contestId int64 0 1.01k | index stringclasses 57
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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
11 | A | Increasing Sequence | PROGRAMMING | 900 | [
"constructive algorithms",
"implementation",
"math"
] | A. Increasing Sequence | 1 | 64 | A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing? | The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106). | Output the minimal number of moves needed to make the sequence increasing. | [
"4 2\n1 3 3 2\n"
] | [
"3\n"
] | none | 0 | [
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
... | 1,635,123,745 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 216 | 19,558,400 | n, d = map(int, input().split())
b, dem = list(map(int, input().split())), 0
for x in range(n - 1):
if b[x] >= b[x + 1]: dem, b[x + 1] = dem + (b[x] - b[x + 1]) // d + 1, b[x + 1] + ((b[x] - b[x + 1]) // d + 1) * d
print(dem) | Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none | ```python
n, d = map(int, input().split())
b, dem = list(map(int, input().split())), 0
for x in range(n - 1):
if b[x] >= b[x + 1]: dem, b[x + 1] = dem + (b[x] - b[x + 1]) // d + 1, b[x + 1] + ((b[x] - b[x + 1]) // d + 1) * d
print(dem)
``` | 3.746279 |
746 | A | Compote | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Nikolay has *a* lemons, *b* apples and *c* pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1:<=2:<=4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can't crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.
Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can't use any fruits, in this case print 0. | The first line contains the positive integer *a* (1<=≤<=*a*<=≤<=1000) — the number of lemons Nikolay has.
The second line contains the positive integer *b* (1<=≤<=*b*<=≤<=1000) — the number of apples Nikolay has.
The third line contains the positive integer *c* (1<=≤<=*c*<=≤<=1000) — the number of pears Nikolay has. | Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. | [
"2\n5\n7\n",
"4\n7\n13\n",
"2\n3\n2\n"
] | [
"7\n",
"21\n",
"0\n"
] | In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.
In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.
In the third example Nikolay don't have enough pears to cook any compote, so the answer is 0. | 500 | [
{
"input": "2\n5\n7",
"output": "7"
},
{
"input": "4\n7\n13",
"output": "21"
},
{
"input": "2\n3\n2",
"output": "0"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "1\n2\n4",
"output": "7"
},
{
"input": "1000\n1000\n1000",
"output": "1750"
}... | 1,619,551,828 | 2,147,483,647 | Python 3 | OK | TESTS | 84 | 108 | 0 | l = int(input())
a = int(input())
p = int(input())
m = min(l, a // 2, p // 4)
print(sum([m, m * 2, m * 4])) | Title: Compote
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nikolay has *a* lemons, *b* apples and *c* pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1:<=2:<=4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can't crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.
Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can't use any fruits, in this case print 0.
Input Specification:
The first line contains the positive integer *a* (1<=≤<=*a*<=≤<=1000) — the number of lemons Nikolay has.
The second line contains the positive integer *b* (1<=≤<=*b*<=≤<=1000) — the number of apples Nikolay has.
The third line contains the positive integer *c* (1<=≤<=*c*<=≤<=1000) — the number of pears Nikolay has.
Output Specification:
Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote.
Demo Input:
['2\n5\n7\n', '4\n7\n13\n', '2\n3\n2\n']
Demo Output:
['7\n', '21\n', '0\n']
Note:
In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.
In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.
In the third example Nikolay don't have enough pears to cook any compote, so the answer is 0. | ```python
l = int(input())
a = int(input())
p = int(input())
m = min(l, a // 2, p // 4)
print(sum([m, m * 2, m * 4]))
``` | 3 | |
205 | A | Little Elephant and Rozdil | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities. | Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes). | [
"2\n7 4\n",
"7\n7 4 47 100 4 9 12\n"
] | [
"2\n",
"Still Rozdil\n"
] | In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | 500 | [
{
"input": "2\n7 4",
"output": "2"
},
{
"input": "7\n7 4 47 100 4 9 12",
"output": "Still Rozdil"
},
{
"input": "1\n47",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "Still Rozdil"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "7"
},
{
... | 1,694,524,058 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | def solve():
size = input()
numbers = sorted(map(int, input().split()))
if numbers[0] == numbers[1]:
print("Still Rozdil")
else:
print(numbers[0])
if __name__ == "__main__":
solve()
| Title: Little Elephant and Rozdil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
Output Specification:
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
Demo Input:
['2\n7 4\n', '7\n7 4 47 100 4 9 12\n']
Demo Output:
['2\n', 'Still Rozdil\n']
Note:
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | ```python
def solve():
size = input()
numbers = sorted(map(int, input().split()))
if numbers[0] == numbers[1]:
print("Still Rozdil")
else:
print(numbers[0])
if __name__ == "__main__":
solve()
``` | 0 | |
86 | A | Reflection | PROGRAMMING | 1,600 | [
"math"
] | A. Reflection | 2 | 256 | For each positive integer *n* consider the integer ψ(*n*) which is obtained from *n* by replacing every digit *a* in the decimal notation of *n* with the digit (9<=<=-<=<=*a*). We say that ψ(*n*) is the reflection of *n*. For example, reflection of 192 equals 807. Note that leading zeros (if any) should be omitted. So reflection of 9 equals 0, reflection of 91 equals 8.
Let us call the weight of the number the product of the number and its reflection. Thus, the weight of the number 10 is equal to 10·89<==<=890.
Your task is to find the maximum weight of the numbers in the given range [*l*,<=*r*] (boundaries are included). | Input contains two space-separated integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=109) — bounds of the range. | Output should contain single integer number: maximum value of the product *n*·ψ(*n*), where *l*<=≤<=*n*<=≤<=*r*.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preferred to use cout (also you may use %I64d). | [
"3 7\n",
"1 1\n",
"8 10\n"
] | [
"20",
"8",
"890"
] | In the third sample weight of 8 equals 8·1 = 8, weight of 9 equals 9·0 = 0, weight of 10 equals 890.
Thus, maximum value of the product is equal to 890. | 500 | [
{
"input": "3 7",
"output": "20"
},
{
"input": "1 1",
"output": "8"
},
{
"input": "8 10",
"output": "890"
},
{
"input": "4 6",
"output": "20"
},
{
"input": "10 100",
"output": "89900"
},
{
"input": "1 999",
"output": "249500"
},
{
"input": ... | 1,658,769,436 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | a,b = map(int,input().split())
la,lb = len(str(a)),len(str(b))
mid = (10**lb)/2
if a<=mid<=b:
print(mid*(mid-1))
else:
if a>mid:
print(a*((10**la)-1-a))
else:
print(b*((10**lb)-1-b)) | Title: Reflection
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
For each positive integer *n* consider the integer ψ(*n*) which is obtained from *n* by replacing every digit *a* in the decimal notation of *n* with the digit (9<=<=-<=<=*a*). We say that ψ(*n*) is the reflection of *n*. For example, reflection of 192 equals 807. Note that leading zeros (if any) should be omitted. So reflection of 9 equals 0, reflection of 91 equals 8.
Let us call the weight of the number the product of the number and its reflection. Thus, the weight of the number 10 is equal to 10·89<==<=890.
Your task is to find the maximum weight of the numbers in the given range [*l*,<=*r*] (boundaries are included).
Input Specification:
Input contains two space-separated integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=109) — bounds of the range.
Output Specification:
Output should contain single integer number: maximum value of the product *n*·ψ(*n*), where *l*<=≤<=*n*<=≤<=*r*.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preferred to use cout (also you may use %I64d).
Demo Input:
['3 7\n', '1 1\n', '8 10\n']
Demo Output:
['20', '8', '890']
Note:
In the third sample weight of 8 equals 8·1 = 8, weight of 9 equals 9·0 = 0, weight of 10 equals 890.
Thus, maximum value of the product is equal to 890. | ```python
a,b = map(int,input().split())
la,lb = len(str(a)),len(str(b))
mid = (10**lb)/2
if a<=mid<=b:
print(mid*(mid-1))
else:
if a>mid:
print(a*((10**la)-1-a))
else:
print(b*((10**lb)-1-b))
``` | 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": "... | 1,660,792,881 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 |
n,m=map(int,input().split())
t=n*(n+1)/2
m=m%t
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())
t=n*(n+1)/2
m=m%t
for i in range(1,n+1):
if(m>=i):
m-=i
else:
break
print(m)
``` | 0 |
371 | C | Hamburgers | PROGRAMMING | 1,600 | [
"binary search",
"brute force"
] | null | null | Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient. | The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0. | [
"BBBSSC\n6 4 1\n1 2 3\n4\n",
"BBC\n1 10 1\n1 10 1\n21\n",
"BSC\n1 1 1\n1 1 3\n1000000000000\n"
] | [
"2\n",
"7\n",
"200000000001\n"
] | none | 1,500 | [
{
"input": "BBBSSC\n6 4 1\n1 2 3\n4",
"output": "2"
},
{
"input": "BBC\n1 10 1\n1 10 1\n21",
"output": "7"
},
{
"input": "BSC\n1 1 1\n1 1 3\n1000000000000",
"output": "200000000001"
},
{
"input": "B\n1 1 1\n1 1 1\n381",
"output": "382"
},
{
"input": "BSC\n3 5 6\n7... | 1,623,754,417 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 93 | 102,400 | import sys
import math
import collections
import heapq
input=sys.stdin.readline
s=input()
k1=s.count('B')
k2=s.count('S')
k3=s.count('C')
a,b,c=(int(i) for i in input().split())
p1,p2,p3=(int(i) for i in input().split())
r=int(input())
if(k1==0):
j1=1000
else:
j1=a//k1
if(k2==0):
j2=1000
else:
j2=b//k2
if(k3==0):
j3=1000
else:
j3=c//k3
c1=min(j1,j2,j3)
a-=c1*k1
b-=c1*k2
c-=c1*k3
a=max(a,0)
b=max(b,0)
c=max(c,0)
k=(max(k1-a,0))*p1+max((k2-b),0)*p2+max((k3-c),0)*p3
if(r<k):
c1+=r//(k1*p1+k2*p2+k3*p3)
else:
j=max(1+(r-k)//(k1*p1+k2*p2+k3*p3),r//(k1*p1+k2*p2+k3*p3))
print(c1+j) | Title: Hamburgers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.
Input Specification:
The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.
Demo Input:
['BBBSSC\n6 4 1\n1 2 3\n4\n', 'BBC\n1 10 1\n1 10 1\n21\n', 'BSC\n1 1 1\n1 1 3\n1000000000000\n']
Demo Output:
['2\n', '7\n', '200000000001\n']
Note:
none | ```python
import sys
import math
import collections
import heapq
input=sys.stdin.readline
s=input()
k1=s.count('B')
k2=s.count('S')
k3=s.count('C')
a,b,c=(int(i) for i in input().split())
p1,p2,p3=(int(i) for i in input().split())
r=int(input())
if(k1==0):
j1=1000
else:
j1=a//k1
if(k2==0):
j2=1000
else:
j2=b//k2
if(k3==0):
j3=1000
else:
j3=c//k3
c1=min(j1,j2,j3)
a-=c1*k1
b-=c1*k2
c-=c1*k3
a=max(a,0)
b=max(b,0)
c=max(c,0)
k=(max(k1-a,0))*p1+max((k2-b),0)*p2+max((k3-c),0)*p3
if(r<k):
c1+=r//(k1*p1+k2*p2+k3*p3)
else:
j=max(1+(r-k)//(k1*p1+k2*p2+k3*p3),r//(k1*p1+k2*p2+k3*p3))
print(c1+j)
``` | 0 | |
712 | C | Memory and De-Evolution | PROGRAMMING | 1,600 | [
"greedy",
"math"
] | null | null | Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*? | The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=<<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively. | Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*. | [
"6 3\n",
"8 5\n",
"22 4\n"
] | [
"4\n",
"3\n",
"6\n"
] | In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,500 | [
{
"input": "6 3",
"output": "4"
},
{
"input": "8 5",
"output": "3"
},
{
"input": "22 4",
"output": "6"
},
{
"input": "4 3",
"output": "3"
},
{
"input": "57 27",
"output": "4"
},
{
"input": "61 3",
"output": "9"
},
{
"input": "5 4",
"out... | 1,473,690,359 | 2,147,483,647 | Python 3 | OK | TESTS | 76 | 77 | 0 | x,y=map(int,input().split(' '))
ans=0
A=[y]*3
while(sum(A)<x*3):
A.sort()
A[0]=min(x,A[1]+A[2]-1)
ans+=1
print(ans)
| Title: Memory and De-Evolution
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Memory is now interested in the de-evolution of objects, specifically triangles. He starts with an equilateral triangle of side length *x*, and he wishes to perform operations to obtain an equilateral triangle of side length *y*.
In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerate triangle (triangle of positive area). At any moment of time, the length of each side should be integer.
What is the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y*?
Input Specification:
The first and only line contains two integers *x* and *y* (3<=≤<=*y*<=<<=*x*<=≤<=100<=000) — the starting and ending equilateral triangle side lengths respectively.
Output Specification:
Print a single integer — the minimum number of seconds required for Memory to obtain the equilateral triangle of side length *y* if he starts with the equilateral triangle of side length *x*.
Demo Input:
['6 3\n', '8 5\n', '22 4\n']
Demo Output:
['4\n', '3\n', '6\n']
Note:
In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides *a*, *b*, and *c* as (*a*, *b*, *c*). Then, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/18af21f738bad490df83097a90e1f2879a4b21c6.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, Memory can do <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcfd51d1b2d764a1cf5fbc255cc02e6f5aaed3b1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test, Memory can do: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0969b7d413854c1e7528991d926bef1f7ffba008.png" style="max-width: 100.0%;max-height: 100.0%;"/>
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/63e9e66b882c03e4c73e93ad92204dc255329309.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
x,y=map(int,input().split(' '))
ans=0
A=[y]*3
while(sum(A)<x*3):
A.sort()
A[0]=min(x,A[1]+A[2]-1)
ans+=1
print(ans)
``` | 3 | |
1,009 | B | Minimum Ternary String | PROGRAMMING | 1,400 | [
"greedy",
"implementation"
] | null | null | You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$. | The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive). | Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero). | [
"100210\n",
"11222121\n",
"20\n"
] | [
"001120\n",
"11112222\n",
"20\n"
] | none | 0 | [
{
"input": "100210",
"output": "001120"
},
{
"input": "11222121",
"output": "11112222"
},
{
"input": "20",
"output": "20"
},
{
"input": "1002",
"output": "0012"
},
{
"input": "10",
"output": "01"
},
{
"input": "000021",
"output": "000012"
},
{
... | 1,532,189,817 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 405 | 1,638,400 | import copy
s = input()
count = s.count('1')
if count == 0:
print(s)
else:
s = s.replace('1','')
if s.count('2') == 0:
print(s,end="")
for j in range(count):
print('1', end="")
else:
index = -1
for i in range(len(s)):
if s[i] == '2':
for j in range(count):
print('1',end="")
print(s[i],end="")
index = i + 1
break
else:
print(s[i],end="")
for i in range(index, len(s)):
print(s[i],end="")
| Title: Minimum Ternary String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
Input Specification:
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
Output Specification:
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
Demo Input:
['100210\n', '11222121\n', '20\n']
Demo Output:
['001120\n', '11112222\n', '20\n']
Note:
none | ```python
import copy
s = input()
count = s.count('1')
if count == 0:
print(s)
else:
s = s.replace('1','')
if s.count('2') == 0:
print(s,end="")
for j in range(count):
print('1', end="")
else:
index = -1
for i in range(len(s)):
if s[i] == '2':
for j in range(count):
print('1',end="")
print(s[i],end="")
index = i + 1
break
else:
print(s[i],end="")
for i in range(index, len(s)):
print(s[i],end="")
``` | 3 | |
21 | B | Intersection | PROGRAMMING | 2,000 | [
"implementation",
"math"
] | B. Intersection | 1 | 256 | You are given two set of points. The first set is determined by the equation *A*1*x*<=+<=*B*1*y*<=+<=*C*1<==<=0, and the second one is determined by the equation *A*2*x*<=+<=*B*2*y*<=+<=*C*2<==<=0.
Write the program which finds the number of points in the intersection of two given sets. | The first line of the input contains three integer numbers *A*1,<=*B*1,<=*C*1 separated by space. The second line contains three integer numbers *A*2,<=*B*2,<=*C*2 separated by space. All the numbers are between -100 and 100, inclusive. | Print the number of points in the intersection or -1 if there are infinite number of points. | [
"1 1 0\n2 2 0\n",
"1 1 0\n2 -2 0\n"
] | [
"-1\n",
"1\n"
] | none | 1,000 | [
{
"input": "1 1 0\n2 2 0",
"output": "-1"
},
{
"input": "1 1 0\n2 -2 0",
"output": "1"
},
{
"input": "0 0 0\n0 0 0",
"output": "-1"
},
{
"input": "1 1 1\n1 1 1",
"output": "-1"
},
{
"input": "8 3 -4\n-5 2 7",
"output": "1"
},
{
"input": "-1 -1 0\n0 -1 ... | 1,572,026,514 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 140 | 0 | a1,b1,c1=list(map(int,input().split()))
a2,b2,c2=list(map(int,input().split()))
b1_zero=False
b2_zero=False
if(b1!=0):
m1=a1/b1
y1=c1/b1
else:
b1_zero=True
if(b2!=0):
y2=c2/b2
m2=a2/b2
else:
b2_zero=True
if ((a1==0 and b1==0 and c1==0) or (a2==0 and b2==0 and c2==0)):
print("-1")
elif(b1_zero==b2_zero):
if(b2_zero==True ):
if(a1==a2 and c1==c2):
print("-1")
else:
print("0")
if(b1_zero==False):#existe m1,m2,y1,y2
if(m1==m2):
if(y1==y2):
print("-1")
else:
print("0")
else:
print("1")
else:
print("1")
| Title: Intersection
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given two set of points. The first set is determined by the equation *A*1*x*<=+<=*B*1*y*<=+<=*C*1<==<=0, and the second one is determined by the equation *A*2*x*<=+<=*B*2*y*<=+<=*C*2<==<=0.
Write the program which finds the number of points in the intersection of two given sets.
Input Specification:
The first line of the input contains three integer numbers *A*1,<=*B*1,<=*C*1 separated by space. The second line contains three integer numbers *A*2,<=*B*2,<=*C*2 separated by space. All the numbers are between -100 and 100, inclusive.
Output Specification:
Print the number of points in the intersection or -1 if there are infinite number of points.
Demo Input:
['1 1 0\n2 2 0\n', '1 1 0\n2 -2 0\n']
Demo Output:
['-1\n', '1\n']
Note:
none | ```python
a1,b1,c1=list(map(int,input().split()))
a2,b2,c2=list(map(int,input().split()))
b1_zero=False
b2_zero=False
if(b1!=0):
m1=a1/b1
y1=c1/b1
else:
b1_zero=True
if(b2!=0):
y2=c2/b2
m2=a2/b2
else:
b2_zero=True
if ((a1==0 and b1==0 and c1==0) or (a2==0 and b2==0 and c2==0)):
print("-1")
elif(b1_zero==b2_zero):
if(b2_zero==True ):
if(a1==a2 and c1==c2):
print("-1")
else:
print("0")
if(b1_zero==False):#existe m1,m2,y1,y2
if(m1==m2):
if(y1==y2):
print("-1")
else:
print("0")
else:
print("1")
else:
print("1")
``` | 0 |
559 | B | Equivalent Strings | PROGRAMMING | 1,700 | [
"divide and conquer",
"hashing",
"sortings",
"strings"
] | null | null | Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn! | The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length. | Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise. | [
"aaba\nabaa\n",
"aabb\nabab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | 1,000 | [
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
... | 1,594,368,187 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 90 | 2,000 | 30,310,400 | def recursion(a, b):
if len(a) != len(b):
return False
elif len(a) == 0:
return True
elif len(a) % 2 == 0:
e = a[:len(a)//2]
f = a[len(a)//2:]
g = b[:len(a)//2]
h = b[len(a)//2:]
return a == b or (recursion(e, g) and recursion(f, h)) or (recursion(e, h) and recursion(f, g))
return a == b
if recursion(input(), input()):
print("YES")
else:
print("NO") | Title: Equivalent Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | ```python
def recursion(a, b):
if len(a) != len(b):
return False
elif len(a) == 0:
return True
elif len(a) % 2 == 0:
e = a[:len(a)//2]
f = a[len(a)//2:]
g = b[:len(a)//2]
h = b[len(a)//2:]
return a == b or (recursion(e, g) and recursion(f, h)) or (recursion(e, h) and recursion(f, g))
return a == b
if recursion(input(), input()):
print("YES")
else:
print("NO")
``` | 0 | |
362 | B | Petya and Staircases | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | null | null | Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order). | Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO". | [
"10 5\n2 4 8 3 6\n",
"10 5\n2 4 5 7 9\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "10 5\n2 4 8 3 6",
"output": "NO"
},
{
"input": "10 5\n2 4 5 7 9",
"output": "YES"
},
{
"input": "10 9\n2 3 4 5 6 7 8 9 10",
"output": "NO"
},
{
"input": "5 2\n4 5",
"output": "NO"
},
{
"input": "123 13\n36 73 111 2 92 5 47 55 48 113 7 78 37",
"outp... | 1,657,760,679 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 30 | 0 | n,m=input().split()
n,m=int(n),int(m)
l=list(map(int,input().split()))[:m]
l=sorted(l)
i=0
while i<m-1:
j1=i+1
j2=i+2
if (l[j1]-l[i]==1) and(l[j2]-l[j1]==1) and j2<m:
print("NO")
exit()
if l[0]==1 and l[m-1]==n:
print("NO")
exit()
i+=1
print("YES") | Title: Petya and Staircases
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).
Output Specification:
Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".
Demo Input:
['10 5\n2 4 8 3 6\n', '10 5\n2 4 5 7 9\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n,m=input().split()
n,m=int(n),int(m)
l=list(map(int,input().split()))[:m]
l=sorted(l)
i=0
while i<m-1:
j1=i+1
j2=i+2
if (l[j1]-l[i]==1) and(l[j2]-l[j1]==1) and j2<m:
print("NO")
exit()
if l[0]==1 and l[m-1]==n:
print("NO")
exit()
i+=1
print("YES")
``` | -1 | |
920 | C | Swap Adjacent Elements | PROGRAMMING | 1,400 | [
"dfs and similar",
"greedy",
"math",
"sortings",
"two pointers"
] | null | null | You have an array *a* consisting of *n* integers. Each integer from 1 to *n* appears exactly once in this array.
For some indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1) it is possible to swap *i*-th element with (*i*<=+<=1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap *i*-th element with (*i*<=+<=1)-th (if the position is not forbidden).
Can you make this array sorted in ascending order performing some sequence of swapping operations? | The first line contains one integer *n* (2<=≤<=*n*<=≤<=200000) — the number of elements in the array.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200000) — the elements of the array. Each integer from 1 to *n* appears exactly once.
The third line contains a string of *n*<=-<=1 characters, each character is either 0 or 1. If *i*-th character is 1, then you can swap *i*-th element with (*i*<=+<=1)-th any number of times, otherwise it is forbidden to swap *i*-th element with (*i*<=+<=1)-th. | If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO. | [
"6\n1 2 5 3 4 6\n01110\n",
"6\n1 2 5 3 4 6\n01010\n"
] | [
"YES\n",
"NO\n"
] | In the first example you may swap *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, and then swap *a*<sub class="lower-index">4</sub> and *a*<sub class="lower-index">5</sub>. | 0 | [
{
"input": "6\n1 2 5 3 4 6\n01110",
"output": "YES"
},
{
"input": "6\n1 2 5 3 4 6\n01010",
"output": "NO"
},
{
"input": "6\n1 6 3 4 5 2\n01101",
"output": "NO"
},
{
"input": "6\n2 3 1 4 5 6\n01111",
"output": "NO"
},
{
"input": "4\n2 3 1 4\n011",
"output": "NO... | 1,646,560,253 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 13 | 217 | 14,233,600 | n = int(input())
arr = list(map(int,input().split()))
lis = input()
broken = False
for i in range(1,n):
if arr[i-1] > arr[i]:
if lis[i-1] == '1':
check = arr[i-1]
arr[i-1] = arr[i]
arr[i] = check
else:
broken = True
if(broken):
print("NO")
else:
print("YES")
| Title: Swap Adjacent Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have an array *a* consisting of *n* integers. Each integer from 1 to *n* appears exactly once in this array.
For some indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1) it is possible to swap *i*-th element with (*i*<=+<=1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap *i*-th element with (*i*<=+<=1)-th (if the position is not forbidden).
Can you make this array sorted in ascending order performing some sequence of swapping operations?
Input Specification:
The first line contains one integer *n* (2<=≤<=*n*<=≤<=200000) — the number of elements in the array.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=200000) — the elements of the array. Each integer from 1 to *n* appears exactly once.
The third line contains a string of *n*<=-<=1 characters, each character is either 0 or 1. If *i*-th character is 1, then you can swap *i*-th element with (*i*<=+<=1)-th any number of times, otherwise it is forbidden to swap *i*-th element with (*i*<=+<=1)-th.
Output Specification:
If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO.
Demo Input:
['6\n1 2 5 3 4 6\n01110\n', '6\n1 2 5 3 4 6\n01010\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example you may swap *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, and then swap *a*<sub class="lower-index">4</sub> and *a*<sub class="lower-index">5</sub>. | ```python
n = int(input())
arr = list(map(int,input().split()))
lis = input()
broken = False
for i in range(1,n):
if arr[i-1] > arr[i]:
if lis[i-1] == '1':
check = arr[i-1]
arr[i-1] = arr[i]
arr[i] = check
else:
broken = True
if(broken):
print("NO")
else:
print("YES")
``` | 0 | |
5 | B | Center Alignment | PROGRAMMING | 1,200 | [
"implementation",
"strings"
] | B. Center Alignment | 1 | 64 | Almost every text editor has a built-in function of center text alignment. The developers of the popular in Berland text editor «Textpad» decided to introduce this functionality into the fourth release of the product.
You are to implement the alignment in the shortest possible time. Good luck! | The input file consists of one or more lines, each of the lines contains Latin letters, digits and/or spaces. The lines cannot start or end with a space. It is guaranteed that at least one of the lines has positive length. The length of each line and the total amount of the lines do not exceed 1000. | Format the given text, aligning it center. Frame the whole text with characters «*» of the minimum size. If a line cannot be aligned perfectly (for example, the line has even length, while the width of the block is uneven), you should place such lines rounding down the distance to the left or to the right edge and bringing them closer left or right alternatively (you should start with bringing left). Study the sample tests carefully to understand the output format better. | [
"This is\n\nCodeforces\nBeta\nRound\n5\n",
"welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck\n"
] | [
"************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************\n",
"****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************\n"
] | none | 0 | [
{
"input": "This is\n\nCodeforces\nBeta\nRound\n5",
"output": "************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************"
},
{
"input": "welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck",
"output": "****************\n*welcome to th... | 1,625,301,631 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 122 | 0 | import sys
l = []
for line in sys.stdin:
l.append(line)
m = max(l, key=len)
max_len = len(m)
if max_len % 2 == 0:
even = True
else:
even = False
print('*'*(max_len + 2))
if max_len % 2 == 0: # max even
for element in l:
element_len = len(element)
diff = max_len - element_len
if element_len % 2 == 0: # element even
print(f"*{' '*int((diff/2))}{element}{' '*int((diff/2))}*")
else: # element odd
print(f"*{' '*int(((diff - 1)/2))}{element}{' '*int(((diff - 1)/2) + 1)}*")
else: # max odd
for element in l:
element_len = len(element)
diff = max_len - element_len
if element_len % 2 == 0: # element even
print(f"*{' '*int(((diff - 1)/2))}{element}{' '*int(((diff - 1)/2) + 1)}*")
else: # element odd
print(f"*{' '*int((diff/2))}{element}{' '*int((diff/2))}*")
print('*'*(max_len + 2))
| Title: Center Alignment
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Almost every text editor has a built-in function of center text alignment. The developers of the popular in Berland text editor «Textpad» decided to introduce this functionality into the fourth release of the product.
