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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
110 | A | Nearly Lucky Number | PROGRAMMING | 800 | [
"implementation"
] | A. Nearly Lucky Number | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number. | The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. | Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes). | [
"40047\n",
"7747774\n",
"1000000000000000000\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO". | 500 | [
{
"input": "40047",
"output": "NO"
},
{
"input": "7747774",
"output": "YES"
},
{
"input": "1000000000000000000",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "4",
"output": "NO"
},
{
"input": "474404774",
"output": "NO"
},
{
... | 1,697,035,415 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<iostream>
using namespace std;
int main() {
string a;
cin >> a;
int counter = 0;
for(int i = 0; i < a.size(); i++)
{
if(a[i] == '4' || '7')
{
counter += 1;
}
}
if(counter == 4 || 7)
{
cout << "YES";
}
else
{
cout << "NO";
}
return 0;
}
| Title: Nearly Lucky Number
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
Demo Input:
['40047\n', '7747774\n', '1000000000000000000\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO". | ```python
#include<iostream>
using namespace std;
int main() {
string a;
cin >> a;
int counter = 0;
for(int i = 0; i < a.size(); i++)
{
if(a[i] == '4' || '7')
{
counter += 1;
}
}
if(counter == 4 || 7)
{
cout << "YES";
}
else
{
cout << "NO";
}
return 0;
}
``` | -1 |
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": "... | 1,587,022,295 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 109 | 0 | a = input()
b = input()
ans = ""
for i in range(0,len(a)):
if a[i]==b[i]=="0" or a[i]==b[i]=="1":
ans += "0"
else:
ans += "1"
print(ans) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
a = input()
b = input()
ans = ""
for i in range(0,len(a)):
if a[i]==b[i]=="0" or a[i]==b[i]=="1":
ans += "0"
else:
ans += "1"
print(ans)
``` | 3.97275 |
47 | B | Coins | PROGRAMMING | 1,200 | [
"implementation"
] | B. Coins | 2 | 256 | One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal. | The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B. | It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights. | [
"A>B\nC<B\nA>C\n",
"A<B\nB>C\nC>A\n"
] | [
"CBA",
"ACB"
] | none | 1,000 | [
{
"input": "A>B\nC<B\nA>C",
"output": "CBA"
},
{
"input": "A<B\nB>C\nC>A",
"output": "ACB"
},
{
"input": "A<C\nB<A\nB>C",
"output": "Impossible"
},
{
"input": "A<B\nA<C\nB>C",
"output": "ACB"
},
{
"input": "B>A\nC<B\nC>A",
"output": "ACB"
},
{
"input":... | 1,643,133,039 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | def Coins():
a ,b, c= 0,0,0
d = {"A":0 , "B":0 , "C":0}
for _ in range(3):
data = input()
if data == "A>B" or data == "B<A" :
d["A"] += 1
d["B"] -= 1
elif data == "A<B" or data == "B>A":
d["A"] -= 1
d["B"] += 1
elif data == "B>C" or data == "C<B":
d["B"] += 1
d["C"] -= 1
elif data == "B<C" or data == "C>B":
d["C"] += 1
d["B"] -= 1
elif data == "A>C" or data == "C<A":
d["A"] += 1
d["C"] -= 1
elif data == "A<C" or data == "C>A":
d["A"] -= 1
d["C"] += 1
sort_d = dict(sorted(d.items(), key=lambda x: x[1], reverse=False))
final = ""
for i in sort_d.keys():
final += i
return final
print(Coins()) | Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
Input Specification:
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
Output Specification:
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
Demo Input:
['A>B\nC<B\nA>C\n', 'A<B\nB>C\nC>A\n']
Demo Output:
['CBA', 'ACB']
Note:
none | ```python
def Coins():
a ,b, c= 0,0,0
d = {"A":0 , "B":0 , "C":0}
for _ in range(3):
data = input()
if data == "A>B" or data == "B<A" :
d["A"] += 1
d["B"] -= 1
elif data == "A<B" or data == "B>A":
d["A"] -= 1
d["B"] += 1
elif data == "B>C" or data == "C<B":
d["B"] += 1
d["C"] -= 1
elif data == "B<C" or data == "C>B":
d["C"] += 1
d["B"] -= 1
elif data == "A>C" or data == "C<A":
d["A"] += 1
d["C"] -= 1
elif data == "A<C" or data == "C>A":
d["A"] -= 1
d["C"] += 1
sort_d = dict(sorted(d.items(), key=lambda x: x[1], reverse=False))
final = ""
for i in sort_d.keys():
final += i
return final
print(Coins())
``` | 0 |
664 | A | Complicated GCD | PROGRAMMING | 800 | [
"math",
"number theory"
