-index">4</sub>(2) < *cnt*<sub class="lower-index">1</sub>(2), so color 4 isn't the winning one for Bob. - All the other colors also have *cnt*<sub class="lower-index">*j*</sub>(2) < *cnt*<sub class="lower-index">1</sub>(2), thus they are not available.
In the third example every color is acceptable except for 10. | 0 | [
{
"input": "4 1\n2 1 4 2",
"output": "2"
},
{
"input": "5 2\n2 2 4 5 3",
"output": "-1"
},
{
"input": "3 10\n1 2 3",
"output": "4"
},
{
"input": "1 1\n2",
"output": "3"
},
{
"input": "1 2\n2",
"output": "-1"
},
{
"input": "10 6\n8 5 1 6 6 5 10 6 9 8",
... | 1,511,182,780 | 2,147,483,647 | PyPy 3 | OK | TESTS | 95 | 265 | 10,649,600 | f = lambda: map(int, input().split())
n, a = f()
d = [0] * 1000001
for q in f():
if d[q] < d[a]: d[q] = -1
d[q] += 1
k = d[a]
d[0] = d[a] = -1
s = max(d)
print(d.index(s) if s >= k else -1) | Title: Multicolored Cars
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice and Bob got very bored during a long car trip so they decided to play a game. From the window they can see cars of different colors running past them. Cars are going one after another.
The game rules are like this. Firstly Alice chooses some color *A*, then Bob chooses some color *B* (*A*<=≠<=*B*). After each car they update the number of cars of their chosen color that have run past them. Let's define this numbers after *i*-th car *cnt**A*(*i*) and *cnt**B*(*i*).
- If *cnt**A*(*i*)<=><=*cnt**B*(*i*) for every *i* then the winner is Alice. - If *cnt**B*(*i*)<=≥<=*cnt**A*(*i*) for every *i* then the winner is Bob. - Otherwise it's a draw.
Bob knows all the colors of cars that they will encounter and order of their appearance. Alice have already chosen her color *A* and Bob now wants to choose such color *B* that he will win the game (draw is not a win). Help him find this color.
If there are multiple solutions, print any of them. If there is no such color then print -1.
Input Specification:
The first line contains two integer numbers *n* and *A* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*A*<=≤<=106) – number of cars and the color chosen by Alice.
The second line contains *n* integer numbers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — colors of the cars that Alice and Bob will encounter in the order of their appearance.
Output Specification:
Output such color *B* (1<=≤<=*B*<=≤<=106) that if Bob chooses it then he will win the game. If there are multiple solutions, print any of them. If there is no such color then print -1.
It is guaranteed that if there exists any solution then there exists solution with (1<=≤<=*B*<=≤<=106).
Demo Input:
['4 1\n2 1 4 2\n', '5 2\n2 2 4 5 3\n', '3 10\n1 2 3\n']
Demo Output:
['2\n', '-1\n', '4\n']
Note:
Let's consider availability of colors in the first example:
- *cnt*<sub class="lower-index">2</sub>(*i*) ≥ *cnt*<sub class="lower-index">1</sub>(*i*) for every *i*, and color 2 can be the answer. - *cnt*<sub class="lower-index">4</sub>(2) < *cnt*<sub class="lower-index">1</sub>(2), so color 4 isn't the winning one for Bob. - All the other colors also have *cnt*<sub class="lower-index">*j*</sub>(2) < *cnt*<sub class="lower-index">1</sub>(2), thus they are not available.
In the third example every color is acceptable except for 10. | ```python
f = lambda: map(int, input().split())
n, a = f()
d = [0] * 1000001
for q in f():
if d[q] < d[a]: d[q] = -1
d[q] += 1
k = d[a]
d[0] = d[a] = -1
s = max(d)
print(d.index(s) if s >= k else -1)
``` | 3 |
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