You are to implement the alignment in the shortest possible time. Good luck!
Input Specification:
The input file consists of one or more lines, each of the lines contains Latin letters, digits and/or spaces. The lines cannot start or end with a space. It is guaranteed that at least one of the lines has positive length. The length of each line and the total amount of the lines do not exceed 1000.
Output Specification:
Format the given text, aligning it center. Frame the whole text with characters «*» of the minimum size. If a line cannot be aligned perfectly (for example, the line has even length, while the width of the block is uneven), you should place such lines rounding down the distance to the left or to the right edge and bringing them closer left or right alternatively (you should start with bringing left). Study the sample tests carefully to understand the output format better.
Demo Input:
['This is\n\nCodeforces\nBeta\nRound\n5\n', 'welcome to the\nCodeforces\nBeta\nRound 5\n\nand\ngood luck\n']
Demo Output:
['************\n* This is *\n* *\n*Codeforces*\n* Beta *\n* Round *\n* 5 *\n************\n', '****************\n*welcome to the*\n* Codeforces *\n* Beta *\n* Round 5 *\n* *\n* and *\n* good luck *\n****************\n']
Note:
none | ```python
import sys
l = []
for line in sys.stdin:
l.append(line)
m = max(l, key=len)
max_len = len(m)
if max_len % 2 == 0:
even = True
else:
even = False
print('*'*(max_len + 2))
if max_len % 2 == 0: # max even
for element in l:
element_len = len(element)
diff = max_len - element_len
if element_len % 2 == 0: # element even
print(f"*{' '*int((diff/2))}{element}{' '*int((diff/2))}*")
else: # element odd
print(f"*{' '*int(((diff - 1)/2))}{element}{' '*int(((diff - 1)/2) + 1)}*")
else: # max odd
for element in l:
element_len = len(element)
diff = max_len - element_len
if element_len % 2 == 0: # element even
print(f"*{' '*int(((diff - 1)/2))}{element}{' '*int(((diff - 1)/2) + 1)}*")
else: # element odd
print(f"*{' '*int((diff/2))}{element}{' '*int((diff/2))}*")
print('*'*(max_len + 2))
``` | 0 |
337 | A | Puzzles | PROGRAMMING | 900 | [
"greedy"
] | null | null | The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. | The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. | Print a single integer — the least possible difference the teacher can obtain. | [
"4 6\n10 12 10 7 5 22\n"
] | [
"5\n"
] | Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | 500 | [
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617... | 1,678,549,121 | 2,147,483,647 | Python 3 | OK | TESTS | 18 | 92 | 0 | n,m = map(int, input().split())
lst = list(map(int, input().split()))
lst.sort()
i = 0
j = n-1
mn = 1000000000
while j<m:
mn = min(mn, lst[j]-lst[i])
i += 1
j += 1
print(mn)
| Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | ```python
n,m = map(int, input().split())
lst = list(map(int, input().split()))
lst.sort()
i = 0
j = n-1
mn = 1000000000
while j<m:
mn = min(mn, lst[j]-lst[i])
i += 1
j += 1
print(mn)
``` | 3 | |
580 | C | Kefa and Park | PROGRAMMING | 1,500 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | Kefa decided to celebrate his first big salary by going to the restaurant.
He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.
The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats.
Your task is to help Kefa count the number of restaurants where he can go. | The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat).
Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge.
It is guaranteed that the given set of edges specifies a tree. | A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats. | [
"4 1\n1 1 0 0\n1 2\n1 3\n1 4\n",
"7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n"
] | [
"2\n",
"2\n"
] | Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.
Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.
Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7. | 1,500 | [
{
"input": "4 1\n1 1 0 0\n1 2\n1 3\n1 4",
"output": "2"
},
{
"input": "7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7",
"output": "2"
},
{
"input": "3 2\n1 1 1\n1 2\n2 3",
"output": "0"
},
{
"input": "5 2\n1 1 0 1 1\n1 2\n2 3\n3 4\n4 5",
"output": "1"
},
{
"inpu... | 1,696,666,021 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 31 | 307,200 | from collections import defaultdict
n, m = map(int, input().split())
arr = list(map(int, input().split()))
adj_list = defaultdict(set)
for _ in range(n-1):
a, b = map(int, input().split())
adj_list[a].add(b)
def dfs(node, i):
if i < 0:
return 0
if node not in adj_list:
return 1
curr = 0
for child in adj_list[node]:
if arr[node-1] == 1:
if arr[child-1] == 1:
curr += dfs(child, i-1)
else:
curr += dfs(child, m-1)
else:
if arr[child-1] == 1:
curr += dfs(child, m-1)
else:
curr += dfs(child, m)
return curr
if arr[0] == 1:
print(dfs(1, m-1))
else:
print(dfs(1, m)) | Title: Kefa and Park
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kefa decided to celebrate his first big salary by going to the restaurant.
He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.
The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats.
Your task is to help Kefa count the number of restaurants where he can go.
Input Specification:
The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat).
Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge.
It is guaranteed that the given set of edges specifies a tree.
Output Specification:
A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats.
Demo Input:
['4 1\n1 1 0 0\n1 2\n1 3\n1 4\n', '7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n']
Demo Output:
['2\n', '2\n']
Note:
Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.
Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.
Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7. | ```python
from collections import defaultdict
n, m = map(int, input().split())
arr = list(map(int, input().split()))
adj_list = defaultdict(set)
for _ in range(n-1):
a, b = map(int, input().split())
adj_list[a].add(b)
def dfs(node, i):
if i < 0:
return 0
if node not in adj_list:
return 1
curr = 0
for child in adj_list[node]:
if arr[node-1] == 1:
if arr[child-1] == 1:
curr += dfs(child, i-1)
else:
curr += dfs(child, m-1)
else:
if arr[child-1] == 1:
curr += dfs(child, m-1)
else:
curr += dfs(child, m)
return curr
if arr[0] == 1:
print(dfs(1, m-1))
else:
print(dfs(1, m))
``` | 0 | |
791 | A | Bear and Big Brother | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob? | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively. | Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob. | [
"4 7\n",
"4 9\n",
"1 1\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | 500 | [
{
"input": "4 7",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output... | 1,695,715,417 | 2,147,483,647 | Python 3 | OK | TESTS | 62 | 46 | 0 | sign = True
a,b = map(int,input().split())
year = 0
while sign:
a *= 3
b *= 2
year += 1
if a > b:
sign = False
print(year) | Title: Bear and Big Brother
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Output Specification:
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
Demo Input:
['4 7\n', '4 9\n', '1 1\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | ```python
sign = True
a,b = map(int,input().split())
year = 0
while sign:
a *= 3
b *= 2
year += 1
if a > b:
sign = False
print(year)
``` | 3 | |
33 | B | String Problem | PROGRAMMING | 1,800 | [
"shortest paths"
] | B. String Problem | 2 | 256 | Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character *A**i* in one of the strings into arbitrary character *B**i*, but he has to pay for every move a particular sum of money, equal to *W**i*. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings. | The first input line contains two initial non-empty strings *s* and *t*, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer *n* (0<=≤<=*n*<=≤<=500) — amount of possible changings. Then follow *n* lines, each containing characters *A**i* and *B**i* (lower case Latin letters) and integer *W**i* (0<=≤<=*W**i*<=≤<=100), saying that it's allowed to change character *A**i* into character *B**i* in any of the strings and spend sum of money *W**i*. | If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any. | [
"uayd\nuxxd\n3\na x 8\nx y 13\nd c 3\n",
"a\nb\n3\na b 2\na b 3\nb a 5\n",
"abc\nab\n6\na b 4\na b 7\nb a 8\nc b 11\nc a 3\na c 0\n"
] | [
"21\nuxyd\n",
"2\nb\n",
"-1\n"
] | none | 1,000 | [
{
"input": "uayd\nuxxd\n3\na x 8\nx y 13\nd c 3",
"output": "21\nuxyd"
},
{
"input": "a\nb\n3\na b 2\na b 3\nb a 5",
"output": "2\nb"
},
{
"input": "abc\nab\n6\na b 4\na b 7\nb a 8\nc b 11\nc a 3\na c 0",
"output": "-1"
},
{
"input": "xhtuopq\nrtutbz\n10\nh x 10\nx d 3\nr u 4... | 1,691,493,056 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 61 | 1,964 | 22,732,800 | import math
import random
import time
from decimal import *
from collections import defaultdict
from bisect import bisect_left as lower_bound
from bisect import bisect_right as upper_bound
import sys,threading
#sys.setrecursionlimit(5*(10**5)+2)
#threading.stack_size(99000000)
alfabet = {'a': 1, 'b': 2,'c': 3,'d': 4,'e': 5,'f': 6,'g': 7,'h': 8,'i': 9,'j': 10,'k': 11,'l': 12,'m': 13,'n': 14,'o': 15,'p': 16,'q': 17,'r': 18,'s': 19,'t': 20,'u': 21,'v': 22,'w': 23,'x': 24,'y': 25,'z': 26}
alfabet_2={'1':"a", '2':"b", '3':"c", '4':"d", '5':"e", '6':"f", '7':"g", '8':"h", '9':"i", '10':"j", '11':"k", '12':"l", '13':"m", '14':"n", '15':"o", '16':"p", '17':"q", '18':"r", '19':"s", '20':"t", '21':"u", '22':"v", '23':"w", '24':"x", '25':"y", '26':"z"}
class FenwickTree:
def __init__(self, x):
bit = self.bit = list(x)
size = self.size = len(bit)
for i in range(size):
j = i | (i + 1)
if j < size:
bit[j] += bit[i]
def update(self, idx, x):
"""updates bit[idx] += x"""
while idx < self.size:
self.bit[idx] += x
idx |= idx + 1
def __call__(self, end):
"""calc sum(bit[:end])"""
x = 0
while end:
x += self.bit[end - 1]
end &= end - 1
return x
def find_kth(self, k):
"""Find largest idx such that sum(bit[:idx]) <= k"""
idx = -1
for d in reversed(range(self.size.bit_length())):
right_idx = idx + (1 << d)
if right_idx < self.size and self.bit[right_idx] <= k:
idx = right_idx
k -= self.bit[idx]
return idx + 1, k
class SortedList:
block_size = 700
def __init__(self, iterable=()):
self.macro = []
self.micros = [[]]
self.micro_size = [0]
self.fenwick = FenwickTree([0])
self.size = 0
for item in iterable:
self.insert(item)
def insert(self, x):
i = lower_bound(self.macro, x)
j = upper_bound(self.micros[i], x)
self.micros[i].insert(j, x)
self.size += 1
self.micro_size[i] += 1
self.fenwick.update(i, 1)
if len(self.micros[i]) >= self.block_size:
self.micros[i:i + 1] = self.micros[i][:self.block_size >> 1], self.micros[i][self.block_size >> 1:]
self.micro_size[i:i + 1] = self.block_size >> 1, self.block_size >> 1
self.fenwick = FenwickTree(self.micro_size)
self.macro.insert(i, self.micros[i + 1][0])
def pop(self, k=-1):
i, j = self._find_kth(k)
self.size -= 1
self.micro_size[i] -= 1
self.fenwick.update(i, -1)
return self.micros[i].pop(j)
def __getitem__(self, k):
i, j = self._find_kth(k)
return self.micros[i][j]
def count(self, x):
return self.upper_bound(x) - self.lower_bound(x)
def __contains__(self, x):
return self.count(x) > 0
def lower_bound(self, x):
i = lower_bound(self.macro, x)
return self.fenwick(i) + lower_bound(self.micros[i], x)
def upper_bound(self, x):
i = upper_bound(self.macro, x)
return self.fenwick(i) + upper_bound(self.micros[i], x)
def _find_kth(self, k):
return self.fenwick.find_kth(k + self.size if k < 0 else k)
def __len__(self):
return self.size
def __iter__(self):
return (x for micro in self.micros for x in micro)
def __repr__(self):
return str(list(self))
#A = SortedList()
#A.insert(30)
#A.insert(50)
#A.insert(20)
#A.insert(30)
#A.insert(30)
#print(A) # prints [20, 30, 30, 30, 50]
#print(A.lower_bound(30), A.upper_bound(30)) # prints 1 4
#print(A[-1]) # prints 50
#print(A.pop(1)) # prints 30
#print(A) # prints [20, 30, 30, 50]
#print(A.count(30)) # prints 2
def binary_search(vector,valoarea):
left=0
right=len(vector)-1
while left<=right:
centru=(left+right)//2
# print(left,right,centru,vector[centru])
if vector[centru]<=valoarea:
left=centru+1
else:
right=centru-1
# print(left,right,centru,vector[centru])
return left
def functie(element,dist,full,graph,vizitat,partial,initialul):
initial=alfabet[element]
vizitat[initial]=1
for elemente in graph[element]:
pozitie=alfabet[elemente]
if vizitat[pozitie]==0:
partial[pozitie]=min(partial[pozitie],partial[initial]+dist[(element,elemente)])
val_partiala=min(partial[pozitie],partial[initial]+dist[(element,elemente)])
# print("element=",element," to =" ,elemente, " part=",val_partiala)
minimul=10**18
target=''
for i in range(1,27):
new_element=alfabet_2[str(i)]
if vizitat[i]==0:
if minimul>partial[i]:
minimul=partial[i]
target=new_element
vizitat[initial]=1
if target!='':
# print("target=",target)
functie(target,dist,full,graph,vizitat,partial,initialul)
else:
# print("part=",partial)
for dd in range(1,27):
if partial[dd]<10**18:
full[(initialul,alfabet_2[str(dd)])]=partial[dd]
def functie_nod(nodul,oprire,distantele,graficul,vizitatul,rezultatul,initial):
pp=10**18
partial=pp
gasit=0
# print("n=",nodul)
ex=0
for vecini in graficul[nodul]:
if vizitatul[vecini]==0:
ex=1
# print("vec=",vecini)
rezultatul[vecini]=min(rezultatul[vecini],rezultatul[nodul]+distantele[(nodul,vecini)])
# print("vec=",vecini,"rez=",rezultatul[vecini])
if partial>=rezultatul[vecini]:
next_one=vecini
partial=rezultatul[vecini]
# print("vec=",next_one, rezultate[next_one],rezultate[nodul],distantele[(nodul,next_one)])
# print(rezultate)
vizitatul[nodul]=1
if ex==0:
# print("ex=",ex,rezultatul[oprire])
#print(rezultatul)
return rezultatul[oprire]
else:
# print("?=",rezultatul[oprire])
functie(next_one,oprire,distantele,graficul,vizitatul,rezultatul)
def main():
#answ=[]
pp=10**18
restul=998244353
# teste=int(input())
#answ=[]
printare=[]
for gg in range(1):
# pp=-1
cate=0
string_unu=input()
string_doi=input()
n=int(input())
distante={}
full_distante={}
sume=0
graficul=defaultdict(list)
for i in range(n):
lista=list(map(str,input().split()))
if (lista[0],lista[1]) not in distante:
distante[(lista[0],lista[1])]=int(lista[2])
graficul[lista[0]].append(lista[1])
else:
distante[(lista[0],lista[1])]=min(distante[(lista[0],lista[1])],int(lista[2]))
# print(graficul)
# print(distante)
#for element in graficul['a']:
# print("el=",element)
for i in range(1,27):
vizitatul=[0] *27
partialul=[pp] *27
partialul[i]=0
functie(alfabet_2[str(i)],distante,full_distante,graficul,vizitatul,partialul,alfabet_2[str(i)])
# print(full_distante)
great_book={}
results={}
for i in range(1,27):
for j in range(1,27):
minimul=pp
targetul=''
for c in range(1,27):
unu=alfabet_2[str(i)]
doi=alfabet_2[str(j)]
trei=alfabet_2[str(c)]
if (unu,trei) in full_distante and (doi,trei) in full_distante:
if full_distante[(unu,trei)]+full_distante[(doi,trei)]<minimul:
minimul=full_distante[(unu,trei)]+full_distante[(doi,trei)]
targetul=trei
if targetul!='':
great_book[(unu,doi)]=targetul
results[(unu,doi)]=minimul
# print(great_book)
#print(results)
sume=0
adev=1
answer=''
if len(string_unu)!=len(string_doi):
adev=0
# print("ff")
else:
for i in range(len(string_unu)):
unu=string_unu[i]
doi=string_doi[i]
if unu!=doi:
if (unu,doi) not in great_book:
adev=0
break
else:
sume+=results[(unu,doi)]
answer+=great_book[(unu,doi)]
else:
answer+=unu
# print("adev=",adev)
if adev==1:
print(sume)
print(answer)
else:
print(-1)
main()
#t=threading.Thread(target=main)
#t.start()
#t.join() | Title: String Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character *A**i* in one of the strings into arbitrary character *B**i*, but he has to pay for every move a particular sum of money, equal to *W**i*. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings.
Input Specification:
The first input line contains two initial non-empty strings *s* and *t*, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer *n* (0<=≤<=*n*<=≤<=500) — amount of possible changings. Then follow *n* lines, each containing characters *A**i* and *B**i* (lower case Latin letters) and integer *W**i* (0<=≤<=*W**i*<=≤<=100), saying that it's allowed to change character *A**i* into character *B**i* in any of the strings and spend sum of money *W**i*.
Output Specification:
If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any.
Demo Input:
['uayd\nuxxd\n3\na x 8\nx y 13\nd c 3\n', 'a\nb\n3\na b 2\na b 3\nb a 5\n', 'abc\nab\n6\na b 4\na b 7\nb a 8\nc b 11\nc a 3\na c 0\n']
Demo Output:
['21\nuxyd\n', '2\nb\n', '-1\n']
Note:
none | ```python
import math
import random
import time
from decimal import *
from collections import defaultdict
from bisect import bisect_left as lower_bound
from bisect import bisect_right as upper_bound
import sys,threading
#sys.setrecursionlimit(5*(10**5)+2)
#threading.stack_size(99000000)
alfabet = {'a': 1, 'b': 2,'c': 3,'d': 4,'e': 5,'f': 6,'g': 7,'h': 8,'i': 9,'j': 10,'k': 11,'l': 12,'m': 13,'n': 14,'o': 15,'p': 16,'q': 17,'r': 18,'s': 19,'t': 20,'u': 21,'v': 22,'w': 23,'x': 24,'y': 25,'z': 26}
alfabet_2={'1':"a", '2':"b", '3':"c", '4':"d", '5':"e", '6':"f", '7':"g", '8':"h", '9':"i", '10':"j", '11':"k", '12':"l", '13':"m", '14':"n", '15':"o", '16':"p", '17':"q", '18':"r", '19':"s", '20':"t", '21':"u", '22':"v", '23':"w", '24':"x", '25':"y", '26':"z"}
class FenwickTree:
def __init__(self, x):
bit = self.bit = list(x)
size = self.size = len(bit)
for i in range(size):
j = i | (i + 1)
if j < size:
bit[j] += bit[i]
def update(self, idx, x):
"""updates bit[idx] += x"""
while idx < self.size:
self.bit[idx] += x
idx |= idx + 1
def __call__(self, end):
"""calc sum(bit[:end])"""
x = 0
while end:
x += self.bit[end - 1]
end &= end - 1
return x
def find_kth(self, k):
"""Find largest idx such that sum(bit[:idx]) <= k"""
idx = -1
for d in reversed(range(self.size.bit_length())):
right_idx = idx + (1 << d)
if right_idx < self.size and self.bit[right_idx] <= k:
idx = right_idx
k -= self.bit[idx]
return idx + 1, k
class SortedList:
block_size = 700
def __init__(self, iterable=()):
self.macro = []
self.micros = [[]]
self.micro_size = [0]
self.fenwick = FenwickTree([0])
self.size = 0
for item in iterable:
self.insert(item)
def insert(self, x):
i = lower_bound(self.macro, x)
j = upper_bound(self.micros[i], x)
self.micros[i].insert(j, x)
self.size += 1
self.micro_size[i] += 1
self.fenwick.update(i, 1)
if len(self.micros[i]) >= self.block_size:
self.micros[i:i + 1] = self.micros[i][:self.block_size >> 1], self.micros[i][self.block_size >> 1:]
self.micro_size[i:i + 1] = self.block_size >> 1, self.block_size >> 1
self.fenwick = FenwickTree(self.micro_size)
self.macro.insert(i, self.micros[i + 1][0])
def pop(self, k=-1):
i, j = self._find_kth(k)
self.size -= 1
self.micro_size[i] -= 1
self.fenwick.update(i, -1)
return self.micros[i].pop(j)
def __getitem__(self, k):
i, j = self._find_kth(k)
return self.micros[i][j]
def count(self, x):
return self.upper_bound(x) - self.lower_bound(x)
def __contains__(self, x):
return self.count(x) > 0
def lower_bound(self, x):
i = lower_bound(self.macro, x)
return self.fenwick(i) + lower_bound(self.micros[i], x)
def upper_bound(self, x):
i = upper_bound(self.macro, x)
return self.fenwick(i) + upper_bound(self.micros[i], x)
def _find_kth(self, k):
return self.fenwick.find_kth(k + self.size if k < 0 else k)
def __len__(self):
return self.size
def __iter__(self):
return (x for micro in self.micros for x in micro)
def __repr__(self):
return str(list(self))
#A = SortedList()
#A.insert(30)
#A.insert(50)
#A.insert(20)
#A.insert(30)
#A.insert(30)
#print(A) # prints [20, 30, 30, 30, 50]
#print(A.lower_bound(30), A.upper_bound(30)) # prints 1 4
#print(A[-1]) # prints 50
#print(A.pop(1)) # prints 30
#print(A) # prints [20, 30, 30, 50]
#print(A.count(30)) # prints 2
def binary_search(vector,valoarea):
left=0
right=len(vector)-1
while left<=right:
centru=(left+right)//2
# print(left,right,centru,vector[centru])
if vector[centru]<=valoarea:
left=centru+1
else:
right=centru-1
# print(left,right,centru,vector[centru])
return left
def functie(element,dist,full,graph,vizitat,partial,initialul):
initial=alfabet[element]
vizitat[initial]=1
for elemente in graph[element]:
pozitie=alfabet[elemente]
if vizitat[pozitie]==0:
partial[pozitie]=min(partial[pozitie],partial[initial]+dist[(element,elemente)])
val_partiala=min(partial[pozitie],partial[initial]+dist[(element,elemente)])
# print("element=",element," to =" ,elemente, " part=",val_partiala)
minimul=10**18
target=''
for i in range(1,27):
new_element=alfabet_2[str(i)]
if vizitat[i]==0:
if minimul>partial[i]:
minimul=partial[i]
target=new_element
vizitat[initial]=1
if target!='':
# print("target=",target)
functie(target,dist,full,graph,vizitat,partial,initialul)
else:
# print("part=",partial)
for dd in range(1,27):
if partial[dd]<10**18:
full[(initialul,alfabet_2[str(dd)])]=partial[dd]
def functie_nod(nodul,oprire,distantele,graficul,vizitatul,rezultatul,initial):
pp=10**18
partial=pp
gasit=0
# print("n=",nodul)
ex=0
for vecini in graficul[nodul]:
if vizitatul[vecini]==0:
ex=1
# print("vec=",vecini)
rezultatul[vecini]=min(rezultatul[vecini],rezultatul[nodul]+distantele[(nodul,vecini)])
# print("vec=",vecini,"rez=",rezultatul[vecini])
if partial>=rezultatul[vecini]:
next_one=vecini
partial=rezultatul[vecini]
# print("vec=",next_one, rezultate[next_one],rezultate[nodul],distantele[(nodul,next_one)])
# print(rezultate)
vizitatul[nodul]=1
if ex==0:
# print("ex=",ex,rezultatul[oprire])
#print(rezultatul)
return rezultatul[oprire]
else:
# print("?=",rezultatul[oprire])
functie(next_one,oprire,distantele,graficul,vizitatul,rezultatul)
def main():
#answ=[]
pp=10**18
restul=998244353
# teste=int(input())
#answ=[]
printare=[]
for gg in range(1):
# pp=-1
cate=0
string_unu=input()
string_doi=input()
n=int(input())
distante={}
full_distante={}
sume=0
graficul=defaultdict(list)
for i in range(n):
lista=list(map(str,input().split()))
if (lista[0],lista[1]) not in distante:
distante[(lista[0],lista[1])]=int(lista[2])
graficul[lista[0]].append(lista[1])
else:
distante[(lista[0],lista[1])]=min(distante[(lista[0],lista[1])],int(lista[2]))
# print(graficul)
# print(distante)
#for element in graficul['a']:
# print("el=",element)
for i in range(1,27):
vizitatul=[0] *27
partialul=[pp] *27
partialul[i]=0
functie(alfabet_2[str(i)],distante,full_distante,graficul,vizitatul,partialul,alfabet_2[str(i)])
# print(full_distante)
great_book={}
results={}
for i in range(1,27):
for j in range(1,27):
minimul=pp
targetul=''
for c in range(1,27):
unu=alfabet_2[str(i)]
doi=alfabet_2[str(j)]
trei=alfabet_2[str(c)]
if (unu,trei) in full_distante and (doi,trei) in full_distante:
if full_distante[(unu,trei)]+full_distante[(doi,trei)]<minimul:
minimul=full_distante[(unu,trei)]+full_distante[(doi,trei)]
targetul=trei
if targetul!='':
great_book[(unu,doi)]=targetul
results[(unu,doi)]=minimul
# print(great_book)
#print(results)
sume=0
adev=1
answer=''
if len(string_unu)!=len(string_doi):
adev=0
# print("ff")
else:
for i in range(len(string_unu)):
unu=string_unu[i]
doi=string_doi[i]
if unu!=doi:
if (unu,doi) not in great_book:
adev=0
break
else:
sume+=results[(unu,doi)]
answer+=great_book[(unu,doi)]
else:
answer+=unu
# print("adev=",adev)
if adev==1:
print(sume)
print(answer)
else:
print(-1)
main()
#t=threading.Thread(target=main)
#t.start()
#t.join()
``` | 3.466657 |
237 | B | Young Table | PROGRAMMING | 1,500 | [
"implementation",
"sortings"
] | null | null | You've got table *a*, consisting of *n* rows, numbered from 1 to *n*. The *i*-th line of table *a* contains *c**i* cells, at that for all *i* (1<=<<=*i*<=≤<=*n*) holds *c**i*<=≤<=*c**i*<=-<=1.