] | null | null | Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type! | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100). | Output one integer — greatest common divisor of all integers from *a* to *b* inclusive. | [
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] | [
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] | none | 500 | [
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000... | 1,631,639,495 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 1,000 | 29,491,200 | def gcd(a,b):
while b!=0:
a, b = b, a%b
return a
a, b = map(int, input().split())
c=0
k=0
numb=[]
for i in range(a, b+1):
for j in range(a, b+1):
numb.append(gcd(i,j))
for elem in numb:
for i in range(a, b+1):
if i%elem!=0:
numb = list(filter((elem).__ne__, numb))
print(*list(set(numb))) | Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none | ```python
def gcd(a,b):
while b!=0:
a, b = b, a%b
return a
a, b = map(int, input().split())
c=0
k=0
numb=[]
for i in range(a, b+1):
for j in range(a, b+1):
numb.append(gcd(i,j))
for elem in numb:
for i in range(a, b+1):
if i%elem!=0:
numb = list(filter((elem).__ne__, numb))
print(*list(set(numb)))
``` | 0 | |
239 | A | Two Bags of Potatoes | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math"
] | null | null | Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. | The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). | Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1. | [
"10 1 10\n",
"10 6 40\n"
] | [
"-1\n",
"2 8 14 20 26 \n"
] | none | 500 | [
{
"input": "10 1 10",
"output": "-1"
},
{
"input": "10 6 40",
"output": "2 8 14 20 26 "
},
{
"input": "10 1 20",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "1 10000 1000000000",
"output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 12999... | 1,676,475,819 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 29 | 248 | 8,806,400 | l=list(map(int,input().split()))
y,k,n=l[0],l[1],l[2]
kittymin=k*int(y/k)+k
kittymax=l[2]
if kittymin>=kittymax:
print(-1)
for i in range(kittymin,kittymax+1,k):
print(i-y,end=" ") | Title: Two Bags of Potatoes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
Input Specification:
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Output Specification:
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
Demo Input:
['10 1 10\n', '10 6 40\n']
Demo Output:
['-1\n', '2 8 14 20 26 \n']
Note:
none | ```python
l=list(map(int,input().split()))
y,k,n=l[0],l[1],l[2]
kittymin=k*int(y/k)+k
kittymax=l[2]
if kittymin>=kittymax:
print(-1)
for i in range(kittymin,kittymax+1,k):
print(i-y,end=" ")
``` | 0 | |
868 | C | Qualification Rounds | PROGRAMMING | 1,500 | [
"bitmasks",
"brute force",
"constructive algorithms",
"dp"
] | null | null | Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset! | The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise. | Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n",
"3 2\n1 0\n1 1\n0 1\n"
] | [
"NO\n",
"YES\n"
] | In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | 1,000 | [
{
"input": "5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0",
"output": "NO"
},
{
"input": "3 2\n1 0\n1 1\n0 1",
"output": "YES"
},
{
"input": "10 2\n1 0\n1 0\n0 0\n1 1\n0 0\n1 1\n0 0\n1 1\n0 1\n0 1",
"output": "YES"
},
{
"input": "10 3\n1 0 0\n0 1 1\n1 0 0\n0 1 0\n0 0 1\n1 0 1\n0 1 1... | 1,507,196,426 | 9,326 | Python 3 | WRONG_ANSWER | PRETESTS | 3 | 46 | 0 | n, k=map(int, input().split())
arr=[]
for i in range(n):
a=[int(x) for x in input().split()]
arr.append(a)
#print(arr)
ans=[]
cnt=0
col=0
for i in range(k):
for j in range(n):
#print(arr[j][i])
if(arr[j][i]==1):
cnt+=1
if(cnt==n):
print("NO")
exit(0)
ans.append(cnt)
cnt=0
#print(ans)
#print(ans.count(ans[0]))
if(ans.count(ans[0])==len(ans)):
print("YES")
else:
print("NO")
| Title: Qualification Rounds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset.
*k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
Determine if Snark and Philip can make an interesting problemset!
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams.
Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise.
Output Specification:
Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n', '3 2\n1 0\n1 1\n0 1\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first example you can't make any interesting problemset, because the first team knows all problems.