Let's denote *s* as the total number of cells of table *a*, that is, . We know that each cell of the table contains a single integer from 1 to *s*, at that all written integers are distinct.
Let's assume that the cells of the *i*-th row of table *a* are numbered from 1 to *c**i*, then let's denote the number written in the *j*-th cell of the *i*-th row as *a**i*,<=*j*. Your task is to perform several swap operations to rearrange the numbers in the table so as to fulfill the following conditions:
1. for all *i*,<=*j* (1<=<<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*c**i*) holds *a**i*,<=*j*<=><=*a**i*<=-<=1,<=*j*; 1. for all *i*,<=*j* (1<=≤<=*i*<=≤<=*n*; 1<=<<=*j*<=≤<=*c**i*) holds *a**i*,<=*j*<=><=*a**i*,<=*j*<=-<=1.
In one swap operation you are allowed to choose two different cells of the table and swap the recorded there numbers, that is the number that was recorded in the first of the selected cells before the swap, is written in the second cell after it. Similarly, the number that was recorded in the second of the selected cells, is written in the first cell after the swap.
Rearrange the numbers in the required manner. Note that you are allowed to perform any number of operations, but not more than *s*. You do not have to minimize the number of operations. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=50) that shows the number of rows in the table. The second line contains *n* space-separated integers *c**i* (1<=≤<=*c**i*<=≤<=50; *c**i*<=≤<=*c**i*<=-<=1) — the numbers of cells on the corresponding rows.
Next *n* lines contain table *а*. The *i*-th of them contains *c**i* space-separated integers: the *j*-th integer in this line represents *a**i*,<=*j*.
It is guaranteed that all the given numbers *a**i*,<=*j* are positive and do not exceed *s*. It is guaranteed that all *a**i*,<=*j* are distinct. | In the first line print a single integer *m* (0<=≤<=*m*<=≤<=*s*), representing the number of performed swaps.
In the next *m* lines print the description of these swap operations. In the *i*-th line print four space-separated integers *x**i*,<=*y**i*,<=*p**i*,<=*q**i* (1<=≤<=*x**i*,<=*p**i*<=≤<=*n*; 1<=≤<=*y**i*<=≤<=*c**x**i*; 1<=≤<=*q**i*<=≤<=*c**p**i*). The printed numbers denote swapping the contents of cells *a**x**i*,<=*y**i* and *a**p**i*,<=*q**i*. Note that a swap operation can change the contents of distinct table cells. Print the swaps in the order, in which they should be executed. | [
"3\n3 2 1\n4 3 5\n6 1\n2\n",
"1\n4\n4 3 2 1\n"
] | [
"2\n1 1 2 2\n2 1 3 1\n",
"2\n1 1 1 4\n1 2 1 3\n"
] | none | 1,000 | [
{
"input": "3\n3 2 1\n4 3 5\n6 1\n2",
"output": "2\n1 1 2 2\n2 1 3 1"
},
{
"input": "1\n4\n4 3 2 1",
"output": "2\n1 1 1 4\n1 2 1 3"
},
{
"input": "5\n4 4 3 3 1\n14 13 4 15\n11 1 2 5\n7 6 10\n8 9 3\n12",
"output": "13\n1 1 2 2\n1 2 2 3\n1 3 4 3\n1 4 4 3\n2 1 2 4\n2 2 3 2\n2 3 3 1\n2 ... | 1,642,165,823 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | import math
import sys
def solve():
n = int(input())
rows = list(map(int, input().split()))
r = []
cords = {}
for i in range(n):
r.append(list(map(int, input().split())))
for j in range(rows[i]):
cords[r[i][j]] = [i, j]
it = 1
ans = []
for i in range(n):
for j in range(rows[i]):
if r[i][j] != it:
r[i][j], r[cords[it][0]][cords[it][1]] = r[cords[it][0]][cords[it][1]], r[i][j]
ans.append([i, j, cords[it][0], cords[it][1]])
it += 1
print(len(ans))
for i in ans:
for j in i:
print(j + 1, end=" ")
print()
if __name__ == '__main__':
solve()
| Title: Young Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got table *a*, consisting of *n* rows, numbered from 1 to *n*. The *i*-th line of table *a* contains *c**i* cells, at that for all *i* (1<=<<=*i*<=≤<=*n*) holds *c**i*<=≤<=*c**i*<=-<=1.
Let's denote *s* as the total number of cells of table *a*, that is, . We know that each cell of the table contains a single integer from 1 to *s*, at that all written integers are distinct.
Let's assume that the cells of the *i*-th row of table *a* are numbered from 1 to *c**i*, then let's denote the number written in the *j*-th cell of the *i*-th row as *a**i*,<=*j*. Your task is to perform several swap operations to rearrange the numbers in the table so as to fulfill the following conditions:
1. for all *i*,<=*j* (1<=<<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*c**i*) holds *a**i*,<=*j*<=><=*a**i*<=-<=1,<=*j*; 1. for all *i*,<=*j* (1<=≤<=*i*<=≤<=*n*; 1<=<<=*j*<=≤<=*c**i*) holds *a**i*,<=*j*<=><=*a**i*,<=*j*<=-<=1.
In one swap operation you are allowed to choose two different cells of the table and swap the recorded there numbers, that is the number that was recorded in the first of the selected cells before the swap, is written in the second cell after it. Similarly, the number that was recorded in the second of the selected cells, is written in the first cell after the swap.
Rearrange the numbers in the required manner. Note that you are allowed to perform any number of operations, but not more than *s*. You do not have to minimize the number of operations.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=50) that shows the number of rows in the table. The second line contains *n* space-separated integers *c**i* (1<=≤<=*c**i*<=≤<=50; *c**i*<=≤<=*c**i*<=-<=1) — the numbers of cells on the corresponding rows.
Next *n* lines contain table *а*. The *i*-th of them contains *c**i* space-separated integers: the *j*-th integer in this line represents *a**i*,<=*j*.
It is guaranteed that all the given numbers *a**i*,<=*j* are positive and do not exceed *s*. It is guaranteed that all *a**i*,<=*j* are distinct.
Output Specification:
In the first line print a single integer *m* (0<=≤<=*m*<=≤<=*s*), representing the number of performed swaps.
In the next *m* lines print the description of these swap operations. In the *i*-th line print four space-separated integers *x**i*,<=*y**i*,<=*p**i*,<=*q**i* (1<=≤<=*x**i*,<=*p**i*<=≤<=*n*; 1<=≤<=*y**i*<=≤<=*c**x**i*; 1<=≤<=*q**i*<=≤<=*c**p**i*). The printed numbers denote swapping the contents of cells *a**x**i*,<=*y**i* and *a**p**i*,<=*q**i*. Note that a swap operation can change the contents of distinct table cells. Print the swaps in the order, in which they should be executed.
Demo Input:
['3\n3 2 1\n4 3 5\n6 1\n2\n', '1\n4\n4 3 2 1\n']
Demo Output:
['2\n1 1 2 2\n2 1 3 1\n', '2\n1 1 1 4\n1 2 1 3\n']
Note:
none | ```python
import math
import sys
def solve():
n = int(input())
rows = list(map(int, input().split()))
r = []
cords = {}
for i in range(n):
r.append(list(map(int, input().split())))
for j in range(rows[i]):
cords[r[i][j]] = [i, j]
it = 1
ans = []
for i in range(n):
for j in range(rows[i]):
if r[i][j] != it:
r[i][j], r[cords[it][0]][cords[it][1]] = r[cords[it][0]][cords[it][1]], r[i][j]
ans.append([i, j, cords[it][0], cords[it][1]])
it += 1
print(len(ans))
for i in ans:
for j in i:
print(j + 1, end=" ")
print()
if __name__ == '__main__':
solve()
``` | 0 | |
832 | A | Sasha and Sticks | PROGRAMMING | 800 | [
"games",
"math"
] | null | null | It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn. | If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower). | [
"1 1\n",
"10 4\n"
] | [
"YES\n",
"NO\n"
] | In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win. | 500 | [
{
"input": "1 1",
"output": "YES"
},
{
"input": "10 4",
"output": "NO"
},
{
"input": "251656215122324104 164397544865601257",
"output": "YES"
},
{
"input": "963577813436662285 206326039287271924",
"output": "NO"
},
{
"input": "1000000000000000000 1",
"output":... | 1,647,573,398 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 93 | 0 | a,b=map(int,input().split())
c=b//a
if c%2!=0:
print("Yes")
if c%2==0:
print("No")
| Title: Sasha and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
Output Specification:
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
Demo Input:
['1 1\n', '10 4\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win. | ```python
a,b=map(int,input().split())
c=b//a
if c%2!=0:
print("Yes")
if c%2==0:
print("No")
``` | 0 | |
340 | A | The Wall | PROGRAMMING | 1,200 | [
"math"
] | null | null | Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question. | The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*). | Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink. | [
"2 3 6 18\n"
] | [
"3"
] | Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | 500 | [
{
"input": "2 3 6 18",
"output": "3"
},
{
"input": "4 6 20 201",
"output": "15"
},
{
"input": "15 27 100 10000",
"output": "74"
},
{
"input": "105 60 3456 78910",
"output": "179"
},
{
"input": "1 1 1000 100000",
"output": "99001"
},
{
"input": "3 2 5 5... | 1,457,883,890 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | x, y, a, b = map(int, input().split())
z = max(x, y)
if a%z != 0:
a = (a+1)*z//a
r = 0
for i in range(a, b+1, z):
if i%x == 0 and i%y == 0:
r += 1
print(r)
| Title: The Wall
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input Specification:
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
Output Specification:
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
Demo Input:
['2 3 6 18\n']
Demo Output:
['3']
Note:
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | ```python
x, y, a, b = map(int, input().split())
z = max(x, y)
if a%z != 0:
a = (a+1)*z//a
r = 0
for i in range(a, b+1, z):
if i%x == 0 and i%y == 0:
r += 1
print(r)
``` | 0 | |
475 | D | CGCDSSQ | PROGRAMMING | 2,000 | [
"brute force",
"data structures",
"math"
] | null | null | Given a sequence of integers *a*1,<=...,<=*a**n* and *q* queries *x*1,<=...,<=*x**q* on it. For each query *x**i* you have to count the number of pairs (*l*,<=*r*) such that 1<=≤<=*l*<=≤<=*r*<=≤<=*n* and *gcd*(*a**l*,<=*a**l*<=+<=1,<=...,<=*a**r*)<==<=*x**i*.
is a greatest common divisor of *v*1,<=*v*2,<=...,<=*v**n*, that is equal to a largest positive integer that divides all *v**i*. | The first line of the input contains integer *n*, (1<=≤<=*n*<=≤<=105), denoting the length of the sequence. The next line contains *n* space separated integers *a*1,<=...,<=*a**n*, (1<=≤<=*a**i*<=≤<=109).
The third line of the input contains integer *q*, (1<=≤<=*q*<=≤<=3<=×<=105), denoting the number of queries. Then follows *q* lines, each contain an integer *x**i*, (1<=≤<=*x**i*<=≤<=109). | For each query print the result in a separate line. | [
"3\n2 6 3\n5\n1\n2\n3\n4\n6\n",
"7\n10 20 3 15 1000 60 16\n10\n1\n2\n3\n4\n5\n6\n10\n20\n60\n1000\n"
] | [
"1\n2\n2\n0\n1\n",
"14\n0\n2\n2\n2\n0\n2\n2\n1\n1\n"
] | none | 2,000 | [
{
"input": "3\n2 6 3\n5\n1\n2\n3\n4\n6",
"output": "1\n2\n2\n0\n1"
},
{
"input": "7\n10 20 3 15 1000 60 16\n10\n1\n2\n3\n4\n5\n6\n10\n20\n60\n1000",
"output": "14\n0\n2\n2\n2\n0\n2\n2\n1\n1"
},
{
"input": "10\n2 2 4 3 2 4 4 2 4 2\n104\n3\n3\n1\n4\n1\n1\n4\n1\n1\n3\n1\n1\n4\n1\n1\n1\n4\n3... | 1,412,523,674 | 9,674 | Python 3 | WRONG_ANSWER | PRETESTS | 1 | 108 | 5,836,800 | from fractions import gcd
from functools import reduce
#input
n=int(input())
alist=list(map(int, input().split()))
q=int(input())
qlist=[int(input()) for i in range(q)]
#variables
countlist=[0]*q
#main
i=0
while i<=n-1:
j=i
while j>=i and j<=n-1:
if reduce(gcd,alist[i:j+1])!=1 and reduce(gcd,alist[i:j+1]) in qlist:
countlist[qlist.index(reduce(gcd,alist[i:j+1]))]+=1
j+=1
else:
if 1 in qlist:
countlist[qlist.index(1)]+=(n-j)
j=n
i+=1
for i in range(q):
print(countlist[i]) | Title: CGCDSSQ
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Given a sequence of integers *a*1,<=...,<=*a**n* and *q* queries *x*1,<=...,<=*x**q* on it. For each query *x**i* you have to count the number of pairs (*l*,<=*r*) such that 1<=≤<=*l*<=≤<=*r*<=≤<=*n* and *gcd*(*a**l*,<=*a**l*<=+<=1,<=...,<=*a**r*)<==<=*x**i*.
is a greatest common divisor of *v*1,<=*v*2,<=...,<=*v**n*, that is equal to a largest positive integer that divides all *v**i*.
Input Specification:
The first line of the input contains integer *n*, (1<=≤<=*n*<=≤<=105), denoting the length of the sequence. The next line contains *n* space separated integers *a*1,<=...,<=*a**n*, (1<=≤<=*a**i*<=≤<=109).
The third line of the input contains integer *q*, (1<=≤<=*q*<=≤<=3<=×<=105), denoting the number of queries. Then follows *q* lines, each contain an integer *x**i*, (1<=≤<=*x**i*<=≤<=109).
Output Specification:
For each query print the result in a separate line.
Demo Input:
['3\n2 6 3\n5\n1\n2\n3\n4\n6\n', '7\n10 20 3 15 1000 60 16\n10\n1\n2\n3\n4\n5\n6\n10\n20\n60\n1000\n']
Demo Output:
['1\n2\n2\n0\n1\n', '14\n0\n2\n2\n2\n0\n2\n2\n1\n1\n']
Note:
none | ```python
from fractions import gcd
from functools import reduce
#input
n=int(input())
alist=list(map(int, input().split()))
q=int(input())
qlist=[int(input()) for i in range(q)]
#variables
countlist=[0]*q
#main
i=0
while i<=n-1:
j=i
while j>=i and j<=n-1:
if reduce(gcd,alist[i:j+1])!=1 and reduce(gcd,alist[i:j+1]) in qlist:
countlist[qlist.index(reduce(gcd,alist[i:j+1]))]+=1
j+=1
else:
if 1 in qlist:
countlist[qlist.index(1)]+=(n-j)
j=n
i+=1
for i in range(q):
print(countlist[i])
``` | 0 | |
489 | B | BerSU Ball | PROGRAMMING | 1,200 | [
"dfs and similar",
"dp",
"graph matchings",
"greedy",
"sortings",
"two pointers"
] | null | null | The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill. | Print a single number — the required maximum possible number of pairs. | [
"4\n1 4 6 2\n5\n5 1 5 7 9\n",
"4\n1 2 3 4\n4\n10 11 12 13\n",
"5\n1 1 1 1 1\n3\n1 2 3\n"
] | [
"3\n",
"0\n",
"2\n"
] | none | 1,000 | [
{
"input": "4\n1 4 6 2\n5\n5 1 5 7 9",
"output": "3"
},
{
"input": "4\n1 2 3 4\n4\n10 11 12 13",
"output": "0"
},
{
"input": "5\n1 1 1 1 1\n3\n1 2 3",
"output": "2"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 10\n1\n9",
"output": "1"
},
{... | 1,675,693,024 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | n=int(input())
listn=list(map(int,input().split()))
m=int(input())
listm=list(map(int,input().split()))
nNum=sorted(listn)
mNum=sorted(listm)
count=0
b=min(n,m)
for i in range(b):
if(abs(nNum[i] - mNum[i]) <= 1):
count+=1
print(count)
| Title: BerSU Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill.
Output Specification:
Print a single number — the required maximum possible number of pairs.
Demo Input:
['4\n1 4 6 2\n5\n5 1 5 7 9\n', '4\n1 2 3 4\n4\n10 11 12 13\n', '5\n1 1 1 1 1\n3\n1 2 3\n']
Demo Output:
['3\n', '0\n', '2\n']
Note:
none | ```python
n=int(input())
listn=list(map(int,input().split()))
m=int(input())
listm=list(map(int,input().split()))
nNum=sorted(listn)
mNum=sorted(listm)
count=0
b=min(n,m)
for i in range(b):
if(abs(nNum[i] - mNum[i]) <= 1):
count+=1
print(count)
``` | 0 | |
439 | A | Devu, the Singer and Churu, the Joker | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event. | The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100). | If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event. | [
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] | [
"5\n",
"-1\n"
] | Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | 500 | [
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 ... | 1,585,458,186 | 2,147,483,647 | PyPy 3 | OK | TESTS | 26 | 140 | 0 | n, d = map(int,input().split())
t = [int(t) for t in input().split()]
s = sum(t) + 10*(n-1)
res = 0
if s > d:
res = -1
else:
s = sum(t)
res = int((d-s)/5)
print(res) | Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | ```python
n, d = map(int,input().split())
t = [int(t) for t in input().split()]
s = sum(t) + 10*(n-1)
res = 0
if s > d:
res = -1
else:
s = sum(t)
res = int((d-s)/5)
print(res)
``` | 3 | |
859 | B | Lazy Security Guard | PROGRAMMING | 1,000 | [
"brute force",
"geometry",
"math"
] | null | null | Your security guard friend recently got a new job at a new security company. The company requires him to patrol an area of the city encompassing exactly *N* city blocks, but they let him choose which blocks. That is, your friend must walk the perimeter of a region whose area is exactly *N* blocks. Your friend is quite lazy and would like your help to find the shortest possible route that meets the requirements. The city is laid out in a square grid pattern, and is large enough that for the sake of the problem it can be considered infinite. | Input will consist of a single integer *N* (1<=≤<=*N*<=≤<=106), the number of city blocks that must be enclosed by the route. | Print the minimum perimeter that can be achieved. | [
"4\n",
"11\n",
"22\n"
] | [
"8\n",
"14\n",
"20\n"
] | Here are some possible shapes for the examples:
<img class="tex-graphics" src="https://espresso.codeforces.com/e11bef2cf82b55dd583cfc97d12b5aee5e483a65.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 750 | [
{
"input": "4",
"output": "8"
},
{
"input": "11",
"output": "14"
},
{
"input": "22",
"output": "20"
},
{
"input": "3",
"output": "8"
},
{
"input": "1024",
"output": "128"
},
{
"input": "101",
"output": "42"
},
{
"input": "30",
"output":... | 1,506,055,549 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 61 | 204,800 | #http://codeforces.com/problemset/problem/859/B
#not done
import math
def GCD(a,b):
#print(a,b)
if a < b:
a, b = b, a
if a%b == 0:
return b
else:
return GCD(b,a%b)
def minLengthsCall(n):
RA = minLengths(n)
#if len(RA) > 2:
#RA.pop()
return RA
def minLengths(n):
#print(n)
if n <= 0:
return []
has = n
while has > n//2:
if math.sqrt(has)%1 == 0:
if has == n:
return [int(math.sqrt(has)),int(math.sqrt(has))]
else:
return [int(math.sqrt(has)),int(math.sqrt(has))]+minLengths(n-has)
has -= 1
else:
maxVol = 1*n
pos = [1,n]
#print(pos)
for i in range(1,n+1):
if i*(n-i) > maxVol:
#print("HERE")
maxVol = i*(n-i)
pos = [i,n-i]
return pos
#RA = [int(math.sqrt(n)),int(math.sqrt(n))]+minLengths(n-int(math.sqrt(n))**2)
#return RA
x = 0
def minStreets(n):
lengths = minLengthsCall(n)
streets = []
i = 0
while i < len(lengths):
#print(lengths[i],lengths[i+1])
streets.append(2*lengths[i]+2*lengths[i+1])
i += 2
i = 0
j = 2
total = 0
if len(streets) == 1:
return streets[0]
currentStreets = 2*lengths[0]+2*lengths[1]
if streets[0] > streets[1]:
maxW = streets[0]
else:
maxW = streets[1]
#print((lengths))
del lengths[0]
del lengths[0]
currentW = 0
while len(lengths) > 0:
if lengths[0] > lengths[1]:
addW = lengths[0]
width = lengths[0]
height = lengths[1]
else:
addW = lengths[1]
width = lengths[1]
height = lengths[0]
#print(currentStreets,height)
if currentW == 0:
currentStreets += 2*height
currentW = addW
elif addW + currentW <= maxW:
currentStreets += 0
currentW += addW
if currentW > maxW:
currentW = 0
else:
currentStreets += height
currentW += addW
if currentW > maxWL:
currentW = 0
del lengths[0]
del lengths[0]
return currentStreets
while True:
try:
case = input()
temp = int(case)
print(minStreets(temp))
except EOFError:
break
| Title: Lazy Security Guard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your security guard friend recently got a new job at a new security company. The company requires him to patrol an area of the city encompassing exactly *N* city blocks, but they let him choose which blocks. That is, your friend must walk the perimeter of a region whose area is exactly *N* blocks. Your friend is quite lazy and would like your help to find the shortest possible route that meets the requirements. The city is laid out in a square grid pattern, and is large enough that for the sake of the problem it can be considered infinite.
Input Specification:
Input will consist of a single integer *N* (1<=≤<=*N*<=≤<=106), the number of city blocks that must be enclosed by the route.
Output Specification:
Print the minimum perimeter that can be achieved.
Demo Input:
['4\n', '11\n', '22\n']
Demo Output:
['8\n', '14\n', '20\n']
Note:
Here are some possible shapes for the examples:
<img class="tex-graphics" src="https://espresso.codeforces.com/e11bef2cf82b55dd583cfc97d12b5aee5e483a65.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
#http://codeforces.com/problemset/problem/859/B
#not done
import math
def GCD(a,b):
#print(a,b)
if a < b:
a, b = b, a
if a%b == 0:
return b
else:
return GCD(b,a%b)
def minLengthsCall(n):
RA = minLengths(n)
#if len(RA) > 2:
#RA.pop()
return RA
def minLengths(n):
#print(n)
if n <= 0:
return []
has = n
while has > n//2:
if math.sqrt(has)%1 == 0:
if has == n:
return [int(math.sqrt(has)),int(math.sqrt(has))]
else:
return [int(math.sqrt(has)),int(math.sqrt(has))]+minLengths(n-has)
has -= 1
else:
maxVol = 1*n
pos = [1,n]
#print(pos)
for i in range(1,n+1):
if i*(n-i) > maxVol:
#print("HERE")
maxVol = i*(n-i)
pos = [i,n-i]
return pos
#RA = [int(math.sqrt(n)),int(math.sqrt(n))]+minLengths(n-int(math.sqrt(n))**2)
#return RA
x = 0
def minStreets(n):
lengths = minLengthsCall(n)
streets = []
i = 0
while i < len(lengths):
#print(lengths[i],lengths[i+1])
streets.append(2*lengths[i]+2*lengths[i+1])
i += 2
i = 0
j = 2
total = 0
if len(streets) == 1:
return streets[0]
currentStreets = 2*lengths[0]+2*lengths[1]
if streets[0] > streets[1]:
maxW = streets[0]
else:
maxW = streets[1]
#print((lengths))
del lengths[0]
del lengths[0]
currentW = 0
while len(lengths) > 0:
if lengths[0] > lengths[1]:
addW = lengths[0]
width = lengths[0]
height = lengths[1]
else:
addW = lengths[1]
width = lengths[1]
height = lengths[0]
#print(currentStreets,height)
if currentW == 0:
currentStreets += 2*height
currentW = addW
elif addW + currentW <= maxW:
currentStreets += 0
currentW += addW
if currentW > maxW:
currentW = 0
else:
currentStreets += height
currentW += addW
if currentW > maxWL:
currentW = 0
del lengths[0]
del lengths[0]
return currentStreets
while True:
try:
case = input()
temp = int(case)
print(minStreets(temp))
except EOFError:
break
``` | 0 | |
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... | 1,669,833,094 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | str = input()
lower, upper = 0, 0
for i in range(len(str)):
if (str[i] >= 'A' and str[i] <= 'Z'):
upper += 1
else:
lower += 1
if (lower < upper):
for j in range(len(str)):
if (str[j] >= 'a' and str[j] <= 'z'):
str = str[:j] + chr(ord(str[j]) - 32) + str[j + 1:]
else:
for j in range(len(str)):
if (str[j] >= 'A' and str[j] <= 'Z'):
str = str[:j] + chr(ord(str[j]) + 32) + str[j + 1:]
print(str)
| 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
str = input()
lower, upper = 0, 0
for i in range(len(str)):
if (str[i] >= 'A' and str[i] <= 'Z'):
upper += 1
else:
lower += 1
if (lower < upper):
for j in range(len(str)):
if (str[j] >= 'a' and str[j] <= 'z'):
str = str[:j] + chr(ord(str[j]) - 32) + str[j + 1:]
else:
for j in range(len(str)):
if (str[j] >= 'A' and str[j] <= 'Z'):
str = str[:j] + chr(ord(str[j]) + 32) + str[j + 1:]
print(str)
``` | 3.977 |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output"... | 1,665,885,549 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | a=input()
a=int(a)
b=list(map(int,input().split()))
x=0
y=0
z=0
for i in b:
if i%2==0:
x+=1
else:
y+=1
if x>y:
while z<a:
if b[z]%2==0
z+=1
else:
print(z+1)
if x<y:
while z<a:
if b[z]%2==0
print(z+1)
else:
z+=1 | Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
a=input()
a=int(a)
b=list(map(int,input().split()))
x=0
y=0
z=0
for i in b:
if i%2==0:
x+=1
else:
y+=1
if x>y:
while z<a:
if b[z]%2==0
z+=1
else:
print(z+1)
if x<y:
while z<a:
if b[z]%2==0
print(z+1)
else:
z+=1
``` | -1 |
6 | E | Exposition | PROGRAMMING | 1,900 | [
"binary search",
"data structures",
"dsu",
"trees",
"two pointers"
] | E. Exposition | 1 | 64 | There are several days left before the fiftieth birthday of a famous Berland's writer Berlbury. In this connection the local library decided to make an exposition of the works of this famous science-fiction writer. It was decided as well that it is necessary to include into the exposition only those books that were published during a particular time period. It is obvious that if the books differ much in size, the visitors will not like it. That was why the organizers came to the opinion, that the difference between the highest and the lowest books in the exposition should be not more than *k* millimeters.
The library has *n* volumes of books by Berlbury, arranged in chronological order of their appearance. The height of each book in millimeters is know, it is *h**i*. As Berlbury is highly respected in the city, the organizers want to include into the exposition as many books as possible, and to find out what periods of his creative work they will manage to cover. You are asked to help the organizers cope with this hard task. | The first line of the input data contains two integer numbers separated by a space *n* (1<=≤<=*n*<=≤<=105) and *k* (0<=≤<=*k*<=≤<=106) — the amount of books by Berlbury in the library, and the maximum allowed height difference between the lowest and the highest books. The second line contains *n* integer numbers separated by a space. Each number *h**i* (1<=≤<=*h**i*<=≤<=106) is the height of the *i*-th book in millimeters. | In the first line of the output data print two numbers *a* and *b* (separate them by a space), where *a* is the maximum amount of books the organizers can include into the exposition, and *b* — the amount of the time periods, during which Berlbury published *a* books, and the height difference between the lowest and the highest among these books is not more than *k* milllimeters.
In each of the following *b* lines print two integer numbers separated by a space — indexes of the first and the last volumes from each of the required time periods of Berlbury's creative work. | [
"3 3\n14 12 10\n",
"2 0\n10 10\n",
"4 5\n8 19 10 13\n"
] | [
"2 2\n1 2\n2 3\n",
"2 1\n1 2\n",
"2 1\n3 4\n"
] | none | 0 | [
{
"input": "3 3\n14 12 10",
"output": "2 2\n1 2\n2 3"
},
{
"input": "2 0\n10 10",
"output": "2 1\n1 2"
},
{
"input": "4 5\n8 19 10 13",
"output": "2 1\n3 4"
},
{
"input": "1 1\n1",
"output": "1 1\n1 1"
},
{
"input": "2 10\n35 45",
"output": "2 1\n1 2"
},
{... | 1,638,599,072 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 343 | 17,305,600 | from sys import stdin
input = stdin.buffer.readline
n, k = map(int, input().split())
a = list(map(int, input().split()))
minq, maxq = [], []
i1, i2, l = 0, 0, 0
ans, sz = [], 0
for r in range(n):
while i1 < len(minq) and a[r] < minq[-1][0]: minq.pop()
minq.append([a[r], r])
while i2 < len(maxq) and a[r] > maxq[-1][0]: maxq.pop()
maxq.append([a[r], r])
while maxq[i2][0] - minq[i1][0] > k:
l += 1
if i1 < len(minq) and minq[i1][1] < l: i1 += 1
if i2 < len(maxq) and maxq[i2][1] < l: i2 += 1
if r - l + 1 > sz: sz = r - l + 1; ans = [(l + 1, r + 1)]
elif r - l + 1 == sz: ans.append((l + 1, r + 1))
print(sz, len(ans))
for i in ans: print(*i) | Title: Exposition
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
There are several days left before the fiftieth birthday of a famous Berland's writer Berlbury. In this connection the local library decided to make an exposition of the works of this famous science-fiction writer. It was decided as well that it is necessary to include into the exposition only those books that were published during a particular time period. It is obvious that if the books differ much in size, the visitors will not like it. That was why the organizers came to the opinion, that the difference between the highest and the lowest books in the exposition should be not more than *k* millimeters.
The library has *n* volumes of books by Berlbury, arranged in chronological order of their appearance. The height of each book in millimeters is know, it is *h**i*. As Berlbury is highly respected in the city, the organizers want to include into the exposition as many books as possible, and to find out what periods of his creative work they will manage to cover. You are asked to help the organizers cope with this hard task.
Input Specification:
The first line of the input data contains two integer numbers separated by a space *n* (1<=≤<=*n*<=≤<=105) and *k* (0<=≤<=*k*<=≤<=106) — the amount of books by Berlbury in the library, and the maximum allowed height difference between the lowest and the highest books. The second line contains *n* integer numbers separated by a space. Each number *h**i* (1<=≤<=*h**i*<=≤<=106) is the height of the *i*-th book in millimeters.
Output Specification:
In the first line of the output data print two numbers *a* and *b* (separate them by a space), where *a* is the maximum amount of books the organizers can include into the exposition, and *b* — the amount of the time periods, during which Berlbury published *a* books, and the height difference between the lowest and the highest among these books is not more than *k* milllimeters.
In each of the following *b* lines print two integer numbers separated by a space — indexes of the first and the last volumes from each of the required time periods of Berlbury's creative work.
Demo Input:
['3 3\n14 12 10\n', '2 0\n10 10\n', '4 5\n8 19 10 13\n']
Demo Output:
['2 2\n1 2\n2 3\n', '2 1\n1 2\n', '2 1\n3 4\n']
Note:
none | ```python
from sys import stdin
input = stdin.buffer.readline
n, k = map(int, input().split())
a = list(map(int, input().split()))
minq, maxq = [], []
i1, i2, l = 0, 0, 0
ans, sz = [], 0
for r in range(n):
while i1 < len(minq) and a[r] < minq[-1][0]: minq.pop()
minq.append([a[r], r])
while i2 < len(maxq) and a[r] > maxq[-1][0]: maxq.pop()
maxq.append([a[r], r])
while maxq[i2][0] - minq[i1][0] > k:
l += 1
if i1 < len(minq) and minq[i1][1] < l: i1 += 1
if i2 < len(maxq) and maxq[i2][1] < l: i2 += 1
if r - l + 1 > sz: sz = r - l + 1; ans = [(l + 1, r + 1)]
elif r - l + 1 == sz: ans.append((l + 1, r + 1))
print(sz, len(ans))
for i in ans: print(*i)
``` | 3.699563 |
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... | 1,643,723,494 | 2,147,483,647 | Python 3 | OK | TESTS | 76 | 92 | 0 | n=int(input())
q={}
l=[]
r=[]
def f(a):
while q[a]!=a:
a=q[a]
return a
for i in range(n):
a,b=map(str,input().split())
o,p="x"+a,"y"+b
l+=[[o,p]]
r+=[o,p]
q[o]=o
q[p]=p
for i in range(n):
l[i][0]=f(l[i][0])
l[i][1]=f(l[i][1])
q[l[i][1]]=q[l[i][0]]
for i in r:
q[i]=f(i)
print(len(set(q.values()))-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())
q={}
l=[]
r=[]
def f(a):
while q[a]!=a:
a=q[a]
return a
for i in range(n):
a,b=map(str,input().split())
o,p="x"+a,"y"+b
l+=[[o,p]]
r+=[o,p]
q[o]=o
q[p]=p
for i in range(n):
l[i][0]=f(l[i][0])
l[i][1]=f(l[i][1])
q[l[i][1]]=q[l[i][0]]
for i in r:
q[i]=f(i)
print(len(set(q.values()))-1)
``` | 3 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,631,712,453 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 154 | 6,758,400 | m, n = map(int, input().split())
if m == 1 and n == 1:
print(0)
else:
t=int(m * n) // 2
print(t) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m, n = map(int, input().split())
if m == 1 and n == 1:
print(0)
else:
t=int(m * n) // 2
print(t)
``` | 3.948911 |
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... | 1,659,228,250 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n=int(input());v=[0]+list(map(int,input().split()));m=int(input());u=sorted(v.copy())
for i in range(1,n+1): v[i]+=v[i-1];u[i]+=u[i-1]
print(u)
print(v)
for i in range(m):
typ,l,r=map(int,input().split())
if typ==2:
print(u[r]-u[l-1])
else:
print(v[r]-v[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=[0]+list(map(int,input().split()));m=int(input());u=sorted(v.copy())
for i in range(1,n+1): v[i]+=v[i-1];u[i]+=u[i-1]
print(u)
print(v)
for i in range(m):
typ,l,r=map(int,input().split())
if typ==2:
print(u[r]-u[l-1])
else:
print(v[r]-v[l-1])
``` | 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",
... | 1,589,797,956 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 248 | 7,065,600 | from collections import Counter
lis = []
for i in range(int(input())):
lis.append(input())
s = Counter(lis)
for a,b in s.most_common(1):
print(a) | 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
from collections import Counter
lis = []
for i in range(int(input())):
lis.append(input())
s = Counter(lis)
for a,b in s.most_common(1):
print(a)
``` | 3.924839 |
165 | A | Supercentral Point | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | [
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] | [
"2\n",
"1\n"
] | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | 500 | [
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n... | 1,595,985,555 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 374 | 22,220,800 | t = int(input())
xs,ys,ps = [],[],[]
for i in range(t):
l1 = [int(x) for x in input().split()]
xs.append(l1[0])
ys.append(l1[1])
for i in range(t):
targx = x[i]
targy = y[i]
l,r,d,u = 0,0,0,0
for j in range(t):
if not l and xs[j]<targx and ys[j]==targy:
l=1
if not r and xs[j]>targx and ys[j]==targy:
r=1
if not u and xs[j]==targx and ys[j]>targy:
u=1
if not d and xs[j]==targx and ys[j]<targy:
d=1
if d and r and u and l:
counter+=1
break
print(counter)
| Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | ```python
t = int(input())
xs,ys,ps = [],[],[]
for i in range(t):
l1 = [int(x) for x in input().split()]
xs.append(l1[0])
ys.append(l1[1])
for i in range(t):
targx = x[i]
targy = y[i]
l,r,d,u = 0,0,0,0
for j in range(t):
if not l and xs[j]<targx and ys[j]==targy:
l=1
if not r and xs[j]>targx and ys[j]==targy:
r=1
if not u and xs[j]==targx and ys[j]>targy:
u=1
if not d and xs[j]==targx and ys[j]<targy:
d=1
if d and r and u and l:
counter+=1
break
print(counter)
``` | -1 | |
136 | A | Presents | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. | Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. | [
"4\n2 3 4 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] | [
"4 1 2 3\n",
"1 3 2\n",
"1 2\n"
] | none | 500 | [
{
"input": "4\n2 3 4 1",
"output": "4 1 2 3"
},
{
"input": "3\n1 3 2",
"output": "1 3 2"
},
{
"input": "2\n1 2",
"output": "1 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 3 2 6 4 5 7 9 8 10",
"output": "1 3 2 5 6 4 7 9 8 10"
},
{
"input"... | 1,689,534,204 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n=int(input())
k=list(map(int,input().split()))
for i in range(n):
print(k.index(i+1)+1,end='') | Title: Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Output Specification:
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
Demo Input:
['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['4 1 2 3\n', '1 3 2\n', '1 2\n']
Note:
none | ```python
n=int(input())
k=list(map(int,input().split()))
for i in range(n):
print(k.index(i+1)+1,end='')
``` | 0 | |
450 | A | Jzzhu and Children | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order? | The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | Output a single integer, representing the number of the last child. | [
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] | [
"4\n",
"6\n"
] | Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | 500 | [
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18... | 1,637,400,231 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | n,cnd = map(int,input().split())
lst = list(map(int,input().split()))
a = max(lst)
if a<=cnd:
print(n)
else :
if lst.count(a)>1:
print(lst[::-1].index(max(lst))+1)
else:
print(lst.index(a)+1) | Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | ```python
n,cnd = map(int,input().split())
lst = list(map(int,input().split()))
a = max(lst)
if a<=cnd:
print(n)
else :
if lst.count(a)>1:
print(lst[::-1].index(max(lst))+1)
else:
print(lst.index(a)+1)
``` | 0 | |
540 | A | Combination Lock | PROGRAMMING | 800 | [
"implementation"
] | null | null | Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock. | Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock. | [
"5\n82195\n64723\n"
] | [
"13\n"
] | In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "5\n82195\n64723",
"output": "13"
},
{
"input": "12\n102021090898\n010212908089",
"output": "16"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "10\n0728592530\n1362615763",
"output": "27"
},
{
... | 1,615,370,566 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 62 | 0 | n=int(input())
t=0
for a,b in zip(input(),input()):
d=abs(int(a)-int(b))
t+=min(d,10-d)
print(t)
| Title: Combination Lock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Output Specification:
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
Demo Input:
['5\n82195\n64723\n']
Demo Output:
['13\n']
Note:
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
n=int(input())
t=0
for a,b in zip(input(),input()):
d=abs(int(a)-int(b))
t+=min(d,10-d)
print(t)
``` | 3 | |
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... | 1,555,928,660 | 2,147,483,647 | PyPy 3 | OK | TESTS | 76 | 278 | 204,800 | nodes = []
n = int(input().strip())
for _ in range(n):
nodes.append(tuple(map(int, input().strip().split())))
visited = [False]*n
def dfs(node):
visited[node] = True
for i in range(n):
if not visited[i] and (nodes[node][0] == nodes[i][0] or nodes[node][1] == nodes[i][1]):
dfs(i)
ans = 0
for i in range(n):
if not visited[i]:
dfs(i)
ans += 1
print(ans-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
nodes = []
n = int(input().strip())
for _ in range(n):
nodes.append(tuple(map(int, input().strip().split())))
visited = [False]*n
def dfs(node):
visited[node] = True
for i in range(n):
if not visited[i] and (nodes[node][0] == nodes[i][0] or nodes[node][1] == nodes[i][1]):
dfs(i)
ans = 0
for i in range(n):
if not visited[i]:
dfs(i)
ans += 1
print(ans-1)
``` | 3 | |
705 | A | Hulk | PROGRAMMING | 800 | [
"implementation"
] | null | null | Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner. | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of layers of love and hate. | Print Dr.Banner's feeling in one line. | [
"1\n",
"2\n",
"3\n"
] | [
"I hate it\n",
"I hate that I love it\n",
"I hate that I love that I hate it\n"
] | none | 500 | [
{
"input": "1",
"output": "I hate it"
},
{
"input": "2",
"output": "I hate that I love it"
},
{
"input": "3",
"output": "I hate that I love that I hate it"
},
{
"input": "4",
"output": "I hate that I love that I hate that I love it"
},
{
"input": "5",
"output"... | 1,698,190,146 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 15 | 0 | if __name__ == "__main__":
n = int(input())
answer = []
for i in range(n, 0, -1):
if(i % 2 == 1):
answer.append("I hate")
else:
answer.append("I love")
if(i != 1):
answer.append("that")
else:
answer.append("it")
print(" ".join(answer)) | Title: Hulk
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of layers of love and hate.
Output Specification:
Print Dr.Banner's feeling in one line.
Demo Input:
['1\n', '2\n', '3\n']
Demo Output:
['I hate it\n', 'I hate that I love it\n', 'I hate that I love that I hate it\n']
Note:
none | ```python
if __name__ == "__main__":
n = int(input())
answer = []
for i in range(n, 0, -1):
if(i % 2 == 1):
answer.append("I hate")
else:
answer.append("I love")
if(i != 1):
answer.append("that")
else:
answer.append("it")
print(" ".join(answer))
``` | 0 | |
544 | B | Sea and Islands | PROGRAMMING | 1,400 | [
"constructive algorithms",
"implementation"
] | null | null | A map of some object is a rectangular field consisting of *n* rows and *n* columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly *k* islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island).
Find a way to cover some cells with sand so that exactly *k* islands appear on the *n*<=×<=*n* map, or determine that no such way exists. | The single line contains two positive integers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*2) — the size of the map and the number of islands you should form. | If the answer doesn't exist, print "NO" (without the quotes) in a single line.
Otherwise, print "YES" in the first line. In the next *n* lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal *n*.
If there are multiple answers, you may print any of them.
You should not maximize the sizes of islands. | [
"5 2\n",
"5 25\n"
] | [
"YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS\n",
"NO\n"
] | none | 1,000 | [
{
"input": "5 2",
"output": "YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS"
},
{
"input": "5 25",
"output": "NO"
},
{
"input": "82 6047",
"output": "NO"
},
{
"input": "6 5",
"output": "YES\nLSLSLS\nSLSLSS\nSSSSSS\nSSSSSS\nSSSSSS\nSSSSSS"
},
{
"input": "10 80",
"outpu... | 1,596,848,102 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 202 | 22,630,400 | import sys
import string
from collections import defaultdict
from functools import lru_cache
from collections import Counter
def mi(s):
return map(int, s.strip().split())
def lmi(s):
return list(mi(s))
def mf(f, s):
return map(f, s)
def lmf(f, s):
return list(mf(f, s))
def main(n, k):
mat = [['S' for _ in range(n)] for _ in range(n)]
s = 0
even = True
for i in range(len(mat)):
for j in range(len(mat[i])):
if even and j % 2 == 0:
mat[i][j] = 'L'
s += 1
if not even and j % 2 == 1:
mat[i][j] == 'L'
s += 1
if s == k:
print("YES")
for r in mat:
print("".join(r))
return
even = not even
print("NO")
if __name__ == "__main__":
for e, line in enumerate(sys.stdin.readlines()):
n, k = mi(line)
main(n, k)
| Title: Sea and Islands
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A map of some object is a rectangular field consisting of *n* rows and *n* columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly *k* islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island).
Find a way to cover some cells with sand so that exactly *k* islands appear on the *n*<=×<=*n* map, or determine that no such way exists.
Input Specification:
The single line contains two positive integers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=*n*2) — the size of the map and the number of islands you should form.
Output Specification:
If the answer doesn't exist, print "NO" (without the quotes) in a single line.
Otherwise, print "YES" in the first line. In the next *n* lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal *n*.
If there are multiple answers, you may print any of them.
You should not maximize the sizes of islands.
Demo Input:
['5 2\n', '5 25\n']
Demo Output:
['YES\nSSSSS\nLLLLL\nSSSSS\nLLLLL\nSSSSS\n', 'NO\n']
Note:
none | ```python
import sys
import string
from collections import defaultdict
from functools import lru_cache
from collections import Counter
def mi(s):
return map(int, s.strip().split())
def lmi(s):
return list(mi(s))
def mf(f, s):
return map(f, s)
def lmf(f, s):
return list(mf(f, s))
def main(n, k):
mat = [['S' for _ in range(n)] for _ in range(n)]
s = 0
even = True
for i in range(len(mat)):
for j in range(len(mat[i])):
if even and j % 2 == 0:
mat[i][j] = 'L'
s += 1
if not even and j % 2 == 1:
mat[i][j] == 'L'
s += 1
if s == k:
print("YES")
for r in mat:
print("".join(r))
return
even = not even
print("NO")
if __name__ == "__main__":
for e, line in enumerate(sys.stdin.readlines()):
n, k = mi(line)
main(n, k)
``` | 0 | |
814 | B | An express train to reveries | PROGRAMMING | 1,300 | [
"constructive algorithms"
] | null | null | Sengoku still remembers the mysterious "colourful meteoroids" she discovered with Lala-chan when they were little. In particular, one of the nights impressed her deeply, giving her the illusion that all her fancies would be realized.
On that night, Sengoku constructed a permutation *p*1,<=*p*2,<=...,<=*p**n* of integers from 1 to *n* inclusive, with each integer representing a colour, wishing for the colours to see in the coming meteor outburst. Two incredible outbursts then arrived, each with *n* meteorids, colours of which being integer sequences *a*1,<=*a*2,<=...,<=*a**n* and *b*1,<=*b*2,<=...,<=*b**n* respectively. Meteoroids' colours were also between 1 and *n* inclusive, and the two sequences were not identical, that is, at least one *i* (1<=≤<=*i*<=≤<=*n*) exists, such that *a**i*<=≠<=*b**i* holds.
Well, she almost had it all — each of the sequences *a* and *b* matched exactly *n*<=-<=1 elements in Sengoku's permutation. In other words, there is exactly one *i* (1<=≤<=*i*<=≤<=*n*) such that *a**i*<=≠<=*p**i*, and exactly one *j* (1<=≤<=*j*<=≤<=*n*) such that *b**j*<=≠<=*p**j*.
For now, Sengoku is able to recover the actual colour sequences *a* and *b* through astronomical records, but her wishes have been long forgotten. You are to reconstruct any possible permutation Sengoku could have had on that night. | The first line of input contains a positive integer *n* (2<=≤<=*n*<=≤<=1<=000) — the length of Sengoku's permutation, being the length of both meteor outbursts at the same time.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of colours in the first meteor outburst.
The third line contains *n* space-separated integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of colours in the second meteor outburst. At least one *i* (1<=≤<=*i*<=≤<=*n*) exists, such that *a**i*<=≠<=*b**i* holds. | Output *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n*, denoting a possible permutation Sengoku could have had. If there are more than one possible answer, output any one of them.
Input guarantees that such permutation exists. | [
"5\n1 2 3 4 3\n1 2 5 4 5\n",
"5\n4 4 2 3 1\n5 4 5 3 1\n",
"4\n1 1 3 4\n1 4 3 4\n"
] | [
"1 2 5 4 3\n",
"5 4 2 3 1\n",
"1 2 3 4\n"
] | In the first sample, both 1, 2, 5, 4, 3 and 1, 2, 3, 4, 5 are acceptable outputs.
In the second sample, 5, 4, 2, 3, 1 is the only permutation to satisfy the constraints. | 1,000 | [
{
"input": "5\n1 2 3 4 3\n1 2 5 4 5",
"output": "1 2 5 4 3"
},
{
"input": "5\n4 4 2 3 1\n5 4 5 3 1",
"output": "5 4 2 3 1"
},
{
"input": "4\n1 1 3 4\n1 4 3 4",
"output": "1 2 3 4"
},
{
"input": "10\n1 2 3 4 7 6 7 8 9 10\n1 2 3 4 5 6 5 8 9 10",
"output": "1 2 3 4 5 6 7 8 9... | 1,497,028,174 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=[]
for i in range(n):
if(a[i] != b[i]):
c.append(i)
d=set()
s1=set(a)
s2=set(b)
s3=set(range(1,n+1))
d=list(s3-s1)
d+=list(s3-s2)
L=[]
if len(c)==1:
a[c[0]] = d[0]
print(a)
exit()
for k in range(2):
L.clear()
L+=a
L[c[k]] = d[0]
if len(set(L)) != n:
continue
tt=0
for m in range(n):
if(L[m] != b[m]):
tt+=1
if(tt>=2):
break;
if(tt==2):
continue
else:
break
print(*L)
| Title: An express train to reveries
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sengoku still remembers the mysterious "colourful meteoroids" she discovered with Lala-chan when they were little. In particular, one of the nights impressed her deeply, giving her the illusion that all her fancies would be realized.
On that night, Sengoku constructed a permutation *p*1,<=*p*2,<=...,<=*p**n* of integers from 1 to *n* inclusive, with each integer representing a colour, wishing for the colours to see in the coming meteor outburst. Two incredible outbursts then arrived, each with *n* meteorids, colours of which being integer sequences *a*1,<=*a*2,<=...,<=*a**n* and *b*1,<=*b*2,<=...,<=*b**n* respectively. Meteoroids' colours were also between 1 and *n* inclusive, and the two sequences were not identical, that is, at least one *i* (1<=≤<=*i*<=≤<=*n*) exists, such that *a**i*<=≠<=*b**i* holds.
Well, she almost had it all — each of the sequences *a* and *b* matched exactly *n*<=-<=1 elements in Sengoku's permutation. In other words, there is exactly one *i* (1<=≤<=*i*<=≤<=*n*) such that *a**i*<=≠<=*p**i*, and exactly one *j* (1<=≤<=*j*<=≤<=*n*) such that *b**j*<=≠<=*p**j*.
For now, Sengoku is able to recover the actual colour sequences *a* and *b* through astronomical records, but her wishes have been long forgotten. You are to reconstruct any possible permutation Sengoku could have had on that night.
Input Specification:
The first line of input contains a positive integer *n* (2<=≤<=*n*<=≤<=1<=000) — the length of Sengoku's permutation, being the length of both meteor outbursts at the same time.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of colours in the first meteor outburst.
The third line contains *n* space-separated integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of colours in the second meteor outburst. At least one *i* (1<=≤<=*i*<=≤<=*n*) exists, such that *a**i*<=≠<=*b**i* holds.
Output Specification:
Output *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n*, denoting a possible permutation Sengoku could have had. If there are more than one possible answer, output any one of them.
Input guarantees that such permutation exists.
Demo Input:
['5\n1 2 3 4 3\n1 2 5 4 5\n', '5\n4 4 2 3 1\n5 4 5 3 1\n', '4\n1 1 3 4\n1 4 3 4\n']
Demo Output:
['1 2 5 4 3\n', '5 4 2 3 1\n', '1 2 3 4\n']
Note:
In the first sample, both 1, 2, 5, 4, 3 and 1, 2, 3, 4, 5 are acceptable outputs.
In the second sample, 5, 4, 2, 3, 1 is the only permutation to satisfy the constraints. | ```python
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=[]
for i in range(n):
if(a[i] != b[i]):
c.append(i)
d=set()
s1=set(a)
s2=set(b)
s3=set(range(1,n+1))
d=list(s3-s1)
d+=list(s3-s2)
L=[]
if len(c)==1:
a[c[0]] = d[0]
print(a)
exit()
for k in range(2):
L.clear()
L+=a
L[c[k]] = d[0]
if len(set(L)) != n:
continue
tt=0
for m in range(n):
if(L[m] != b[m]):
tt+=1
if(tt>=2):
break;
if(tt==2):
continue
else:
break
print(*L)
``` | 0 | |
491 | A | Up the hill | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation"
] | null | null | Hiking club "Up the hill" just returned from a walk. Now they are trying to remember which hills they've just walked through.
It is known that there were *N* stops, all on different integer heights between 1 and *N* kilometers (inclusive) above the sea level. On the first day they've traveled from the first stop to the second stop, on the second day they've traveled from the second to the third and so on, and on the last day they've traveled from the stop *N*<=-<=1 to the stop *N* and successfully finished their expedition.
They are trying to find out which heights were their stops located at. They have an entry in a travel journal specifying how many days did they travel up the hill, and how many days did they walk down the hill.
Help them by suggesting some possible stop heights satisfying numbers from the travel journal. | In the first line there is an integer non-negative number *A* denoting the number of days of climbing up the hill. Second line contains an integer non-negative number *B* — the number of days of walking down the hill (*A*<=+<=*B*<=+<=1<==<=*N*, 1<=≤<=*N*<=≤<=100<=000). | Output *N* space-separated distinct integers from 1 to *N* inclusive, denoting possible heights of the stops in order of visiting. | [
"0\n1\n",
"2\n1"
] | [
"2 1 \n",
"1 3 4 2"
] | none | 500 | [
{
"input": "0\n1",
"output": "2 1 "
},
{
"input": "2\n1",
"output": "2 3 4 1 "
},
{
"input": "0\n3",
"output": "4 3 2 1 "
},
{
"input": "1\n1",
"output": "2 3 1 "
},
{
"input": "3\n7",
"output": "8 9 10 11 7 6 5 4 3 2 1 "
},
{
"input": "700\n300",
... | 1,589,457,688 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 124 | 20,172,800 | h = int(input())
d = int(input())
t = h + d + 1
up_hill = []
down_hill = []
if h < 1:
up_hill = [t]
else:
up_hill = [1]
for i in range(h):
up_hill.append(t - i)
if d < 1:
down_hill = []
else:
for i in range(1, t):
if i not in up_hill:
down_hill.append(i)
up_hill = up_hill + down_hill
for n in up_hill:
print(n, end=" ")
| Title: Up the hill
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hiking club "Up the hill" just returned from a walk. Now they are trying to remember which hills they've just walked through.