In the second example you can choose the first and the third problems. | ```python
n, k=map(int, input().split())
arr=[]
for i in range(n):
a=[int(x) for x in input().split()]
arr.append(a)
#print(arr)
ans=[]
cnt=0
col=0
for i in range(k):
for j in range(n):
#print(arr[j][i])
if(arr[j][i]==1):
cnt+=1
if(cnt==n):
print("NO")
exit(0)
ans.append(cnt)
cnt=0
#print(ans)
#print(ans.count(ans[0]))
if(ans.count(ans[0])==len(ans)):
print("YES")
else:
print("NO")
``` | 0 | |
935 | A | Fafa and his Company | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly. | The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company. | Print a single integer representing the answer to the problem. | [
"2\n",
"10\n"
] | [
"1\n",
"3\n"
] | In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | 500 | [
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
... | 1,648,161,539 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | n=int(input())
nb=bin(n).replace("0b","")
lnb=len(nb)
x=lnb-1
print(x)
| Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | ```python
n=int(input())
nb=bin(n).replace("0b","")
lnb=len(nb)
x=lnb-1
print(x)
``` | 0 | |
161 | A | Dress'em in Vests! | PROGRAMMING | 1,300 | [
"binary search",
"brute force",
"greedy",
"two pointers"
] | null | null | The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*. | The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests. | In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them. | [
"5 3 0 0\n1 2 3 3 4\n1 3 5\n",
"3 3 2 2\n1 5 9\n3 5 7\n"
] | [
"2\n1 1\n3 2\n",
"3\n1 1\n2 2\n3 3\n"
] | In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped. | 1,000 | [
{
"input": "5 3 0 0\n1 2 3 3 4\n1 3 5",
"output": "2\n1 1\n3 2"
},
{
"input": "3 3 2 2\n1 5 9\n3 5 7",
"output": "3\n1 1\n2 2\n3 3"
},
{
"input": "1 1 0 0\n1\n1",
"output": "1\n1 1"
},
{
"input": "1 1 0 0\n1\n2",
"output": "0"
},
{
"input": "2 3 1 4\n1 5\n1 2 2",
... | 1,546,267,457 | 2,147,483,647 | Python 3 | OK | TESTS | 84 | 1,028 | 15,155,200 | # 161A-sua bai >> accepted
# n: num of solder
# m: num of vest
# x: margin duoi
# y: margin tren
# chu giai giai thuat: i chay tu 0>>n, neu ao vest b[j] nho hon (a[i]-x), thi tang j (xet bo ao vest tiep theo) cho toi khi tim
# thay j phu hop, sau do xet xem ung voi j do, ao vest co phu hop voi soldier i ko (b[j]<(a[i]+y)), neu phu hop thi luu cap i,j
# do lai, neu khong phu hop thi tang i (bang vong for)
n,m,x,y = map(int,input().split())
a = list(map(int,input().split())) # list size soldier
b = list(map(int,input().split())) # list cac size ao
j=0
u = []
for i in range (n):
if (j==m):
break
# print (b[j])
# print (a[i]-x)
while ((b[j]<(a[i]-x))):
# print ('i=',i)
# print ('j=',j)
j+=1
if (j==m):
break
if (j==m):
break
if (b[j]<=(a[i]+y)):
u.append([i+1,j+1])
j+=1
print (len(u))
for i in range (len(u)):
print ('%d %d'%(u[i][0],u[i][1])) | Title: Dress'em in Vests!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
Input Specification:
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
Output Specification:
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
Demo Input:
['5 3 0 0\n1 2 3 3 4\n1 3 5\n', '3 3 2 2\n1 5 9\n3 5 7\n']
Demo Output:
['2\n1 1\n3 2\n', '3\n1 1\n2 2\n3 3\n']
Note:
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped. | ```python
# 161A-sua bai >> accepted
# n: num of solder
# m: num of vest
# x: margin duoi
# y: margin tren
# chu giai giai thuat: i chay tu 0>>n, neu ao vest b[j] nho hon (a[i]-x), thi tang j (xet bo ao vest tiep theo) cho toi khi tim
# thay j phu hop, sau do xet xem ung voi j do, ao vest co phu hop voi soldier i ko (b[j]<(a[i]+y)), neu phu hop thi luu cap i,j
# do lai, neu khong phu hop thi tang i (bang vong for)
n,m,x,y = map(int,input().split())
a = list(map(int,input().split())) # list size soldier
b = list(map(int,input().split())) # list cac size ao
j=0
u = []
for i in range (n):
if (j==m):
break
# print (b[j])
# print (a[i]-x)
while ((b[j]<(a[i]-x))):
# print ('i=',i)
# print ('j=',j)
j+=1
if (j==m):
break
if (j==m):
break
if (b[j]<=(a[i]+y)):
u.append([i+1,j+1])
j+=1
print (len(u))
for i in range (len(u)):
print ('%d %d'%(u[i][0],u[i][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... | 1,663,362,589 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | m,n = map(int,input().split())
aria = m * n
if aria %2 != 0:
aria-=1
print(int(aria/2)) | 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())
aria = m * n
if aria %2 != 0:
aria-=1
print(int(aria/2))
``` | 3.977 |
177 | A1 | Good Matrix Elements | PROGRAMMING | 800 | [
"implementation"
] | null | null | The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix. | The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101 | Print a single integer — the sum of good matrix elements. | [
"3\n1 2 3\n4 5 6\n7 8 9\n",