It is known that there were *N* stops, all on different integer heights between 1 and *N* kilometers (inclusive) above the sea level. On the first day they've traveled from the first stop to the second stop, on the second day they've traveled from the second to the third and so on, and on the last day they've traveled from the stop *N*<=-<=1 to the stop *N* and successfully finished their expedition.
They are trying to find out which heights were their stops located at. They have an entry in a travel journal specifying how many days did they travel up the hill, and how many days did they walk down the hill.
Help them by suggesting some possible stop heights satisfying numbers from the travel journal.
Input Specification:
In the first line there is an integer non-negative number *A* denoting the number of days of climbing up the hill. Second line contains an integer non-negative number *B* — the number of days of walking down the hill (*A*<=+<=*B*<=+<=1<==<=*N*, 1<=≤<=*N*<=≤<=100<=000).
Output Specification:
Output *N* space-separated distinct integers from 1 to *N* inclusive, denoting possible heights of the stops in order of visiting.
Demo Input:
['0\n1\n', '2\n1']
Demo Output:
['2 1 \n', '1 3 4 2']
Note:
none | ```python
h = int(input())
d = int(input())
t = h + d + 1
up_hill = []
down_hill = []
if h < 1:
up_hill = [t]
else:
up_hill = [1]
for i in range(h):
up_hill.append(t - i)
if d < 1:
down_hill = []
else:
for i in range(1, t):
if i not in up_hill:
down_hill.append(i)
up_hill = up_hill + down_hill
for n in up_hill:
print(n, end=" ")
``` | 0 | |
792 | A | New Bus Route | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | null | null | There are *n* cities situated along the main road of Berland. Cities are represented by their coordinates — integer numbers *a*1,<=*a*2,<=...,<=*a**n*. All coordinates are pairwise distinct.
It is possible to get from one city to another only by bus. But all buses and roads are very old, so the Minister of Transport decided to build a new bus route. The Minister doesn't want to spend large amounts of money — he wants to choose two cities in such a way that the distance between them is minimal possible. The distance between two cities is equal to the absolute value of the difference between their coordinates.
It is possible that there are multiple pairs of cities with minimal possible distance, so the Minister wants to know the quantity of such pairs.
Your task is to write a program that will calculate the minimal possible distance between two pairs of cities and the quantity of pairs which have this distance. | The first line contains one integer number *n* (2<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). All numbers *a**i* are pairwise distinct. | Print two integer numbers — the minimal distance and the quantity of pairs with this distance. | [
"4\n6 -3 0 4\n",
"3\n-2 0 2\n"
] | [
"2 1\n",
"2 2\n"
] | In the first example the distance between the first city and the fourth city is |4 - 6| = 2, and it is the only pair with this distance. | 0 | [
{
"input": "4\n6 -3 0 4",
"output": "2 1"
},
{
"input": "3\n-2 0 2",
"output": "2 2"
},
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "2\n1000000000 -1000000000",
"output": "2000000000 1"
},
{
"input": "5\n-979619606 -979619602 -979619604 -979619605 -97961960... | 1,490,625,693 | 393 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 77 | 23,142,400 | n = int(str(input()).strip())
a = [int(x) for x in str(input()).split(' ')]
md = min([abs(a1 - a0) for a0, a1 in zip(a, a[1:])])
print(sum([md == abs(a1 - a0) for a0, a1 in zip(a, a[1:])]))
| Title: New Bus Route
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities situated along the main road of Berland. Cities are represented by their coordinates — integer numbers *a*1,<=*a*2,<=...,<=*a**n*. All coordinates are pairwise distinct.
It is possible to get from one city to another only by bus. But all buses and roads are very old, so the Minister of Transport decided to build a new bus route. The Minister doesn't want to spend large amounts of money — he wants to choose two cities in such a way that the distance between them is minimal possible. The distance between two cities is equal to the absolute value of the difference between their coordinates.
It is possible that there are multiple pairs of cities with minimal possible distance, so the Minister wants to know the quantity of such pairs.
Your task is to write a program that will calculate the minimal possible distance between two pairs of cities and the quantity of pairs which have this distance.
Input Specification:
The first line contains one integer number *n* (2<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). All numbers *a**i* are pairwise distinct.
Output Specification:
Print two integer numbers — the minimal distance and the quantity of pairs with this distance.
Demo Input:
['4\n6 -3 0 4\n', '3\n-2 0 2\n']
Demo Output:
['2 1\n', '2 2\n']
Note:
In the first example the distance between the first city and the fourth city is |4 - 6| = 2, and it is the only pair with this distance. | ```python
n = int(str(input()).strip())
a = [int(x) for x in str(input()).split(' ')]
md = min([abs(a1 - a0) for a0, a1 in zip(a, a[1:])])
print(sum([md == abs(a1 - a0) for a0, a1 in zip(a, a[1:])]))
``` | 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": "... | 1,658,253,580 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 46 | 0 | s=input()
d=input()
o=''
for i in range(len(s)):
if s[i]==d[i]:
o+='0'
else:o+='1'
print(o) | 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=input()
d=input()
o=''
for i in range(len(s)):
if s[i]==d[i]:
o+='0'
else:o+='1'
print(o)
``` | 3.9885 |
166 | C | Median | PROGRAMMING | 1,500 | [
"greedy",
"math",
"sortings"
] | null | null | A median in an array with the length of *n* is an element which occupies position number after we sort the elements in the non-decreasing order (the array elements are numbered starting with 1). A median of an array (2,<=6,<=1,<=2,<=3) is the number 2, and a median of array (0,<=96,<=17,<=23) — the number 17.
We define an expression as the integer part of dividing number *a* by number *b*.
One day Vasya showed Petya an array consisting of *n* integers and suggested finding the array's median. Petya didn't even look at the array and said that it equals *x*. Petya is a very honest boy, so he decided to add several numbers to the given array so that the median of the resulting array would be equal to *x*.
Petya can add any integers from 1 to 105 to the array, including the same numbers. Of course, he can add nothing to the array. If a number is added multiple times, then we should consider it the number of times it occurs. It is not allowed to delete of change initial numbers of the array.
While Petya is busy distracting Vasya, your task is to find the minimum number of elements he will need. | The first input line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=500, 1<=≤<=*x*<=≤<=105) — the initial array's length and the required median's value. The second line contains *n* space-separated numbers — the initial array. The elements of the array are integers from 1 to 105. The array elements are not necessarily different. | Print the only integer — the minimum number of elements Petya needs to add to the array so that its median equals *x*. | [
"3 10\n10 20 30\n",
"3 4\n1 2 3\n"
] | [
"1\n",
"4\n"
] | In the first sample we can add number 9 to array (10, 20, 30). The resulting array (9, 10, 20, 30) will have a median in position <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7dd92241318a531b780c7783dfa446a3e413115e.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 10.
In the second sample you should add numbers 4, 5, 5, 5. The resulting array has median equal to 4. | 1,000 | [
{
"input": "3 10\n10 20 30",
"output": "1"
},
{
"input": "3 4\n1 2 3",
"output": "4"
},
{
"input": "2 2\n3 2",
"output": "0"
},
{
"input": "5 1\n1 1 2 1 2",
"output": "0"
},
{
"input": "5 4\n5 5 4 3 5",
"output": "1"
},
{
"input": "10 2\n2 2 1 3 2 1 2 ... | 1,687,585,950 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 124 | 0 | import sys
input = sys.stdin.readline
n, x = map(int, input().split())
a = list(map(int, input().split()))
f = 0
if(x not in a):
a.append(x)
n += 1
f = 1
a.sort()
ans, m, c = n, n, 0
while(c <= m and a[(n + 1) // 2 - 1] != x):
c += 1
a.insert(0, a[0])
n += 1
# print(a)
# print(c)
ans = min(ans, c)
a = a[c:]
c = 0
n = m
while(c <= m and a[(n + 1) // 2 - 1] != x):
c += 1
a.append(a[-1])
n += 1
ans = min(ans, c)
print(ans + f)
| Title: Median
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A median in an array with the length of *n* is an element which occupies position number after we sort the elements in the non-decreasing order (the array elements are numbered starting with 1). A median of an array (2,<=6,<=1,<=2,<=3) is the number 2, and a median of array (0,<=96,<=17,<=23) — the number 17.
We define an expression as the integer part of dividing number *a* by number *b*.
One day Vasya showed Petya an array consisting of *n* integers and suggested finding the array's median. Petya didn't even look at the array and said that it equals *x*. Petya is a very honest boy, so he decided to add several numbers to the given array so that the median of the resulting array would be equal to *x*.
Petya can add any integers from 1 to 105 to the array, including the same numbers. Of course, he can add nothing to the array. If a number is added multiple times, then we should consider it the number of times it occurs. It is not allowed to delete of change initial numbers of the array.
While Petya is busy distracting Vasya, your task is to find the minimum number of elements he will need.
Input Specification:
The first input line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=500, 1<=≤<=*x*<=≤<=105) — the initial array's length and the required median's value. The second line contains *n* space-separated numbers — the initial array. The elements of the array are integers from 1 to 105. The array elements are not necessarily different.
Output Specification:
Print the only integer — the minimum number of elements Petya needs to add to the array so that its median equals *x*.
Demo Input:
['3 10\n10 20 30\n', '3 4\n1 2 3\n']
Demo Output:
['1\n', '4\n']
Note:
In the first sample we can add number 9 to array (10, 20, 30). The resulting array (9, 10, 20, 30) will have a median in position <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7dd92241318a531b780c7783dfa446a3e413115e.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 10.
In the second sample you should add numbers 4, 5, 5, 5. The resulting array has median equal to 4. | ```python
import sys
input = sys.stdin.readline
n, x = map(int, input().split())
a = list(map(int, input().split()))
f = 0
if(x not in a):
a.append(x)
n += 1
f = 1
a.sort()
ans, m, c = n, n, 0
while(c <= m and a[(n + 1) // 2 - 1] != x):
c += 1
a.insert(0, a[0])
n += 1
# print(a)
# print(c)
ans = min(ans, c)
a = a[c:]
c = 0
n = m
while(c <= m and a[(n + 1) // 2 - 1] != x):
c += 1
a.append(a[-1])
n += 1
ans = min(ans, c)
print(ans + f)
``` | 3 | |
729 | B | Spotlights | PROGRAMMING | 1,200 | [
"dp",
"implementation"
] | null | null | Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ. | The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan. | Print one integer — the number of good positions for placing the spotlight. | [
"2 4\n0 1 0 0\n1 0 1 0\n",
"4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n"
] | [
"9\n",
"20\n"
] | In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example. | 1,000 | [
{
"input": "2 4\n0 1 0 0\n1 0 1 0",
"output": "9"
},
{
"input": "4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0",
"output": "20"
},
{
"input": "1 5\n1 1 0 0 0",
"output": "3"
},
{
"input": "2 10\n0 0 0 0 0 0 0 1 0 0\n1 0 0 0 0 0 0 0 0 0",
"output": "20"
},
{
"input": "3 ... | 1,653,129,723 | 2,147,483,647 | Python 3 | OK | TESTS | 72 | 936 | 6,348,800 | n, m = [int(x) for x in input().split()]
plan = []
for line in range(n):
plan.append([int(x) for x in input().split(" ")])
total_positions = 0
for i in range(n):
actor_before = False
free_spot_counter = 0
for j in range(m):
if(plan[i][j] == 1):
actor_before = True
total_positions += free_spot_counter
free_spot_counter = 0
else:
free_spot_counter += 1
if(actor_before):
total_positions += 1
for j in range(m):
actor_before = False
free_spot_counter = 0
for i in range(n):
if(plan[i][j] == 1):
actor_before = True
total_positions += free_spot_counter
free_spot_counter = 0
else:
free_spot_counter += 1
if(actor_before):
total_positions += 1
print(total_positions)
| Title: Spotlights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Theater stage is a rectangular field of size *n*<=×<=*m*. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to; - there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
Input Specification:
The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the plan.
The next *n* lines contain *m* integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
Output Specification:
Print one integer — the number of good positions for placing the spotlight.
Demo Input:
['2 4\n0 1 0 0\n1 0 1 0\n', '4 4\n0 0 0 0\n1 0 0 1\n0 1 1 0\n0 1 0 0\n']
Demo Output:
['9\n', '20\n']
Note:
In the first example the following positions are good:
1. the (1, 1) cell and right direction; 1. the (1, 1) cell and down direction; 1. the (1, 3) cell and left direction; 1. the (1, 3) cell and down direction; 1. the (1, 4) cell and left direction; 1. the (2, 2) cell and left direction; 1. the (2, 2) cell and up direction; 1. the (2, 2) and right direction; 1. the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example. | ```python
n, m = [int(x) for x in input().split()]
plan = []
for line in range(n):
plan.append([int(x) for x in input().split(" ")])
total_positions = 0
for i in range(n):
actor_before = False
free_spot_counter = 0
for j in range(m):
if(plan[i][j] == 1):
actor_before = True
total_positions += free_spot_counter
free_spot_counter = 0
else:
free_spot_counter += 1
if(actor_before):
total_positions += 1
for j in range(m):
actor_before = False
free_spot_counter = 0
for i in range(n):
if(plan[i][j] == 1):
actor_before = True
total_positions += free_spot_counter
free_spot_counter = 0
else:
free_spot_counter += 1
if(actor_before):
total_positions += 1
print(total_positions)
``` | 3 | |
976 | A | Minimum Binary Number | PROGRAMMING | 800 | [
"implementation"
] | null | null | String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all). | The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct. | Print one string — the minimum correct string that you can obtain from the given one. | [
"4\n1001\n",
"1\n1\n"
] | [
"100\n",
"1\n"
] | In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | 0 | [
{
"input": "4\n1001",
"output": "100"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "100\n1110111100001111011111111010110011111111011110000111101101011100110110001011000000101010110101011100",
"output": "1000000000000000000000000000000000000000"
},
{
"input": "100\n100000... | 1,588,309,683 | 2,147,483,647 | PyPy 3 | OK | TESTS | 14 | 140 | 0 | n=int(input())
s=input()
while('01' in s or '11' in s):
if('01' in s):
s=s.replace('01','10')
elif('11' in s):
s=s.replace('11','1')
print(s) | Title: Minimum Binary Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
Output Specification:
Print one string — the minimum correct string that you can obtain from the given one.
Demo Input:
['4\n1001\n', '1\n1\n']
Demo Output:
['100\n', '1\n']
Note:
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | ```python
n=int(input())
s=input()
while('01' in s or '11' in s):
if('01' in s):
s=s.replace('01','10')
elif('11' in s):
s=s.replace('11','1')
print(s)
``` | 3 | |
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"
},
... | 1,697,486,026 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | n=int(input())
solved_problems=0
for i in range(n):
mo=input()
mo2=input()
mo3=input()
if (mo == "1" and mo2 == "1") or (mo2 == "1" and mo3 == "1") or (mo == "1" and mo3 == "1"):
solved_problems+=1
print(solved_problems) | 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
n=int(input())
solved_problems=0
for i in range(n):
mo=input()
mo2=input()
mo3=input()
if (mo == "1" and mo2 == "1") or (mo2 == "1" and mo3 == "1") or (mo == "1" and mo3 == "1"):
solved_problems+=1
print(solved_problems)
``` | -1 | |
478 | A | Initial Bet | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | There are five people playing a game called "Generosity". Each person gives some non-zero number of coins *b* as an initial bet. After all players make their bets of *b* coins, the following operation is repeated for several times: a coin is passed from one player to some other player.
Your task is to write a program that can, given the number of coins each player has at the end of the game, determine the size *b* of the initial bet or find out that such outcome of the game cannot be obtained for any positive number of coins *b* in the initial bet. | The input consists of a single line containing five integers *c*1,<=*c*2,<=*c*3,<=*c*4 and *c*5 — the number of coins that the first, second, third, fourth and fifth players respectively have at the end of the game (0<=≤<=*c*1,<=*c*2,<=*c*3,<=*c*4,<=*c*5<=≤<=100). | Print the only line containing a single positive integer *b* — the number of coins in the initial bet of each player. If there is no such value of *b*, then print the only value "-1" (quotes for clarity). | [
"2 5 4 0 4\n",
"4 5 9 2 1\n"
] | [
"3\n",
"-1\n"
] | In the first sample the following sequence of operations is possible:
1. One coin is passed from the fourth player to the second player; 1. One coin is passed from the fourth player to the fifth player; 1. One coin is passed from the first player to the third player; 1. One coin is passed from the fourth player to the second player. | 500 | [
{
"input": "2 5 4 0 4",
"output": "3"
},
{
"input": "4 5 9 2 1",
"output": "-1"
},
{
"input": "0 0 0 0 0",
"output": "-1"
},
{
"input": "1 2 1 2 3",
"output": "-1"
},
{
"input": "100 0 0 0 0",
"output": "20"
},
{
"input": "2 3 4 5 6",
"output": "4"... | 1,698,419,788 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | n=list(map(int,input().split()))
m=sum(n)
if m%5==0:
print(m//5)
else:
print(-1)
| Title: Initial Bet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are five people playing a game called "Generosity". Each person gives some non-zero number of coins *b* as an initial bet. After all players make their bets of *b* coins, the following operation is repeated for several times: a coin is passed from one player to some other player.
Your task is to write a program that can, given the number of coins each player has at the end of the game, determine the size *b* of the initial bet or find out that such outcome of the game cannot be obtained for any positive number of coins *b* in the initial bet.
Input Specification:
The input consists of a single line containing five integers *c*1,<=*c*2,<=*c*3,<=*c*4 and *c*5 — the number of coins that the first, second, third, fourth and fifth players respectively have at the end of the game (0<=≤<=*c*1,<=*c*2,<=*c*3,<=*c*4,<=*c*5<=≤<=100).
Output Specification:
Print the only line containing a single positive integer *b* — the number of coins in the initial bet of each player. If there is no such value of *b*, then print the only value "-1" (quotes for clarity).
Demo Input:
['2 5 4 0 4\n', '4 5 9 2 1\n']
Demo Output:
['3\n', '-1\n']
Note:
In the first sample the following sequence of operations is possible:
1. One coin is passed from the fourth player to the second player; 1. One coin is passed from the fourth player to the fifth player; 1. One coin is passed from the first player to the third player; 1. One coin is passed from the fourth player to the second player. | ```python
n=list(map(int,input().split()))
m=sum(n)
if m%5==0:
print(m//5)
else:
print(-1)
``` | 0 | |
500 | A | New Year Transportation | PROGRAMMING | 1,000 | [
"dfs and similar",
"graphs",
"implementation"
] | null | null | New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. | If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". | [
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | 500 | [
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 ... | 1,672,473,350 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 1,945,600 | n,t=map(int,input().split())
ai=list(map(int,input().split()))
ai.insert(0,0)
visited=False
currentcell=1
while currentcell<=t:
currentcell=currentcell+ai[currentcell]
if currentcell==t:
visited=True
break
if visited:
print("YES")
else:
print("NO")
| Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | ```python
n,t=map(int,input().split())
ai=list(map(int,input().split()))
ai.insert(0,0)
visited=False
currentcell=1
while currentcell<=t:
currentcell=currentcell+ai[currentcell]
if currentcell==t:
visited=True
break
if visited:
print("YES")
else:
print("NO")
``` | 3 | |
277 | A | Learning Languages | PROGRAMMING | 1,400 | [
"dfs and similar",
"dsu"
] | null | null | The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating). | The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces. | Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating). | [
"5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n",
"8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n",
"2 2\n1 2\n0\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2. | 500 | [
{
"input": "5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5",
"output": "0"
},
{
"input": "8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1",
"output": "2"
},
{
"input": "2 2\n1 2\n0",
"output": "1"
},
{
"input": "2 2\n0\n0",
"output": "2"
},
{
"input": "5 5\n1 3\n0\n0\n2 4... | 1,553,788,622 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 218 | 307,200 | def isCommon(l1,l2):
for i in l1:
if i in l2:
return True
return False
def BFS(V,E,start,visited):
q = []
q.append(start)
vis[start] = True
#print(start , end = ' ')
while len(q)!=0:
v = q.pop(0)
for i in E[v]:
if not vis[i]:
vis[i] = True
q.append(i)
# print(i , end = ' ')
#print()
return vis
n,m =map(int,input().split())
lang = []
k = [0]*n
for i in range(n):
ls = list(map(int,input().split()))
k[i] = ls.pop(0)
lang.append(ls)
v=[]
for i in range(n):
v.append(i)
adj =[0]*n
for i in range(n):
adj[i] = []
for i in range(n):
for j in range(n):
if isCommon(lang[i],lang[j]):
adj[i].append(j)
adj[j].append(i)
vis = [False]*n
vis = BFS(v,adj,0,vis)
count = 0
while False in vis:
ind = vis.index(False)
vis = BFS(v,adj,ind,vis)
count+=1
print(count)
| Title: Learning Languages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).
Demo Input:
['5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n', '8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n', '2 2\n1 2\n0\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2. | ```python
def isCommon(l1,l2):
for i in l1:
if i in l2:
return True
return False
def BFS(V,E,start,visited):
q = []
q.append(start)
vis[start] = True
#print(start , end = ' ')
while len(q)!=0:
v = q.pop(0)
for i in E[v]:
if not vis[i]:
vis[i] = True
q.append(i)
# print(i , end = ' ')
#print()
return vis
n,m =map(int,input().split())
lang = []
k = [0]*n
for i in range(n):
ls = list(map(int,input().split()))
k[i] = ls.pop(0)
lang.append(ls)
v=[]
for i in range(n):
v.append(i)
adj =[0]*n
for i in range(n):
adj[i] = []
for i in range(n):
for j in range(n):
if isCommon(lang[i],lang[j]):
adj[i].append(j)
adj[j].append(i)
vis = [False]*n
vis = BFS(v,adj,0,vis)
count = 0
while False in vis:
ind = vis.index(False)
vis = BFS(v,adj,ind,vis)
count+=1
print(count)
``` | 0 | |
279 | B | Books | PROGRAMMING | 1,400 | [
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null | When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. | Print a single integer — the maximum number of books Valera can read. | [
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] | [
"3\n",
"1\n"
] | none | 1,000 | [
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4... | 1,671,123,529 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 5,939,200 | n, t = map(int, input().split())
minutes = list(map(int, input().split()))
max_col = 0
for i in range(n):
a = t
cur_col = 0
while i < n:
if a >= minutes[i]:
cur_col += 1
a -= minutes[i]
i += 1
else:
break
max_col = max(cur_col, max_col)
print(max_col) | Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
n, t = map(int, input().split())
minutes = list(map(int, input().split()))
max_col = 0
for i in range(n):
a = t
cur_col = 0
while i < n:
if a >= minutes[i]:
cur_col += 1
a -= minutes[i]
i += 1
else:
break
max_col = max(cur_col, max_col)
print(max_col)
``` | 0 | |
233 | A | Perfect Permutation | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*. | A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size. | If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces. | [
"1\n",
"2\n",
"4\n"
] | [
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] | none | 500 | [
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
... | 1,637,462,372 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 10,035,200 | def permutations(ch, curr_index=0):
if curr_index == len(ch) - 1:
pr = True
for i in range(1, len(ch) + 1):
if ch[ch[i - 1] - 1] != i or ch[i - 1] == i:
pr = False
if pr:
print(" ".join([str(i) for i in ch]))
exit()
for i in range(curr_index, len(ch)):
ch[curr_index], ch[i] = ch[i], ch[curr_index]
permutations(ch, curr_index + 1)
ch[curr_index], ch[i] = ch[i], ch[curr_index]
n = int(input())
permutations(list(range(1, n+1)))
print(-1)
| Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none | ```python
def permutations(ch, curr_index=0):
if curr_index == len(ch) - 1:
pr = True
for i in range(1, len(ch) + 1):
if ch[ch[i - 1] - 1] != i or ch[i - 1] == i:
pr = False
if pr:
print(" ".join([str(i) for i in ch]))
exit()
for i in range(curr_index, len(ch)):
ch[curr_index], ch[i] = ch[i], ch[curr_index]
permutations(ch, curr_index + 1)
ch[curr_index], ch[i] = ch[i], ch[curr_index]
n = int(input())
permutations(list(range(1, n+1)))
print(-1)
``` | 0 | |
722 | A | Broken Clock | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.
You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format.
For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39. | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively.
The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | [
"24\n17:30\n",
"12\n17:30\n",
"24\n99:99\n"
] | [
"17:30\n",
"07:30\n",
"09:09\n"
] | none | 500 | [
{
"input": "24\n17:30",
"output": "17:30"
},
{
"input": "12\n17:30",
"output": "07:30"
},
{
"input": "24\n99:99",
"output": "09:09"
},
{
"input": "12\n05:54",
"output": "05:54"
},
{
"input": "12\n00:05",
"output": "01:05"
},
{
"input": "24\n23:80",
... | 1,475,334,024 | 3,324 | Python 3 | WRONG_ANSWER | PRETESTS | 4 | 46 | 0 | n=int(input());
x=list(input());
h=int(x[3])*10+int(x[4]);
m=int(x[0])*10+int(x[1]);
if(m>=60):
x[3]='1';
if(n==12):
if(h<1):
x[1]='1';
if h>12:
x[0]='0';
if(n==24):
if(h>23):
x[0]='1';
print(''.join(x));
| Title: Broken Clock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.
You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format.
For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39.
Input Specification:
The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively.
The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes.
Output Specification:
The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them.
Demo Input:
['24\n17:30\n', '12\n17:30\n', '24\n99:99\n']
Demo Output:
['17:30\n', '07:30\n', '09:09\n']
Note:
none | ```python
n=int(input());
x=list(input());
h=int(x[3])*10+int(x[4]);
m=int(x[0])*10+int(x[1]);
if(m>=60):
x[3]='1';
if(n==12):
if(h<1):
x[1]='1';
if h>12:
x[0]='0';
if(n==24):
if(h>23):
x[0]='1';
print(''.join(x));
``` | 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... | 1,620,239,280 | 2,147,483,647 | Python 3 | OK | TESTS | 82 | 530 | 20,582,400 | n, p, q, r = map(int, input().split())
nums = list(map(int, input().split()))
# make a dp with each row as the running max
# of p, q and r product with current element
# add the value from the row above
dp = [[0 for _ in range(n+1)] for _ in range(4)]
dp[0][0] = 0
coef = [p, q, r]
for i in range(1, 4):
for j in range(1, n+1):
if j==1:
dp[i][j] = dp[i-1][j] + coef[i-1]*nums[j-1]
else:
dp[i][j] = max(dp[i-1][j] + coef[i-1]*nums[j-1], dp[i][j-1])
#for d in dp:
# print(d)
print(dp[3][n])
| 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
n, p, q, r = map(int, input().split())
nums = list(map(int, input().split()))
# make a dp with each row as the running max
# of p, q and r product with current element
# add the value from the row above
dp = [[0 for _ in range(n+1)] for _ in range(4)]
dp[0][0] = 0
coef = [p, q, r]
for i in range(1, 4):
for j in range(1, n+1):
if j==1:
dp[i][j] = dp[i-1][j] + coef[i-1]*nums[j-1]
else:
dp[i][j] = max(dp[i-1][j] + coef[i-1]*nums[j-1], dp[i][j-1])
#for d in dp:
# print(d)
print(dp[3][n])
``` | 3 | |
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... | 1,612,721,777 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 124 | 0 | string=input()
i=0
while i!=len(string):
if string[i]=='.':
print(0,end="")
i=i+1
elif string[i]=='-':
if string[i+1]=='.':
print(1,end="")
i=i+2
elif string[i+1]=='-':
print(2,end="")
i=i+2
| 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()
i=0
while i!=len(string):
if string[i]=='.':
print(0,end="")
i=i+1
elif string[i]=='-':
if string[i+1]=='.':
print(1,end="")
i=i+2
elif string[i+1]=='-':
print(2,end="")
i=i+2
``` | 3.969 |
158 | A | Next Round | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | "Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round. | The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1). | Output the number of participants who advance to the next round. | [
"8 5\n10 9 8 7 7 7 5 5\n",
"4 2\n0 0 0 0\n"
] | [
"6\n",
"0\n"
] | In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | 500 | [
{
"input": "8 5\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "4 2\n0 0 0 0",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "1 1\n10",
"output": "1"
},
{
"input": "17 14\n16 15... | 1,699,798,232 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 60 | 0 | n,k = map(int, input().split())
scores = list(map(int, input().split()))
passedScores = []
for i in range(n):
if scores[i] >= scores[k-1]:
passedScores.append(scores[i])
print(len(passedScores)) | Title: Next Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output Specification:
Output the number of participants who advance to the next round.
Demo Input:
['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n']
Demo Output:
['6\n', '0\n']
Note:
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | ```python
n,k = map(int, input().split())
scores = list(map(int, input().split()))
passedScores = []
for i in range(n):
if scores[i] >= scores[k-1]:
passedScores.append(scores[i])
print(len(passedScores))
``` | 0 | |
845 | C | Two TVs | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"sortings"
] | null | null | Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains *n* shows, *i*-th of them starts at moment *l**i* and ends at moment *r**i*.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all *n* shows. Are two TVs enough to do so? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of shows.
Each of the next *n* lines contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=109) — starting and ending time of *i*-th show. | If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes). | [
"3\n1 2\n2 3\n4 5\n",
"4\n1 2\n2 3\n2 3\n1 2\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "3\n1 2\n2 3\n4 5",
"output": "YES"
},
{
"input": "4\n1 2\n2 3\n2 3\n1 2",
"output": "NO"
},
{
"input": "4\n0 1\n1 2\n2 3\n3 4",
"output": "YES"
},
{
"input": "3\n1 2\n2 3\n2 4",
"output": "NO"
},
{
"input": "3\n0 100\n0 100\n0 100",
"output": "NO"
... | 1,650,205,675 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 560 | 16,281,600 | def can_all_shows_be_watched (show_times, n):
reserved_before = 0
show_times.sort(key=lambda elem: elem[0])
for i in range(n):
current_show_time = show_times[i]
if current_show_time [0] < reserved_before:
return "NO"
reserved_before = current_show_time[1]
return "YES"
n_shows_input = int(input())
show_times_input = []
for i_input in range(n_shows_input):
l_curr_input, r_curr_input = input().split(" ")
show_times_input.append((int(l_curr_input), int(r_curr_input)))
print(can_all_shows_be_watched(show_times_input, n_shows_input))
| Title: Two TVs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains *n* shows, *i*-th of them starts at moment *l**i* and ends at moment *r**i*.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all *n* shows. Are two TVs enough to do so?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of shows.
Each of the next *n* lines contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=109) — starting and ending time of *i*-th show.
Output Specification:
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
Demo Input:
['3\n1 2\n2 3\n4 5\n', '4\n1 2\n2 3\n2 3\n1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
def can_all_shows_be_watched (show_times, n):
reserved_before = 0
show_times.sort(key=lambda elem: elem[0])
for i in range(n):
current_show_time = show_times[i]
if current_show_time [0] < reserved_before:
return "NO"
reserved_before = current_show_time[1]
return "YES"
n_shows_input = int(input())
show_times_input = []
for i_input in range(n_shows_input):
l_curr_input, r_curr_input = input().split(" ")
show_times_input.append((int(l_curr_input), int(r_curr_input)))
print(can_all_shows_be_watched(show_times_input, n_shows_input))
``` | 0 | |
573 | A | Bear and Poker | PROGRAMMING | 1,300 | [
"implementation",
"math",
"number theory"
] | null | null | Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot? | First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players. | Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise. | [
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] | [
"Yes\n",
"No\n"
] | In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | 500 | [
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724... | 1,588,005,462 | 2,147,483,647 | Python 3 | OK | TESTS | 70 | 919 | 9,318,400 | from collections import Counter,defaultdict,deque
import heapq as hq
from itertools import count, islice
#alph = 'abcdefghijklmnopqrstuvwxyz'
#from math import factorial as fact
#a,b = [int(x) for x in input().split()]
import math
import sys
input=sys.stdin.readline
n = int(input())
arr = [int(x) for x in input().split()]
for i in range(n):
while (arr[i]%3)==0:
arr[i]//=3
while (arr[i]%2)==0:
arr[i]//=2
q = arr[0]
for i in range(1,n):
if arr[i]!=q:
print('No')
exit()
print('Yes')
| Title: Bear and Poker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | ```python
from collections import Counter,defaultdict,deque
import heapq as hq
from itertools import count, islice
#alph = 'abcdefghijklmnopqrstuvwxyz'
#from math import factorial as fact
#a,b = [int(x) for x in input().split()]
import math
import sys
input=sys.stdin.readline
n = int(input())
arr = [int(x) for x in input().split()]
for i in range(n):
while (arr[i]%3)==0:
arr[i]//=3
while (arr[i]%2)==0:
arr[i]//=2
q = arr[0]
for i in range(1,n):
if arr[i]!=q:
print('No')
exit()
print('Yes')
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | One day, Hongcow goes to the store and sees a brand new deck of *n* special cards. Each individual card is either red or blue. He decides he wants to buy them immediately. To do this, he needs to play a game with the owner of the store.
This game takes some number of turns to complete. On a turn, Hongcow may do one of two things:
- Collect tokens. Hongcow collects 1 red token and 1 blue token by choosing this option (thus, 2 tokens in total per one operation). - Buy a card. Hongcow chooses some card and spends tokens to purchase it as specified below.
The *i*-th card requires *r**i* red resources and *b**i* blue resources. Suppose Hongcow currently has *A* red cards and *B* blue cards. Then, the *i*-th card will require Hongcow to spend *max*(*r**i*<=-<=*A*,<=0) red tokens, and *max*(*b**i*<=-<=*B*,<=0) blue tokens. Note, only tokens disappear, but the cards stay with Hongcow forever. Each card can be bought only once.
Given a description of the cards and their costs determine the minimum number of turns Hongcow needs to purchase all cards. | The first line of input will contain a single integer *n* (1<=≤<=*n*<=≤<=16).
The next *n* lines of input will contain three tokens *c**i*, *r**i* and *b**i*. *c**i* will be 'R' or 'B', denoting the color of the card as red or blue. *r**i* will be an integer denoting the amount of red resources required to obtain the card, and *b**i* will be an integer denoting the amount of blue resources required to obtain the card (0<=≤<=*r**i*,<=*b**i*<=≤<=107). | Output a single integer, denoting the minimum number of turns needed to acquire all the cards. | [
"3\nR 0 1\nB 1 0\nR 1 1\n",
"3\nR 3 0\nR 2 0\nR 1 0\n"
] | [
"4\n",
"6\n"
] | For the first sample, Hongcow's four moves are as follows:
1. Collect tokens 1. Buy card 1 1. Buy card 2 1. Buy card 3
For the second sample, one optimal strategy is as follows:
1. Collect tokens 1. Collect tokens 1. Buy card 2 1. Collect tokens 1. Buy card 3 1. Buy card 1 | 0 | [] | 1,482,019,060 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 61 | 204,800 | import math
def sum_num(n):
sum = 0
for i in range(1,n+1):
sum += i
return sum
def edit(l):
global red_card, blue_card
print("edit %d %d %d %d" % (red,blue,red_card,blue_card))
return l[0], l[1]-red_card, l[2]-blue_card
def get(n):
global red, blue, count
print("get %d %d %d %d" % (red,blue,red_card,blue_card))
red += n
blue += n
count += n
print("after get %d %d %d %d" % (red,blue,red_card,blue_card))
def buy(card):
global red,blue, red_card, blue_card, count
print("buy %d %d %d %d" % (red,blue,red_card,blue_card))
if card[0] == 'R':
red -= card[1]
blue -= card[2]
red_card += 1
elif card[0] == 'B':
red -= card[1]
blue -= card[2]
blue_card += 1
count += 1
print("after buy %d %d %d %d" % (red,blue,red_card,blue_card))
n = int(input())
ori_cards = dict()
blue = 0
red = 0
red_card = 0
blue_card = 0
string_card = ""
count = 0
for i in range(n):
s, r, b = input().split()
blue += int(b)
red += int(r)
string_card += s
red_card = string_card.count('R')
blue_card = string_card.count('B')
count_1 = max(red-sum_num(red_card-1), blue-sum_num(blue_card))
count_2 = max(red-sum_num(red_card), blue-sum_num(blue_card-1))
if blue_card == 0:
count_2 = 10**10
elif red_card == 0:
count_1 = 10**10
count = min(count_1,count_2)
count += red_card + blue_card
print(count) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, Hongcow goes to the store and sees a brand new deck of *n* special cards. Each individual card is either red or blue. He decides he wants to buy them immediately. To do this, he needs to play a game with the owner of the store.
This game takes some number of turns to complete. On a turn, Hongcow may do one of two things:
- Collect tokens. Hongcow collects 1 red token and 1 blue token by choosing this option (thus, 2 tokens in total per one operation). - Buy a card. Hongcow chooses some card and spends tokens to purchase it as specified below.
The *i*-th card requires *r**i* red resources and *b**i* blue resources. Suppose Hongcow currently has *A* red cards and *B* blue cards. Then, the *i*-th card will require Hongcow to spend *max*(*r**i*<=-<=*A*,<=0) red tokens, and *max*(*b**i*<=-<=*B*,<=0) blue tokens. Note, only tokens disappear, but the cards stay with Hongcow forever. Each card can be bought only once.
Given a description of the cards and their costs determine the minimum number of turns Hongcow needs to purchase all cards.
Input Specification:
The first line of input will contain a single integer *n* (1<=≤<=*n*<=≤<=16).
The next *n* lines of input will contain three tokens *c**i*, *r**i* and *b**i*. *c**i* will be 'R' or 'B', denoting the color of the card as red or blue. *r**i* will be an integer denoting the amount of red resources required to obtain the card, and *b**i* will be an integer denoting the amount of blue resources required to obtain the card (0<=≤<=*r**i*,<=*b**i*<=≤<=107).
Output Specification:
Output a single integer, denoting the minimum number of turns needed to acquire all the cards.
Demo Input:
['3\nR 0 1\nB 1 0\nR 1 1\n', '3\nR 3 0\nR 2 0\nR 1 0\n']
Demo Output:
['4\n', '6\n']
Note:
For the first sample, Hongcow's four moves are as follows:
1. Collect tokens 1. Buy card 1 1. Buy card 2 1. Buy card 3
For the second sample, one optimal strategy is as follows:
1. Collect tokens 1. Collect tokens 1. Buy card 2 1. Collect tokens 1. Buy card 3 1. Buy card 1 | ```python
import math
def sum_num(n):
sum = 0
for i in range(1,n+1):
sum += i
return sum
def edit(l):
global red_card, blue_card
print("edit %d %d %d %d" % (red,blue,red_card,blue_card))
return l[0], l[1]-red_card, l[2]-blue_card
def get(n):
global red, blue, count
print("get %d %d %d %d" % (red,blue,red_card,blue_card))
red += n
blue += n
count += n
print("after get %d %d %d %d" % (red,blue,red_card,blue_card))
def buy(card):
global red,blue, red_card, blue_card, count
print("buy %d %d %d %d" % (red,blue,red_card,blue_card))
if card[0] == 'R':
red -= card[1]
blue -= card[2]
red_card += 1
elif card[0] == 'B':
red -= card[1]
blue -= card[2]
blue_card += 1
count += 1
print("after buy %d %d %d %d" % (red,blue,red_card,blue_card))
n = int(input())
ori_cards = dict()
blue = 0
red = 0
red_card = 0
blue_card = 0
string_card = ""
count = 0
for i in range(n):
s, r, b = input().split()
blue += int(b)
red += int(r)
string_card += s
red_card = string_card.count('R')
blue_card = string_card.count('B')
count_1 = max(red-sum_num(red_card-1), blue-sum_num(blue_card))
count_2 = max(red-sum_num(red_card), blue-sum_num(blue_card-1))
if blue_card == 0:
count_2 = 10**10
elif red_card == 0:
count_1 = 10**10
count = min(count_1,count_2)
count += red_card + blue_card
print(count)
``` | 0 | |
989 | A | A Blend of Springtime | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | "What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order. | The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively. | Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower). | [
".BAC.\n",
"AA..CB\n"
] | [
"Yes\n",
"No\n"
] | In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | 500 | [
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
... | 1,675,907,068 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 46 | 0 | # LUOGU_RID: 101744204
s = input()
print(any(x in s for x in ['ABC', 'ACB', 'CAB', 'CBA', 'BCA', 'BAC']) and 'Yes' or 'No') | Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | ```python
# LUOGU_RID: 101744204
s = input()
print(any(x in s for x in ['ABC', 'ACB', 'CAB', 'CBA', 'BCA', 'BAC']) and 'Yes' or 'No')
``` | 3 | |
920 | B | Tea Queue | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Recently *n* students from city S moved to city P to attend a programming camp.
They moved there by train. In the evening, all students in the train decided that they want to drink some tea. Of course, no two people can use the same teapot simultaneously, so the students had to form a queue to get their tea.
*i*-th student comes to the end of the queue at the beginning of *l**i*-th second. If there are multiple students coming to the queue in the same moment, then the student with greater index comes after the student with lesser index. Students in the queue behave as follows: if there is nobody in the queue before the student, then he uses the teapot for exactly one second and leaves the queue with his tea; otherwise the student waits for the people before him to get their tea. If at the beginning of *r**i*-th second student *i* still cannot get his tea (there is someone before him in the queue), then he leaves the queue without getting any tea.
For each student determine the second he will use the teapot and get his tea (if he actually gets it). | The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=1000).
Then *t* test cases follow. The first line of each test case contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of students.
Then *n* lines follow. Each line contains two integer *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=5000) — the second *i*-th student comes to the end of the queue, and the second he leaves the queue if he still cannot get his tea.
It is guaranteed that for every condition *l**i*<=-<=1<=≤<=*l**i* holds.
The sum of *n* over all test cases doesn't exceed 1000.
Note that in hacks you have to set *t*<==<=1. | For each test case print *n* integers. *i*-th of them must be equal to the second when *i*-th student gets his tea, or 0 if he leaves without tea. | [
"2\n2\n1 3\n1 4\n3\n1 5\n1 1\n2 3\n"
] | [
"1 2 \n1 0 2 \n"
] | The example contains 2 tests:
1. During 1-st second, students 1 and 2 come to the queue, and student 1 gets his tea. Student 2 gets his tea during 2-nd second. 1. During 1-st second, students 1 and 2 come to the queue, student 1 gets his tea, and student 2 leaves without tea. During 2-nd second, student 3 comes and gets his tea. | 0 | [
{
"input": "2\n2\n1 3\n1 4\n3\n1 5\n1 1\n2 3",
"output": "1 2 \n1 0 2 "
},
{
"input": "19\n1\n1 1\n1\n1 2\n1\n1 1000\n1\n1 2000\n1\n2 2\n1\n2 3\n1\n2 1000\n1\n2 2000\n1\n1999 1999\n1\n1999 2000\n1\n2000 2000\n2\n1 1\n1 1\n2\n1 1\n1 2\n2\n1 2\n1 1\n2\n1 2000\n1 1\n2\n1 1\n1 2000\n2\n1 2000\n2 2\n2\n2... | 1,517,584,285 | 2,185 | Python 3 | OK | TESTS | 8 | 77 | 5,734,400 | #!/usr/bin/env python3
from sys import stdin, stdout
def rint():
return map(int, stdin.readline().split())
#lines = stdin.readlines()
t = int(input())
for _ in range(t):
n = int(input())
lb, rb = 0, 0
ct = 0
ans = []
for __ in range(n):
lc, rc = rint()
if lc >= ct:
ans.append(lc)
ct = lc+1
elif ct <= rc:
ans.append(ct)
ct += 1
else:
ans.append(0)
print(*ans)
| Title: Tea Queue
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently *n* students from city S moved to city P to attend a programming camp.
They moved there by train. In the evening, all students in the train decided that they want to drink some tea. Of course, no two people can use the same teapot simultaneously, so the students had to form a queue to get their tea.
*i*-th student comes to the end of the queue at the beginning of *l**i*-th second. If there are multiple students coming to the queue in the same moment, then the student with greater index comes after the student with lesser index. Students in the queue behave as follows: if there is nobody in the queue before the student, then he uses the teapot for exactly one second and leaves the queue with his tea; otherwise the student waits for the people before him to get their tea. If at the beginning of *r**i*-th second student *i* still cannot get his tea (there is someone before him in the queue), then he leaves the queue without getting any tea.
For each student determine the second he will use the teapot and get his tea (if he actually gets it).
Input Specification:
The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=1000).
Then *t* test cases follow. The first line of each test case contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of students.
Then *n* lines follow. Each line contains two integer *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=5000) — the second *i*-th student comes to the end of the queue, and the second he leaves the queue if he still cannot get his tea.
It is guaranteed that for every condition *l**i*<=-<=1<=≤<=*l**i* holds.
The sum of *n* over all test cases doesn't exceed 1000.
Note that in hacks you have to set *t*<==<=1.
Output Specification:
For each test case print *n* integers. *i*-th of them must be equal to the second when *i*-th student gets his tea, or 0 if he leaves without tea.
Demo Input:
['2\n2\n1 3\n1 4\n3\n1 5\n1 1\n2 3\n']
Demo Output:
['1 2 \n1 0 2 \n']
Note:
The example contains 2 tests:
1. During 1-st second, students 1 and 2 come to the queue, and student 1 gets his tea. Student 2 gets his tea during 2-nd second. 1. During 1-st second, students 1 and 2 come to the queue, student 1 gets his tea, and student 2 leaves without tea. During 2-nd second, student 3 comes and gets his tea. | ```python
#!/usr/bin/env python3
from sys import stdin, stdout
def rint():
return map(int, stdin.readline().split())
#lines = stdin.readlines()
t = int(input())
for _ in range(t):
n = int(input())
lb, rb = 0, 0
ct = 0
ans = []
for __ in range(n):
lc, rc = rint()
if lc >= ct:
ans.append(lc)
ct = lc+1
elif ct <= rc:
ans.append(ct)
ct += 1
else:
ans.append(0)
print(*ans)
``` | 3 | |
526 | A | King of Thieves | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | In this problem you will meet the simplified model of game King of Thieves.
In a new ZeptoLab game called "King of Thieves" your aim is to reach a chest with gold by controlling your character, avoiding traps and obstacles on your way.
An interesting feature of the game is that you can design your own levels that will be available to other players. Let's consider the following simple design of a level.
A dungeon consists of *n* segments located at a same vertical level, each segment is either a platform that character can stand on, or a pit with a trap that makes player lose if he falls into it. All segments have the same length, platforms on the scheme of the level are represented as '*' and pits are represented as '.'.
One of things that affects speedrun characteristics of the level is a possibility to perform a series of consecutive jumps of the same length. More formally, when the character is on the platform number *i*1, he can make a sequence of jumps through the platforms *i*1<=<<=*i*2<=<<=...<=<<=*i**k*, if *i*2<=-<=*i*1<==<=*i*3<=-<=*i*2<==<=...<==<=*i**k*<=-<=*i**k*<=-<=1. Of course, all segments *i*1,<=*i*2,<=... *i**k* should be exactly the platforms, not pits.
Let's call a level to be good if you can perform a sequence of four jumps of the same length or in the other words there must be a sequence *i*1,<=*i*2,<=...,<=*i*5, consisting of five platforms so that the intervals between consecutive platforms are of the same length. Given the scheme of the level, check if it is good. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of segments on the level.
Next line contains the scheme of the level represented as a string of *n* characters '*' and '.'. | If the level is good, print the word "yes" (without the quotes), otherwise print the word "no" (without the quotes). | [
"16\n.**.*..*.***.**.\n",
"11\n.*.*...*.*.\n"
] | [
"yes",
"no"
] | In the first sample test you may perform a sequence of jumps through platforms 2, 5, 8, 11, 14. | 500 | [
{
"input": "16\n.**.*..*.***.**.",
"output": "yes"
},
{
"input": "11\n.*.*...*.*.",
"output": "no"
},
{
"input": "53\n*.*.****.*.*......**....**.***.*.*.**.*.*.***...*..*.",
"output": "yes"
},
{
"input": "71\n**.**..*****.*.*.*.********.....*****.****.*..***...*.*.*.**.****.*... | 1,428,823,474 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 61 | 0 | # coding: utf-8
n = int(input())
s = input()
for i in range(n//3):
for j in range(i+1,n//3):
dis = j-i
if i+2*dis < n and s[i+2*dis]=='*' and\
i+3*dis < n and s[i+3*dis]=='*' and\
i+4*dis < n and s[i+4*dis]=='*':
print('yes')
exit()
print('no')
| Title: King of Thieves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you will meet the simplified model of game King of Thieves.
In a new ZeptoLab game called "King of Thieves" your aim is to reach a chest with gold by controlling your character, avoiding traps and obstacles on your way.
An interesting feature of the game is that you can design your own levels that will be available to other players. Let's consider the following simple design of a level.
A dungeon consists of *n* segments located at a same vertical level, each segment is either a platform that character can stand on, or a pit with a trap that makes player lose if he falls into it. All segments have the same length, platforms on the scheme of the level are represented as '*' and pits are represented as '.'.
One of things that affects speedrun characteristics of the level is a possibility to perform a series of consecutive jumps of the same length. More formally, when the character is on the platform number *i*1, he can make a sequence of jumps through the platforms *i*1<=<<=*i*2<=<<=...<=<<=*i**k*, if *i*2<=-<=*i*1<==<=*i*3<=-<=*i*2<==<=...<==<=*i**k*<=-<=*i**k*<=-<=1. Of course, all segments *i*1,<=*i*2,<=... *i**k* should be exactly the platforms, not pits.
Let's call a level to be good if you can perform a sequence of four jumps of the same length or in the other words there must be a sequence *i*1,<=*i*2,<=...,<=*i*5, consisting of five platforms so that the intervals between consecutive platforms are of the same length. Given the scheme of the level, check if it is good.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of segments on the level.
Next line contains the scheme of the level represented as a string of *n* characters '*' and '.'.
Output Specification:
If the level is good, print the word "yes" (without the quotes), otherwise print the word "no" (without the quotes).
Demo Input:
['16\n.**.*..*.***.**.\n', '11\n.*.*...*.*.\n']
Demo Output:
['yes', 'no']
Note:
In the first sample test you may perform a sequence of jumps through platforms 2, 5, 8, 11, 14. | ```python
# coding: utf-8
n = int(input())
s = input()
for i in range(n//3):
for j in range(i+1,n//3):
dis = j-i
if i+2*dis < n and s[i+2*dis]=='*' and\
i+3*dis < n and s[i+3*dis]=='*' and\
i+4*dis < n and s[i+4*dis]=='*':
print('yes')
exit()
print('no')
``` | 0 | |
242 | B | Big Segment | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | null | null | A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide. | Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input. | [
"3\n1 1\n2 2\n3 3\n",
"6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n"
] | [
"-1\n",
"3\n"
] | none | 1,000 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "-1"
},
{
"input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10",
"output": "3"
},
{
"input": "4\n1 5\n2 2\n2 4\n2 5",
"output": "1"
},
{
"input": "5\n3 3\n1 3\n2 2\n2 3\n1 2",
"output": "2"
},
{
"input": "7\n7 7\n8 8\n3 7\n1 6\n1 ... | 1,562,125,416 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 30 | 872 | 5,427,200 | n = int(input())
x = []
y = []
for i in range(n):
a, b = map(int, input().split())
x.append(a)
y.append(b)
min = 10**5
max = 0
for i in range(n):
if x[i] < min:
min = x[i]
for j in range(n):
if y[j] > max:
max = y[j]
tf = False
for i in range(n):
if x[i] == min and y[i] == max:
tf = True
print(i + 1)
if tf == False:
print('-1')
| Title: Big Segment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
Output Specification:
Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n']
Demo Output:
['-1\n', '3\n']
Note:
none | ```python
n = int(input())
x = []
y = []
for i in range(n):
a, b = map(int, input().split())
x.append(a)
y.append(b)
min = 10**5
max = 0
for i in range(n):
if x[i] < min:
min = x[i]
for j in range(n):
if y[j] > max:
max = y[j]
tf = False
for i in range(n):
if x[i] == min and y[i] == max:
tf = True
print(i + 1)
if tf == False:
print('-1')
``` | 0 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
... | 1,665,547,491 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | rolls = input
ywakko = max(rolls[0],rolls[2])
if ywakko = 1:
print("1/1")
if ywakko = 2:
print("5/6")
if ywakko = 3:
print("2/3")
if ywakko = 4:
print("1/2")
if ywakko = 5:
print("1/3")
if ywakko = 6:
print("1/6") | Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
rolls = input
ywakko = max(rolls[0],rolls[2])
if ywakko = 1:
print("1/1")
if ywakko = 2:
print("5/6")
if ywakko = 3:
print("2/3")
if ywakko = 4:
print("1/2")
if ywakko = 5:
print("1/3")
if ywakko = 6:
print("1/6")
``` | -1 |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk... | 1,616,389,991 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 124 | 0 | n=input()
f=input()
n=list(n)
f=list(f)
c=0
i=0
j=len(f)-1
for k in range(len(n)):
if n[i]==f[j]:
i+=1
j-=1
c+=1
if c==len(n):
print("YES")
else:
print("NO") | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
n=input()
f=input()
n=list(n)
f=list(f)
c=0
i=0
j=len(f)-1
for k in range(len(n)):
if n[i]==f[j]:
i+=1
j-=1
c+=1
if c==len(n):
print("YES")
else:
print("NO")
``` | 3.969 |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,689,284,088 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 33 | 139 | 22,323,200 | import sys
input=lambda:sys.stdin.readline().strip()
mapin=lambda:map(int,input().split())
n=int(input())
a=list(mapin())
b=list(mapin())
c=list(mapin())
x=0
for i in a:x^=i
for i in b:x^=i
print(x)
x=0
for i in b:x^=i
for i in c:x^=i
print(x) | Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
import sys
input=lambda:sys.stdin.readline().strip()
mapin=lambda:map(int,input().split())
n=int(input())
a=list(mapin())
b=list(mapin())
c=list(mapin())
x=0
for i in a:x^=i
for i in b:x^=i
print(x)
x=0
for i in b:x^=i
for i in c:x^=i
print(x)
``` | 3 | |
344 | A | Magnets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | On the single line of the output print the number of groups of magnets. | [
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] | [
"3\n",
"2\n"
] | The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | 500 | [
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",... | 1,691,577,009 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 154 | 3,072,000 | t=int(input())
for i in range(t):
count=1
old=int(input())
for i in range(1,n):
s=int(input())
if old!=s:
old=s
count+=1
print(count) | Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | ```python
t=int(input())
for i in range(t):
count=1
old=int(input())
for i in range(1,n):
s=int(input())
if old!=s:
old=s
count+=1
print(count)
``` | -1 | |
300 | A | Array | PROGRAMMING | 1,100 | [
"brute force",
"constructive algorithms",
"implementation"
] | null | null | Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array. | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements. | In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them. | [
"3\n-1 2 0\n",
"4\n-1 -2 -3 0\n"
] | [
"1 -1\n1 2\n1 0\n",
"1 -1\n2 -3 -2\n1 0\n"
] | none | 500 | [
{
"input": "3\n-1 2 0",
"output": "1 -1\n1 2\n1 0"
},
{
"input": "4\n-1 -2 -3 0",
"output": "1 -1\n2 -3 -2\n1 0"
},
{
"input": "5\n-1 -2 1 2 0",
"output": "1 -1\n2 1 2\n2 0 -2"
},
{
"input": "100\n-64 -51 -75 -98 74 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -1... | 1,653,166,063 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 60 | 0 | n = int(input())
impar = 0
showImpar = ''
par = 0
showPar = ''
tercer = 0
showTercer = ''
a = list(map(int, input().split()))
for elem in a:
if elem < 0 and impar == 0:
impar = 1
showImpar = '1 '+str(elem)
elif elem > 0 and par != 2:
showPar = '1 '+str(elem)
par = 2
elif elem < 0 and impar == 1 and par < 2:
if par == 0:
showPar = str(elem)
par = 1
else:
showPar = '2 ' + showPar + str(elem)
par = 2
else:
tercer += 1
showTercer += ' '+str(elem)
print(showImpar)
print(showPar)
print(str(tercer)+showTercer) | Title: Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
Output Specification:
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Demo Input:
['3\n-1 2 0\n', '4\n-1 -2 -3 0\n']
Demo Output:
['1 -1\n1 2\n1 0\n', '1 -1\n2 -3 -2\n1 0\n']
Note:
none | ```python
n = int(input())
impar = 0
showImpar = ''
par = 0
showPar = ''
tercer = 0
showTercer = ''
a = list(map(int, input().split()))
for elem in a:
if elem < 0 and impar == 0:
impar = 1
showImpar = '1 '+str(elem)
elif elem > 0 and par != 2:
showPar = '1 '+str(elem)
par = 2
elif elem < 0 and impar == 1 and par < 2:
if par == 0:
showPar = str(elem)
par = 1
else:
showPar = '2 ' + showPar + str(elem)
par = 2
else:
tercer += 1
showTercer += ' '+str(elem)
print(showImpar)
print(showPar)
print(str(tercer)+showTercer)
``` | 0 | |
651 | A | Joysticks | PROGRAMMING | 1,100 | [
"dp",
"greedy",
"implementation",
"math"
] | null | null | Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent. | The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively. | Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged. | [
"3 5\n",
"4 4\n"
] | [
"6\n",
"5\n"
] | In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped. | 500 | [
{
"input": "3 5",
"output": "6"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 100",
"output": "197"
},
{
"input": "1 100",
"output": "98"
},
{
"input": "100 1",
"output": "98"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "1 1",
... | 1,567,057,024 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 140 | 0 | a,b=[int(x) for x in input().split()]
count=0
while a*b >1:
a,b=min(a,b),max(a,b)
a+=1
b-=2
count+=1
print(count)
| Title: Joysticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
Input Specification:
The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.
Output Specification:
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
Demo Input:
['3 5\n', '4 4\n']
Demo Output:
['6\n', '5\n']
Note:
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped. | ```python
a,b=[int(x) for x in input().split()]
count=0
while a*b >1:
a,b=min(a,b),max(a,b)
a+=1
b-=2
count+=1
print(count)
``` | 3 | |
48 | A | Rock-paper-scissors | PROGRAMMING | 900 | [
"implementation",
"schedules"
] | A. Rock-paper-scissors | 2 | 256 | Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown. | The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture. | Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?". | [
"rock\nrock\nrock\n",
"paper\nrock\nrock\n",
"scissors\nrock\nrock\n",
"scissors\npaper\nrock\n"
] | [
"?\n",
"F\n",
"?\n",
"?\n"
] | none | 0 | [
{
"input": "rock\nrock\nrock",
"output": "?"
},
{
"input": "paper\nrock\nrock",
"output": "F"
},
{
"input": "scissors\nrock\nrock",
"output": "?"
},
{
"input": "scissors\npaper\nrock",
"output": "?"
},
{
"input": "paper\npaper\nrock",
"output": "?"
},
{
... | 1,691,083,455 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | def rpc(f,m,s):
if f == "cisors" and m == s == "paper":
return f
elif s == "cisors" and m == f == "paper":
return s
elif m == "cisors" and f == s == "paper":
return m
elif f=="rock" and m==s=="cisors":
return f
elif s=="rock" and m==f=="cisors":
return s
elif m=="rock" and f==s=="cisors":
return m
elif f=="paper" and m==s=="rock":
return f
elif f=="paper" and m==s=="rock":
return s
elif f=="paper" and m==s=="rock":
return m
return "?" | Title: Rock-paper-scissors
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Uncle Fyodor, Matroskin the Cat and Sharic the Dog live their simple but happy lives in Prostokvashino. Sometimes they receive parcels from Uncle Fyodor’s parents and sometimes from anonymous benefactors, in which case it is hard to determine to which one of them the package has been sent. A photographic rifle is obviously for Sharic who loves hunting and fish is for Matroskin, but for whom was a new video game console meant? Every one of the three friends claimed that the present is for him and nearly quarreled. Uncle Fyodor had an idea how to solve the problem justly: they should suppose that the console was sent to all three of them and play it in turns. Everybody got relieved but then yet another burning problem popped up — who will play first? This time Matroskin came up with a brilliant solution, suggesting the most fair way to find it out: play rock-paper-scissors together. The rules of the game are very simple. On the count of three every player shows a combination with his hand (or paw). The combination corresponds to one of three things: a rock, scissors or paper. Some of the gestures win over some other ones according to well-known rules: the rock breaks the scissors, the scissors cut the paper, and the paper gets wrapped over the stone. Usually there are two players. Yet there are three friends, that’s why they decided to choose the winner like that: If someone shows the gesture that wins over the other two players, then that player wins. Otherwise, another game round is required. Write a program that will determine the winner by the gestures they have shown.
Input Specification:
The first input line contains the name of the gesture that Uncle Fyodor showed, the second line shows which gesture Matroskin showed and the third line shows Sharic’s gesture.
Output Specification:
Print "F" (without quotes) if Uncle Fyodor wins. Print "M" if Matroskin wins and "S" if Sharic wins. If it is impossible to find the winner, print "?".
Demo Input:
['rock\nrock\nrock\n', 'paper\nrock\nrock\n', 'scissors\nrock\nrock\n', 'scissors\npaper\nrock\n']
Demo Output:
['?\n', 'F\n', '?\n', '?\n']
Note:
none | ```python
def rpc(f,m,s):
if f == "cisors" and m == s == "paper":
return f
elif s == "cisors" and m == f == "paper":
return s
elif m == "cisors" and f == s == "paper":
return m
elif f=="rock" and m==s=="cisors":
return f
elif s=="rock" and m==f=="cisors":
return s
elif m=="rock" and f==s=="cisors":
return m
elif f=="paper" and m==s=="rock":
return f
elif f=="paper" and m==s=="rock":
return s
elif f=="paper" and m==s=="rock":
return m
return "?"
``` | 0 |
285 | B | Find Marble | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Petya and Vasya are playing a game. Petya's got *n* non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to *n* from left to right. Note that the positions are indexed but the glasses are not.
First Petya puts a marble under the glass in position *s*. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position *p*1, the glass from the second position to position *p*2 and so on. That is, a glass goes from position *i* to position *p**i*. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.
After all shuffling operations Petya shows Vasya that the ball has moved to position *t*. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position *s* to position *t*. | The first line contains three integers: *n*,<=*s*,<=*t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*s*,<=*t*<=≤<=*n*) — the number of glasses, the ball's initial and final position. The second line contains *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the shuffling operation parameters. It is guaranteed that all *p**i*'s are distinct.
Note that *s* can equal *t*. | If the marble can move from position *s* to position *t*, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position *t*. If it is impossible, print number -1. | [
"4 2 1\n2 3 4 1\n",
"4 3 3\n4 1 3 2\n",
"4 3 4\n1 2 3 4\n",
"3 1 3\n2 1 3\n"
] | [
"3\n",
"0\n",
"-1\n",
"-1\n"
] | none | 1,000 | [
{
"input": "4 2 1\n2 3 4 1",
"output": "3"
},
{
"input": "4 3 3\n4 1 3 2",
"output": "0"
},
{
"input": "4 3 4\n1 2 3 4",
"output": "-1"
},
{
"input": "3 1 3\n2 1 3",
"output": "-1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "10 6 7\n10 7 8 1... | 1,539,454,719 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 248 | 102,400 | from sys import stdin, stdout
from collections import OrderedDict
def solve(n, s, t, positions):
count = 0
while count < n:
if positions[s-1] == t:
return count +1
if positions[s-1] == s:
return -1
s = positions[s-1]
count += 1
if count == n:
count = -1
return count
if __name__ == "__main__":
n, s, t = list(map(int, stdin.readline().rstrip().split()))
positions = list(map(int, stdin.readline().rstrip().split()))
stdout.write(str(solve(n, s, t, positions)))
| Title: Find Marble
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya are playing a game. Petya's got *n* non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to *n* from left to right. Note that the positions are indexed but the glasses are not.
First Petya puts a marble under the glass in position *s*. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position *p*1, the glass from the second position to position *p*2 and so on. That is, a glass goes from position *i* to position *p**i*. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.
After all shuffling operations Petya shows Vasya that the ball has moved to position *t*. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position *s* to position *t*.
Input Specification:
The first line contains three integers: *n*,<=*s*,<=*t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*s*,<=*t*<=≤<=*n*) — the number of glasses, the ball's initial and final position. The second line contains *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the shuffling operation parameters. It is guaranteed that all *p**i*'s are distinct.
Note that *s* can equal *t*.
Output Specification:
If the marble can move from position *s* to position *t*, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position *t*. If it is impossible, print number -1.
Demo Input:
['4 2 1\n2 3 4 1\n', '4 3 3\n4 1 3 2\n', '4 3 4\n1 2 3 4\n', '3 1 3\n2 1 3\n']
Demo Output:
['3\n', '0\n', '-1\n', '-1\n']
Note:
none | ```python
from sys import stdin, stdout
from collections import OrderedDict
def solve(n, s, t, positions):
count = 0
while count < n:
if positions[s-1] == t:
return count +1
if positions[s-1] == s:
return -1
s = positions[s-1]
count += 1
if count == n:
count = -1
return count
if __name__ == "__main__":
n, s, t = list(map(int, stdin.readline().rstrip().split()))
positions = list(map(int, stdin.readline().rstrip().split()))
stdout.write(str(solve(n, s, t, positions)))
``` | 0 | |
659 | C | Tanya and Toys | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles.
Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.
Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this. | The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has. | In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*.
In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose.
If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order. | [
"3 7\n1 3 4\n",
"4 14\n4 6 12 8\n"
] | [
"2\n2 5 \n",
"4\n7 2 3 1\n"
] | In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys. | 1,000 | [
{
"input": "3 7\n1 3 4",
"output": "2\n2 5 "
},
{
"input": "4 14\n4 6 12 8",
"output": "4\n1 2 3 5 "
},
{
"input": "5 6\n97746 64770 31551 96547 65684",
"output": "3\n1 2 3 "
},
{
"input": "10 10\n94125 56116 29758 94024 29289 31663 99794 35076 25328 58656",
"output": "4\... | 1,578,841,526 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 1,000 | 1,536,000 | X, Toys, Answer, i = list(map(int, input().split())), list(map(int, input().split())), [], 1
while i <= X[1]:
if i not in Toys: Answer.append(i);X[1] -= i
i += 1
print(len(Answer))
print(*Answer)
# Come together for getting better !!!!
| Title: Tanya and Toys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. A toy from the new collection of the *i*-th type costs *i* bourles.
Tania has managed to collect *n* different types of toys *a*1,<=*a*2,<=...,<=*a**n* from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than *m* bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.
Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.
Input Specification:
The first line contains two integers *n* (1<=≤<=*n*<=≤<=100<=000) and *m* (1<=≤<=*m*<=≤<=109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.
The next line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the types of toys that Tanya already has.
Output Specification:
In the first line print a single integer *k* — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed *m*.
In the second line print *k* distinct space-separated integers *t*1,<=*t*2,<=...,<=*t**k* (1<=≤<=*t**i*<=≤<=109) — the types of toys that Tanya should choose.
If there are multiple answers, you may print any of them. Values of *t**i* can be printed in any order.
Demo Input:
['3 7\n1 3 4\n', '4 14\n4 6 12 8\n']
Demo Output:
['2\n2 5 \n', '4\n7 2 3 1\n']
Note:
In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys. | ```python
X, Toys, Answer, i = list(map(int, input().split())), list(map(int, input().split())), [], 1
while i <= X[1]:
if i not in Toys: Answer.append(i);X[1] -= i
i += 1
print(len(Answer))
print(*Answer)
# Come together for getting better !!!!
``` | 0 | |
339 | A | Helpful Maths | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings",
"strings"
] | null | null | Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum. | The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long. | Print the new sum that Xenia can count. | [
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] | [
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] | none | 500 | [
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": ... | 1,696,251,064 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | s = input() # 读入加数序列
# 统计数字1、2、3的出现次数
count = [0, 0, 0]
for c in s:
if c == '1':
count[0] += 1
elif c == '2':
count[1] += 1
else:
count[2] += 1
# 按照非降序输出加数序列
result = ''
for i in range(3):
result += str(i+1) + '+'
result += str(i+1) + '+' * (count[i] - 1)
# 去掉最后一个加号并输出结果
print(result[:-1]) | Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none | ```python
s = input() # 读入加数序列
# 统计数字1、2、3的出现次数
count = [0, 0, 0]
for c in s:
if c == '1':
count[0] += 1
elif c == '2':
count[1] += 1
else:
count[2] += 1
# 按照非降序输出加数序列
result = ''
for i in range(3):
result += str(i+1) + '+'
result += str(i+1) + '+' * (count[i] - 1)
# 去掉最后一个加号并输出结果
print(result[:-1])
``` | 0 | |
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"
},
... | 1,685,478,500 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 31 | 0 | n,m=map(int,input().split())
f=n*m
if f%2==0:
print('Malvika')
elif f%2!=0:
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())
f=n*m
if f%2==0:
print('Malvika')
elif f%2!=0:
print('Akshat')
``` | 0 | |
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... | 1,542,996,933 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 218 | 0 | s1=input()
s2=s1.upper()
a=0
for i in range(len(s1)):
if s1[i]==s2[i]:
a+=1
if 2*a>len(s1):
print(s2)
else:
print(s1.lower()) | 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
s1=input()
s2=s1.upper()
a=0
for i in range(len(s1)):
if s1[i]==s2[i]:
a+=1
if 2*a>len(s1):
print(s2)
else:
print(s1.lower())
``` | 3.9455 |
831 | B | Keyboard Layouts | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters. | The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string *s* consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of *s* does not exceed 1000. | Print the text if the same keys were pressed in the second layout. | [
"qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017\n",
"mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7\n"
] | [
"HelloVKCup2017\n",
"7uduGUDUUDUgudu7\n"
] | none | 750 | [
{
"input": "qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017",
"output": "HelloVKCup2017"
},
{
"input": "mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7",
"output": "7uduGUDUUDUgudu7"
},
{
"input": "ayvguplhjsoiencbkxdrfwmqtz\nkhzvtbspcndier... | 1,673,459,983 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 19 | 77 | 614,400 | from sys import stdin; inp = stdin.readline
from math import dist, ceil, floor, sqrt, log
from collections import defaultdict, Counter, deque
def IA(sep=' '): return list(map(int, inp().split(sep)))
def FA(): return list(map(float, inp().split()))
def SA(): return inp().split()
def I(): return int(inp())
def F(): return float(inp())
def S(): return input()
def O(l:list): return ' '.join(map(str, l))
def main():
a = S()
b = S()
s = S()
caps = [c.isupper() for c in s]
s = s.lower()
d = {}
for i in range(26):
d[a[i]] = b[i]
res = ''
for i in range(len(s)):
char = s[i]
if char in d:
char = d[s[i]]
if caps[i]:
char = char.upper()
res += char
return res
if __name__ == '__main__':
print(main()) | Title: Keyboard Layouts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
Input Specification:
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string *s* consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of *s* does not exceed 1000.
Output Specification:
Print the text if the same keys were pressed in the second layout.
Demo Input:
['qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017\n', 'mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7\n']
Demo Output:
['HelloVKCup2017\n', '7uduGUDUUDUgudu7\n']
Note:
none | ```python
from sys import stdin; inp = stdin.readline
from math import dist, ceil, floor, sqrt, log
from collections import defaultdict, Counter, deque
def IA(sep=' '): return list(map(int, inp().split(sep)))
def FA(): return list(map(float, inp().split()))
def SA(): return inp().split()
def I(): return int(inp())
def F(): return float(inp())
def S(): return input()
def O(l:list): return ' '.join(map(str, l))
def main():
a = S()
b = S()
s = S()
caps = [c.isupper() for c in s]
s = s.lower()
d = {}
for i in range(26):
d[a[i]] = b[i]
res = ''
for i in range(len(s)):
char = s[i]
if char in d:
char = d[s[i]]
if caps[i]:
char = char.upper()
res += char
return res
if __name__ == '__main__':
print(main())
``` | 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... | 1,655,208,062 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 17,100,800 | n = int(input())
l = [int(i) for i in input().split()]
m = int(input())
t = [int(i) for i in input().split()]
a = b = 0
for i in range(m):
a += l.index(t[i]) + 1
b += n - l.index(t[i])
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())
l = [int(i) for i in input().split()]
m = int(input())
t = [int(i) for i in input().split()]
a = b = 0
for i in range(m):
a += l.index(t[i]) + 1
b += n - l.index(t[i])
print(a, b)
``` | 0 | |
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 ... | 1,693,760,468 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | n = 7
lis = [1, 2, 3, 4 ,5 ,6, 7 ]
sum_Sereja = []
sum_Dima = []
for i in range(len(lis)):
if i % 2 == 0:
sum_Sereja.append(max(lis))
lis.remove(max(lis))
else:
sum_Dima.append(max(lis))
lis.remove(max(lis))
print(sum(sum_Sereja), sum(sum_Dima)) | 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 = 7
lis = [1, 2, 3, 4 ,5 ,6, 7 ]
sum_Sereja = []
sum_Dima = []
for i in range(len(lis)):
if i % 2 == 0:
sum_Sereja.append(max(lis))
lis.remove(max(lis))
else:
sum_Dima.append(max(lis))
lis.remove(max(lis))
print(sum(sum_Sereja), sum(sum_Dima))
``` | 0 | |
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 58697... | 1,698,834,936 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | a=input().split()
n=0
for i in a:
if i==a[0]:n+=1
if i==a[1]:n+=1
if i==a[2]:n+=1
if i==a[3]:n+=1
print(n) | Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
a=input().split()
n=0
for i in a:
if i==a[0]:n+=1
if i==a[1]:n+=1
if i==a[2]:n+=1
if i==a[3]:n+=1
print(n)
``` | 0 | |
758 | A | Holiday Of Equality | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. | In the only line print the integer *S* — the minimum number of burles which are had to spend. | [
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] | [
"10",
"1",
"4",
"0"
] | In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | 500 | [
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853... | 1,671,722,853 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 46 | 0 |
def solve():
n = int(input())
numbers = input().split()
# Convert the list of strings to a list of integers
numbers = [int(x) for x in numbers]
biggest = max(numbers)
equality = n * biggest
print(equality - sum(numbers))
def main():
solve()
main() | Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | ```python
def solve():
n = int(input())
numbers = input().split()
# Convert the list of strings to a list of integers
numbers = [int(x) for x in numbers]
biggest = max(numbers)
equality = n * biggest
print(equality - sum(numbers))
def main():
solve()
main()
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Dima came to the horse land. There are *n* horses living in the land. Each horse in the horse land has several enemies (enmity is a symmetric relationship). The horse land isn't very hostile, so the number of enemies of each horse is at most 3.
Right now the horse land is going through an election campaign. So the horses trusted Dima to split them into two parts. At that the horses want the following condition to hold: a horse shouldn't have more than one enemy in its party.
Help Dima split the horses into parties. Note that one of the parties can turn out to be empty. | The first line contains two integers *n*,<=*m* — the number of horses in the horse land and the number of enemy pairs.
Next *m* lines define the enemy pairs. The *i*-th line contains integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*), which mean that horse *a**i* is the enemy of horse *b**i*.
Consider the horses indexed in some way from 1 to *n*. It is guaranteed that each horse has at most three enemies. No pair of enemies occurs more than once in the input. | Print a line, consisting of *n* characters: the *i*-th character of the line must equal "0", if the horse number *i* needs to go to the first party, otherwise this character should equal "1".
If there isn't a way to divide the horses as required, print -1. | [
"3 3\n1 2\n3 2\n3 1\n",
"2 1\n2 1\n",
"10 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n"
] | [
"100\n",
"00\n",
"0110000000\n"
] | none | 0 | [] | 1,690,589,722 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | print("_RANDOM_GUESS_1690589721.9361548")# 1690589721.9361763 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima came to the horse land. There are *n* horses living in the land. Each horse in the horse land has several enemies (enmity is a symmetric relationship). The horse land isn't very hostile, so the number of enemies of each horse is at most 3.
Right now the horse land is going through an election campaign. So the horses trusted Dima to split them into two parts. At that the horses want the following condition to hold: a horse shouldn't have more than one enemy in its party.
Help Dima split the horses into parties. Note that one of the parties can turn out to be empty.
Input Specification:
The first line contains two integers *n*,<=*m* — the number of horses in the horse land and the number of enemy pairs.
Next *m* lines define the enemy pairs. The *i*-th line contains integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*), which mean that horse *a**i* is the enemy of horse *b**i*.
Consider the horses indexed in some way from 1 to *n*. It is guaranteed that each horse has at most three enemies. No pair of enemies occurs more than once in the input.
Output Specification:
Print a line, consisting of *n* characters: the *i*-th character of the line must equal "0", if the horse number *i* needs to go to the first party, otherwise this character should equal "1".
If there isn't a way to divide the horses as required, print -1.
Demo Input:
['3 3\n1 2\n3 2\n3 1\n', '2 1\n2 1\n', '10 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n']
Demo Output:
['100\n', '00\n', '0110000000\n']
Note:
none | ```python
print("_RANDOM_GUESS_1690589721.9361548")# 1690589721.9361763
``` | 0 | |
260 | A | Adding Digits | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times. | The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105). | In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them. | [
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] | [
"524848\n",
"121\n",
"-1\n"
] | none | 500 | [
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479"... | 1,593,973,860 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 93 | 6,963,200 | # https://codeforces.com/problemset/problem/260/A
"""
Two numbers a and b
You can perform an operation of lengthening by:
* adding one digit to the number to the right. The new number must be divisible by 5.
Determine whether it is possible to lengthen a n times. If yes print any lengthening.
If not print -1.
Try adding digits from 0 to 9. If none of those work then it is impossible.
After that just chuck in a load of zeros as those don't change divisibility.
"""
a, b, n = map(int, input().split())
flag = 0 # Initially assume impossible
for i in range(10):
if int(str(a) + str(i)) % b == 0:
flag = 1
break
if flag:
print(str(a) + str(i) + '0'*(n-1))
else:
print(-1)
| Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none | ```python
# https://codeforces.com/problemset/problem/260/A
"""
Two numbers a and b
You can perform an operation of lengthening by:
* adding one digit to the number to the right. The new number must be divisible by 5.
Determine whether it is possible to lengthen a n times. If yes print any lengthening.
If not print -1.
Try adding digits from 0 to 9. If none of those work then it is impossible.
After that just chuck in a load of zeros as those don't change divisibility.
"""
a, b, n = map(int, input().split())
flag = 0 # Initially assume impossible
for i in range(10):
if int(str(a) + str(i)) % b == 0:
flag = 1
break
if flag:
print(str(a) + str(i) + '0'*(n-1))
else:
print(-1)
``` | 3 | |
50 | B | Choosing Symbol Pairs | PROGRAMMING | 1,500 | [
"strings"
] | B. Choosing Symbol Pairs | 2 | 256 | There is a given string *S* consisting of *N* symbols. Your task is to find the number of ordered pairs of integers *i* and *j* such that
1. 1<=≤<=*i*,<=*j*<=≤<=*N*
2. *S*[*i*]<==<=*S*[*j*], that is the *i*-th symbol of string *S* is equal to the *j*-th. | The single input line contains *S*, consisting of lowercase Latin letters and digits. It is guaranteed that string *S* in not empty and its length does not exceed 105. | Print a single number which represents the number of pairs *i* and *j* with the needed property. Pairs (*x*,<=*y*) and (*y*,<=*x*) should be considered different, i.e. the ordered pairs count. | [
"great10\n",
"aaaaaaaaaa\n"
] | [
"7\n",
"100\n"
] | none | 1,000 | [
{
"input": "great10",
"output": "7"
},
{
"input": "aaaaaaaaaa",
"output": "100"
},
{
"input": "great10",
"output": "7"
},
{
"input": "aaaaaaaaaa",
"output": "100"
},
{
"input": "aabb",
"output": "8"
},
{
"input": "w",
"output": "1"
},
{
"in... | 1,646,210,108 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | s=input()
lens = len(s)
cnt =0
for i in range(lens) :
for j in range (lens) :
if s[i]=s[j]:
cnt =1
print(cnt)
| Title: Choosing Symbol Pairs
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given string *S* consisting of *N* symbols. Your task is to find the number of ordered pairs of integers *i* and *j* such that
1. 1<=≤<=*i*,<=*j*<=≤<=*N*
2. *S*[*i*]<==<=*S*[*j*], that is the *i*-th symbol of string *S* is equal to the *j*-th.
Input Specification:
The single input line contains *S*, consisting of lowercase Latin letters and digits. It is guaranteed that string *S* in not empty and its length does not exceed 105.
Output Specification:
Print a single number which represents the number of pairs *i* and *j* with the needed property. Pairs (*x*,<=*y*) and (*y*,<=*x*) should be considered different, i.e. the ordered pairs count.
Demo Input:
['great10\n', 'aaaaaaaaaa\n']
Demo Output:
['7\n', '100\n']
Note:
none | ```python
s=input()
lens = len(s)
cnt =0
for i in range(lens) :
for j in range (lens) :
if s[i]=s[j]:
cnt =1
print(cnt)
``` | -1 |
584 | A | Olesya and Rodion | PROGRAMMING | 1,000 | [
"math"
] | null | null | Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. | The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. | Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. | [
"3 2\n"
] | [
"712"
] | none | 500 | [
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input"... | 1,692,901,045 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 46 | 0 | n,t = map(int,input().split())
if n == 1 and t > 10:
print(-1)
sys.exit()
print(t*10**(n-1))
| Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none | ```python
n,t = map(int,input().split())
if n == 1 and t > 10:
print(-1)
sys.exit()
print(t*10**(n-1))
``` | 0 | |
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... | 1,645,657,705 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | s= input()
counter = 0
for i in s:
if i>= 'A' and i<='Z':
counter =counter + 1
if counter > (1/2):
print(s.upper())
else:
print(s.lower()) | 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
s= input()
counter = 0
for i in s:
if i>= 'A' and i<='Z':
counter =counter + 1
if counter > (1/2):
print(s.upper())
else:
print(s.lower())
``` | 0 |
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... | 1,660,238,602 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | def dist(x1, y1, x2, y2):
return abs(x1 - x2) + abs(y1 - y2)
a, b, s = map(int, input().split())
if dist(0, 0, a, b) <= s:
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
def dist(x1, y1, x2, y2):
return abs(x1 - x2) + abs(y1 - y2)
a, b, s = map(int, input().split())
if dist(0, 0, a, b) <= s:
print('Yes')
else:
print('No')
``` | 0 | |
805 | A | Fake NP | PROGRAMMING | 1,000 | [
"greedy",
"math"
] | null | null | Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.
You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.
Solve the problem to show that it's not a NP problem. | The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109). | Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them. | [
"19 29\n",
"3 6\n"
] | [
"2\n",
"3\n"
] | Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html)
The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}. | 500 | [
{
"input": "19 29",
"output": "2"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "39 91",
"output": "2"
},
{
"input": "76 134",
"output": "2"
},
{
"input": "93 95",
"output": "2"
},
{
"input": "17 35",
"output": "2"
},
{
"input": "94 95",... | 1,641,376,187 | 2,147,483,647 | Python 3 | OK | TESTS | 93 | 46 | 0 | l, r= map(int,input().split())
if l!=r: print(2)
else: print(l)
# Wed Jan 05 2022 09:49:47 GMT+0000 (Coordinated Universal Time)
| Title: Fake NP
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.
You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.
Solve the problem to show that it's not a NP problem.
Input Specification:
The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109).
Output Specification:
Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them.
Demo Input:
['19 29\n', '3 6\n']
Demo Output:
['2\n', '3\n']
Note:
Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html)
The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}. | ```python
l, r= map(int,input().split())
if l!=r: print(2)
else: print(l)
# Wed Jan 05 2022 09:49:47 GMT+0000 (Coordinated Universal Time)
``` | 3 | |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk... | 1,650,555,026 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 92 | 0 | a = input()
b = input()
def reverse(s):
return s[::-1]
if b == reverse(a):
print('YES')
else:
print('NO')
| Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
a = input()
b = input()
def reverse(s):
return s[::-1]
if b == reverse(a):
print('YES')
else:
print('NO')
``` | 3.977 |
33 | A | What is for dinner? | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | A. What is for dinner? | 2 | 256 | In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one. | The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth. | In the first line output the maximum amount of crucians that Valerie can consume for dinner. | [
"4 3 18\n2 3\n1 2\n3 6\n2 3\n",
"2 2 13\n1 13\n2 12\n"
] | [
"11\n",
"13\n"
] | none | 500 | [
{
"input": "4 3 18\n2 3\n1 2\n3 6\n2 3",
"output": "11"
},
{
"input": "2 2 13\n1 13\n2 12",
"output": "13"
},
{
"input": "5 4 8\n4 6\n4 5\n1 3\n2 0\n3 3",
"output": "8"
},
{
"input": "1 1 0\n1 3",
"output": "0"
},
{
"input": "7 1 30\n1 8\n1 15\n1 5\n1 17\n1 9\n1 1... | 1,643,048,168 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 124 | 0 | n, m, k = map(int, input().split())
d = [-1 for i in range(m)]
for i in range(n):
r, c = map(int, input().split())
r -= 1
d[r] = c if d[r] == -1 else min(d[r], c)
ans = 0
for i in d:
ans += i
print(min(ans, k)) | Title: What is for dinner?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output Specification:
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Demo Input:
['4 3 18\n2 3\n1 2\n3 6\n2 3\n', '2 2 13\n1 13\n2 12\n']
Demo Output:
['11\n', '13\n']
Note:
none | ```python
n, m, k = map(int, input().split())
d = [-1 for i in range(m)]
for i in range(n):
r, c = map(int, input().split())
r -= 1
d[r] = c if d[r] == -1 else min(d[r], c)
ans = 0
for i in d:
ans += i
print(min(ans, k))
``` | 3.969 |
411 | A | Password Check | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q. | The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_". | If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes). | [
"abacaba\n",
"X12345\n",
"CONTEST_is_STARTED!!11\n"
] | [
"Too weak\n",
"Too weak\n",
"Correct\n"
] | none | 0 | [
{
"input": "abacaba",
"output": "Too weak"
},
{
"input": "X12345",
"output": "Too weak"
},
{
"input": "CONTEST_is_STARTED!!11",
"output": "Correct"
},
{
"input": "1zA__",
"output": "Correct"
},
{
"input": "1zA_",
"output": "Too weak"
},
{
"input": "zA_... | 1,638,945,102 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 81 | 140 | 2,355,200 | pw = input()
strong = False
if len(pw) >= 5:
for i in pw:
if i.isupper() == True:
for y in pw:
if y.islower() == True:
for x in pw:
if x.isdigit() == True:
strong = True
if strong == True:
print("Correct")
else:
print("Too weak")
| Title: Password Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input Specification:
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output Specification:
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Demo Input:
['abacaba\n', 'X12345\n', 'CONTEST_is_STARTED!!11\n']
Demo Output:
['Too weak\n', 'Too weak\n', 'Correct\n']
Note:
none | ```python
pw = input()
strong = False
if len(pw) >= 5:
for i in pw:
if i.isupper() == True:
for y in pw:
if y.islower() == True:
for x in pw:
if x.isdigit() == True:
strong = True
if strong == True:
print("Correct")
else:
print("Too weak")
``` | 3 | |
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 58697... | 1,687,056,618 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | s = input().split()
print (len(s)-len(set(s)))
| Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
s = input().split()
print (len(s)-len(set(s)))
``` | 3 | |
628 | B | New Skateboard | PROGRAMMING | 1,300 | [
"dp"
] | null | null | Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. | The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9. | Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | [
"124\n",
"04\n",
"5810438174\n"
] | [
"4\n",
"3\n",
"9\n"
] | none | 0 | [
{
"input": "124",
"output": "4"
},
{
"input": "04",
"output": "3"
},
{
"input": "5810438174",
"output": "9"
},
{
"input": "1",
"output": "0"
},
{
"input": "039",
"output": "1"
},
{
"input": "97247",
"output": "6"
},
{
"input": "5810438174",... | 1,608,977,365 | 2,147,483,647 | PyPy 3 | OK | TESTS | 22 | 264 | 4,403,200 | import sys
import os
import math
import copy
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations,combinations,accumulate
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Mat2x2(n): return [List() for _ in range(n)]
def Lcm(x,y): return (x*y)//gcd(x,y)
def dtob(n): return bin(n).replace("0b","")
def btod(n): return int(n,2)
def watch(x): return print(x)
def common(l1, l2): return set(l1).intersection(l2)
def Most_frequent(list): return max(set(list), key = list.count)
def solution():
#for _ in range(Int()):
s=Str()
ans=0
for i in range(len(s)):
if(int(s[i])%4==0):
ans+=1
for i in range(len(s)-1):
if(int(s[i]+s[i+1])%4==0):
ans+=(i+1)
print(ans)
if __name__ == "__main__":
solution() | Title: New Skateboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string *s* consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string *s* is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input Specification:
The only line contains string *s* (1<=≤<=|*s*|<=≤<=3·105). The string *s* contains only digits from 0 to 9.
Output Specification:
Print integer *a* — the number of substrings of the string *s* that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Demo Input:
['124\n', '04\n', '5810438174\n']
Demo Output:
['4\n', '3\n', '9\n']
Note:
none | ```python
import sys
import os
import math
import copy
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations,combinations,accumulate
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Mat2x2(n): return [List() for _ in range(n)]
def Lcm(x,y): return (x*y)//gcd(x,y)
def dtob(n): return bin(n).replace("0b","")
def btod(n): return int(n,2)
def watch(x): return print(x)
def common(l1, l2): return set(l1).intersection(l2)
def Most_frequent(list): return max(set(list), key = list.count)
def solution():
#for _ in range(Int()):
s=Str()
ans=0
for i in range(len(s)):
if(int(s[i])%4==0):
ans+=1
for i in range(len(s)-1):
if(int(s[i]+s[i+1])%4==0):
ans+=(i+1)
print(ans)
if __name__ == "__main__":
solution()
``` | 3 | |
198 | A | About Bacteria | PROGRAMMING | 1,700 | [
"implementation",
"math"
] | null | null | Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into *k* bacteria. After that some abnormal effects create *b* more bacteria in the test tube. Thus, if at the beginning of some second the test tube had *x* bacteria, then at the end of the second it will have *kx*<=+<=*b* bacteria.
The experiment showed that after *n* seconds there were exactly *z* bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there *t* bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least *z* bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least *z* bacteria in the second experiment. | The first line contains four space-separated integers *k*, *b*, *n* and *t* (1<=≤<=*k*,<=*b*,<=*n*,<=*t*<=≤<=106) — the parameters of bacterial growth, the time Qwerty needed to grow *z* bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly. | Print a single number — the minimum number of seconds Qwerty needs to grow at least *z* bacteria in the tube. | [
"3 1 3 5\n",
"1 4 4 7\n",
"2 2 4 100\n"
] | [
"2",
"3",
"0"
] | none | 500 | [
{
"input": "3 1 3 5",
"output": "2"
},
{
"input": "1 4 4 7",
"output": "3"
},
{
"input": "2 2 4 100",
"output": "0"
},
{
"input": "1 2 3 100",
"output": "0"
},
{
"input": "10 10 10 123456",
"output": "6"
},
{
"input": "847 374 283 485756",
"output"... | 1,672,031,593 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | import math
k,b,n,t = map(int,input().split())
m = 0
if k==1:
z = 1+n*b
m = (z-x)/b
else:
m = n+ ((k-1+b)/((k-1)*t+b))*math.log(k)
if m<0:
print(0)
elif m%10>0:
print(int(m)+1)
else:
print(int(m)) | Title: About Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't difficult, but the salary is high.
At the beginning of the first experiment there is a single bacterium in the test tube. Every second each bacterium in the test tube divides itself into *k* bacteria. After that some abnormal effects create *b* more bacteria in the test tube. Thus, if at the beginning of some second the test tube had *x* bacteria, then at the end of the second it will have *kx*<=+<=*b* bacteria.
The experiment showed that after *n* seconds there were exactly *z* bacteria and the experiment ended at this point.
For the second experiment Qwerty is going to sterilize the test tube and put there *t* bacteria. He hasn't started the experiment yet but he already wonders, how many seconds he will need to grow at least *z* bacteria. The ranger thinks that the bacteria will divide by the same rule as in the first experiment.
Help Qwerty and find the minimum number of seconds needed to get a tube with at least *z* bacteria in the second experiment.
Input Specification:
The first line contains four space-separated integers *k*, *b*, *n* and *t* (1<=≤<=*k*,<=*b*,<=*n*,<=*t*<=≤<=106) — the parameters of bacterial growth, the time Qwerty needed to grow *z* bacteria in the first experiment and the initial number of bacteria in the second experiment, correspondingly.
Output Specification:
Print a single number — the minimum number of seconds Qwerty needs to grow at least *z* bacteria in the tube.
Demo Input:
['3 1 3 5\n', '1 4 4 7\n', '2 2 4 100\n']
Demo Output:
['2', '3', '0']
Note:
none | ```python
import math
k,b,n,t = map(int,input().split())
m = 0
if k==1:
z = 1+n*b
m = (z-x)/b
else:
m = n+ ((k-1+b)/((k-1)*t+b))*math.log(k)
if m<0:
print(0)
elif m%10>0:
print(int(m)+1)
else:
print(int(m))
``` | 0 | |
1,011 | A | Stages | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once. | The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once. | Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all. | [
"5 3\nxyabd\n",
"7 4\nproblem\n",
"2 2\nab\n",
"12 1\nabaabbaaabbb\n"
] | [
"29",
"34",
"-1",
"1"
] | In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | 500 | [
{
"input": "5 3\nxyabd",
"output": "29"
},
{
"input": "7 4\nproblem",
"output": "34"
},
{
"input": "2 2\nab",
"output": "-1"
},
{
"input": "12 1\nabaabbaaabbb",
"output": "1"
},
{
"input": "50 13\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": ... | 1,597,586,970 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 124 | 0 | n, k = map(int, input().split())
s = list(input())
s.sort()
e = 1
ans = (ord(s[0])-96)
for i in range(1, n):
if ord(s[i]) - ord(s[i-1]) > 1:
e += 1
ans += (ord(s[i])-96)
if e == k:
break
print(ans if e == k else -1) | Title: Stages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
Input Specification:
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
Output Specification:
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
Demo Input:
['5 3\nxyabd\n', '7 4\nproblem\n', '2 2\nab\n', '12 1\nabaabbaaabbb\n']
Demo Output:
['29', '34', '-1', '1']
Note:
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | ```python
n, k = map(int, input().split())
s = list(input())
s.sort()
e = 1
ans = (ord(s[0])-96)
for i in range(1, n):
if ord(s[i]) - ord(s[i-1]) > 1:
e += 1
ans += (ord(s[i])-96)
if e == k:
break
print(ans if e == k else -1)
``` | 0 | |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,635,941,511 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 53,043,200 | n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
def insertion(data):
for i in range(len(data)):
j = i - 1
key = data[i]
while data[j] > key and j >= 0:
data[j + 1] = data[j]
j -= 1
data[j + 1] = key
return data
a = insertion(a)
b = insertion(b)
c = insertion(c)
def razn(a, b):
n = len(a)
t = -1
index = 0
while index < n-1:
if a[index] != b[index]:
break
index += 1
return a[index]
print(razn(a, b), '\n', razn(b, c), sep = '')
| Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
def insertion(data):
for i in range(len(data)):
j = i - 1
key = data[i]
while data[j] > key and j >= 0:
data[j + 1] = data[j]
j -= 1
data[j + 1] = key
return data
a = insertion(a)
b = insertion(b)
c = insertion(c)
def razn(a, b):
n = len(a)
t = -1
index = 0
while index < n-1:
if a[index] != b[index]:
break
index += 1
return a[index]
print(razn(a, b), '\n', razn(b, c), sep = '')
``` | 0 | |
673 | B | Problems for Round | PROGRAMMING | 1,300 | [
"greedy",
"implementation"
] | null | null | There are *n* problems prepared for the next Codeforces round. They are arranged in ascending order by their difficulty, and no two problems have the same difficulty. Moreover, there are *m* pairs of similar problems. Authors want to split problems between two division according to the following rules:
- Problemset of each division should be non-empty. - Each problem should be used in exactly one division (yes, it is unusual requirement). - Each problem used in division 1 should be harder than any problem used in division 2. - If two problems are similar, they should be used in different divisions.
Your goal is count the number of ways to split problem between two divisions and satisfy all the rules. Two ways to split problems are considered to be different if there is at least one problem that belongs to division 1 in one of them and to division 2 in the other.
Note, that the relation of similarity is not transitive. That is, if problem *i* is similar to problem *j* and problem *j* is similar to problem *k*, it doesn't follow that *i* is similar to *k*. | The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100<=000, 0<=≤<=*m*<=≤<=100<=000) — the number of problems prepared for the round and the number of pairs of similar problems, respectively.
Each of the following *m* lines contains a pair of similar problems *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*,<=*u**i*<=≠<=*v**i*). It's guaranteed, that no pair of problems meets twice in the input. | Print one integer — the number of ways to split problems in two divisions. | [
"5 2\n1 4\n5 2\n",
"3 3\n1 2\n2 3\n1 3\n",
"3 2\n3 1\n3 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first sample, problems 1 and 2 should be used in division 2, while problems 4 and 5 in division 1. Problem 3 may be used either in division 1 or in division 2.
In the second sample, all pairs of problems are similar and there is no way to split problem between two divisions without breaking any rules.
Third sample reminds you that the similarity relation is not transitive. Problem 3 is similar to both 1 and 2, but 1 is not similar to 2, so they may be used together. | 750 | [
{
"input": "5 2\n1 4\n5 2",
"output": "2"
},
{
"input": "3 3\n1 2\n2 3\n1 3",
"output": "0"
},
{
"input": "3 2\n3 1\n3 2",
"output": "1"
},
{
"input": "2 0",
"output": "1"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "3 0",
"output": "2"
... | 1,462,635,746 | 2,246 | Python 3 | WRONG_ANSWER | PRETESTS | 3 | 46 | 5,120,000 |
def start() :
a = input().split()
n = int(a[0])
p = int(a[1])
small1 = n
large2 = 0
arr = []
for i in range(0, n):
arr.append(0)
pairs = []
for i in range(0, p):
pairs.append(input().split())
for i in range(0, p):
s = pairs[i]
a1 = int(s[0])
a2 = int(s[1])
if(a1>a2):
if(arr[a1-1]==2):
print("0")
return
else:
arr[a1-1] = 1
if(small1>a1):
small1 = a1
if(arr[a2-1]==1):
print("0")
return
else:
arr[a2-1] = 2
if(large2<a2):
large2 = a2
else:
if(arr[a1-1]==1):
print("0")
return
else:
arr[a1-1] = 2
if(large2<a1):
large2 = a1
if(arr[a2-1]==2):
print("0")
return
else:
arr[a2-1] = 1
if(small1>a2):
small1 = a2
if(small1<=large2):
print("0")
return
w = 1
for i in range(0, n):
if(arr[i]==0):
if(small1>(i+1)):
if(large2<(i+1)):
w += 1
print(str(w))
return
start() | Title: Problems for Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* problems prepared for the next Codeforces round. They are arranged in ascending order by their difficulty, and no two problems have the same difficulty. Moreover, there are *m* pairs of similar problems. Authors want to split problems between two division according to the following rules:
- Problemset of each division should be non-empty. - Each problem should be used in exactly one division (yes, it is unusual requirement). - Each problem used in division 1 should be harder than any problem used in division 2. - If two problems are similar, they should be used in different divisions.
Your goal is count the number of ways to split problem between two divisions and satisfy all the rules. Two ways to split problems are considered to be different if there is at least one problem that belongs to division 1 in one of them and to division 2 in the other.
Note, that the relation of similarity is not transitive. That is, if problem *i* is similar to problem *j* and problem *j* is similar to problem *k*, it doesn't follow that *i* is similar to *k*.
Input Specification:
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100<=000, 0<=≤<=*m*<=≤<=100<=000) — the number of problems prepared for the round and the number of pairs of similar problems, respectively.
Each of the following *m* lines contains a pair of similar problems *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*,<=*u**i*<=≠<=*v**i*). It's guaranteed, that no pair of problems meets twice in the input.
Output Specification:
Print one integer — the number of ways to split problems in two divisions.
Demo Input:
['5 2\n1 4\n5 2\n', '3 3\n1 2\n2 3\n1 3\n', '3 2\n3 1\n3 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first sample, problems 1 and 2 should be used in division 2, while problems 4 and 5 in division 1. Problem 3 may be used either in division 1 or in division 2.
In the second sample, all pairs of problems are similar and there is no way to split problem between two divisions without breaking any rules.
Third sample reminds you that the similarity relation is not transitive. Problem 3 is similar to both 1 and 2, but 1 is not similar to 2, so they may be used together. | ```python
def start() :
a = input().split()
n = int(a[0])
p = int(a[1])
small1 = n
large2 = 0
arr = []
for i in range(0, n):
arr.append(0)
pairs = []
for i in range(0, p):
pairs.append(input().split())
for i in range(0, p):
s = pairs[i]
a1 = int(s[0])
a2 = int(s[1])
if(a1>a2):
if(arr[a1-1]==2):
print("0")
return
else:
arr[a1-1] = 1
if(small1>a1):
small1 = a1
if(arr[a2-1]==1):
print("0")
return
else:
arr[a2-1] = 2
if(large2<a2):
large2 = a2
else:
if(arr[a1-1]==1):
print("0")
return
else:
arr[a1-1] = 2
if(large2<a1):
large2 = a1
if(arr[a2-1]==2):
print("0")
return
else:
arr[a2-1] = 1
if(small1>a2):
small1 = a2
if(small1<=large2):
print("0")
return
w = 1
for i in range(0, n):
if(arr[i]==0):
if(small1>(i+1)):
if(large2<(i+1)):
w += 1
print(str(w))
return
start()
``` | 0 | |
979 | A | Pizza, Pizza, Pizza!!! | PROGRAMMING | 1,000 | [
"math"
] | null | null | Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem? | A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces. | A single integer — the number of straight cuts Shiro needs. | [
"3\n",
"4\n"
] | [
"2",
"5"
] | To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts. | 500 | [
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "5"
},
{
"input": "10",
"output": "11"
},
{
"input": "10000000000",
"output": "10000000001"
},
{
"input": "1234567891",
"output": "617283946"
},
{
"input": "7509213957",
"output": "37546069... | 1,559,043,509 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 108 | 0 | n=int(input())
n+=1
if n%4==0:
p=n//2
print(min(p,n))
else:
print(n) | Title: Pizza, Pizza, Pizza!!!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
Input Specification:
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
Output Specification:
A single integer — the number of straight cuts Shiro needs.
Demo Input:
['3\n', '4\n']
Demo Output:
['2', '5']
Note:
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts. | ```python
n=int(input())
n+=1
if n%4==0:
p=n//2
print(min(p,n))
else:
print(n)
``` | 0 | |
701 | A | Cards | PROGRAMMING | 800 | [
"greedy",
"implementation"
] | null | null | There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible. | The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card. | Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them. | [
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] | [
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] | In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable. | 500 | [
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ... | 1,672,324,032 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | n= int(input())
mass = list(map(int,input().split()))
edinoe_sum= sum(mass)//(n//2)
count = 0
for i in range(n-1):
for j in range(i+1,n):
if mass[i]+mass[j]==edinoe_sum and count<n//2:
print(i+1,j+1)
count+=1 | Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable. | ```python
n= int(input())
mass = list(map(int,input().split()))
edinoe_sum= sum(mass)//(n//2)
count = 0
for i in range(n-1):
for j in range(i+1,n):
if mass[i]+mass[j]==edinoe_sum and count<n//2:
print(i+1,j+1)
count+=1
``` | 0 | |
989 | B | A Tide of Riverscape | PROGRAMMING | 1,200 | [
"constructive algorithms",
"strings"
] | null | null | "Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$. | The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character. | Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)). | [
"10 7\n1.0.1.0.1.\n",
"10 6\n1.0.1.1000\n",
"10 9\n1........1\n"
] | [
"1000100010\n",
"1001101000\n",
"No\n"
] | In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them. | 1,000 | [
{
"input": "10 7\n1.0.1.0.1.",
"output": "1000100010"
},
{
"input": "10 6\n1.0.1.1000",
"output": "1001101000"
},
{
"input": "10 9\n1........1",
"output": "No"
},
{
"input": "1 1\n.",
"output": "No"
},
{
"input": "5 1\n0...1",
"output": "00001"
},
{
"i... | 1,529,077,810 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 92 | 204,800 | n, p = [int(i) for i in input().split()]
s = list(input())
i = 0
j = p
c = 0
while j != n and not c:
if j < 2 * p:
if s[i] != s[j]:
if s[i] == '.' and s[j] == '1':
s[i] = '0'
elif s[i] == '.' and s[j] == '0':
s[i] = '1'
if s[j] == '.' and s[i] == '1':
s[j] = '0'
elif s[j] == '.' and s[i] == '0':
s[j] = '1'
c = 1
if s[i] == '.' and s[j] == '.':
s[i] = '0'
s[j] = '1'
c = 1
else:
if i == p:
i = 0
if s[i] != s[j]:
c = 1
i += 1
j += 1
if c:
for i in range(i, len(s)):
if s[i] == '.':
s[i] = '0'
for item in s:
print(item, end='')
else:
print('No') | Title: A Tide of Riverscape
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
Input Specification:
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
Output Specification:
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
Demo Input:
['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n']
Demo Output:
['1000100010\n', '1001101000\n', 'No\n']
Note:
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them. | ```python
n, p = [int(i) for i in input().split()]
s = list(input())
i = 0
j = p
c = 0
while j != n and not c:
if j < 2 * p:
if s[i] != s[j]:
if s[i] == '.' and s[j] == '1':
s[i] = '0'
elif s[i] == '.' and s[j] == '0':
s[i] = '1'
if s[j] == '.' and s[i] == '1':
s[j] = '0'
elif s[j] == '.' and s[i] == '0':
s[j] = '1'
c = 1
if s[i] == '.' and s[j] == '.':
s[i] = '0'
s[j] = '1'
c = 1
else:
if i == p:
i = 0
if s[i] != s[j]:
c = 1
i += 1
j += 1
if c:
for i in range(i, len(s)):
if s[i] == '.':
s[i] = '0'
for item in s:
print(item, end='')
else:
print('No')
``` | 0 | |
791 | A | Bear and Big Brother | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob? | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively. | Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob. | [
"4 7\n",
"4 9\n",
"1 1\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | 500 | [
{
"input": "4 7",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output... | 1,694,086,879 | 2,147,483,647 | Python 3 | OK | TESTS | 62 | 46 | 0 | n1 , n2 = map(int,input().split())
count=0
while True:
if n1>n2:
break
else:
n1=n1*3
n2=n2*2
count+=1
print(count) | Title: Bear and Big Brother
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Output Specification:
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
Demo Input:
['4 7\n', '4 9\n', '1 1\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | ```python
n1 , n2 = map(int,input().split())
count=0
while True:
if n1>n2:
break
else:
n1=n1*3
n2=n2*2
count+=1
print(count)
``` | 3 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
... | 1,656,796,859 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 46 | 0 | n,m = map(int,input().split())
bw = '#Black&White'
for i in range(n):
l = list(input().split())
for c in l:
if c != 'W' and c != 'G'and c != 'B':
bw = '#Color'
print(bw) | Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
n,m = map(int,input().split())
bw = '#Black&White'
for i in range(n):
l = list(input().split())
for c in l:
if c != 'W' and c != 'G'and c != 'B':
bw = '#Color'
print(bw)
``` | 3 |
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