"5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n"
] | [
"45\n",
"17\n"
] | In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure. | 30 | [
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "45"
},
{
"input": "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "17"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "5\n27 7 3 11 72\n19 49 68 19 59\n41 25 37 64 65\n8 39 96 62 90\n13 37 43 26 33",
... | 1,563,706,195 | 2,147,483,647 | Python 3 | OK | TESTS1 | 17 | 218 | 0 | #import sys
#sys.stdin = open("input.in","r")
#sys.stdout = open("test.out","w")
n=int(input())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
c=0
for i in range(n):
for j in range(n):
if(i==j or i==n//2 or j==n//2 or i==n-1-j):
c+=l[i][j]
print(c) | Title: Good Matrix Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
Input Specification:
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
Output Specification:
Print a single integer — the sum of good matrix elements.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n', '5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n']
Demo Output:
['45\n', '17\n']
Note:
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure. | ```python
#import sys
#sys.stdin = open("input.in","r")
#sys.stdout = open("test.out","w")
n=int(input())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
c=0
for i in range(n):
for j in range(n):
if(i==j or i==n//2 or j==n//2 or i==n-1-j):
c+=l[i][j]
print(c)
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,614,718,348 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 0 | s=[100]
for i in range(3):
s=input()
c = s.lower()
print(c)
print() | 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=[100]
for i in range(3):
s=input()
c = s.lower()
print(c)
print()
``` | -1 |
577 | B | Modulo Sum | PROGRAMMING | 1,900 | [
"combinatorics",
"data structures",
"dp",
"two pointers"
] | null | null | You are given a sequence of numbers *a*1,<=*a*2,<=...,<=*a**n*, and a number *m*.
Check if it is possible to choose a non-empty subsequence *a**i**j* such that the sum of numbers in this subsequence is divisible by *m*. | The first line contains two numbers, *n* and *m* (1<=≤<=*n*<=≤<=106, 2<=≤<=*m*<=≤<=103) — the size of the original sequence and the number such that sum should be divisible by it.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109). | In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist. | [
"3 5\n1 2 3\n",
"1 6\n5\n",
"4 6\n3 1 1 3\n",
"6 6\n5 5 5 5 5 5\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"YES\n"
] | In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.
In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
In the third sample test you need to choose two numbers 3 on the ends.
In the fourth sample test you can take the whole subsequence. | 1,250 | [
{
"input": "3 5\n1 2 3",
"output": "YES"
},
{
"input": "1 6\n5",
"output": "NO"
},
{
"input": "4 6\n3 1 1 3",
"output": "YES"
},
{
"input": "6 6\n5 5 5 5 5 5",
"output": "YES"
},
{
"input": "4 5\n1 1 1 1",
"output": "NO"
},
{
"input": "5 5\n1 1 1 1 1",... | 1,691,375,165 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 93 | 3,072,000 | def subset_sum(nums, total):
dp = [False] * (total + 1)
dp[0] = True
n = len(nums)
for i in range(1, n + 1):
for j in range(total, -1, -1):
if j - nums[i - 1] >= 0:
dp[j] |= dp[j - nums[i - 1]]
return dp[total]
def solve(tc):
n, m = map(int, input().split())
v = list(map(int, input().split()))
def sum_vector(vec):
return sum(vec)
sum_val = sum_vector(v)
ans = False
v.sort()
if v[0] == 0:
print("yes")
return
subset_sum(v, m * (sum_val // m))
for i in range(1, (sum_val // m) + 1):
if i * m < sum_val:
ans |= dp[i * m]
if ans:
print("yes")
else:
print("no")
def main():
t = 1
# t = int(input())
for i in range(1, t + 1):
solve(i)
if __name__ == "__main__":
main()
| Title: Modulo Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence of numbers *a*1,<=*a*2,<=...,<=*a**n*, and a number *m*.
Check if it is possible to choose a non-empty subsequence *a**i**j* such that the sum of numbers in this subsequence is divisible by *m*.
Input Specification:
The first line contains two numbers, *n* and *m* (1<=≤<=*n*<=≤<=106, 2<=≤<=*m*<=≤<=103) — the size of the original sequence and the number such that sum should be divisible by it.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109).
Output Specification:
In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.
Demo Input:
['3 5\n1 2 3\n', '1 6\n5\n', '4 6\n3 1 1 3\n', '6 6\n5 5 5 5 5 5\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'YES\n']
Note:
In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.
In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.
In the third sample test you need to choose two numbers 3 on the ends.
In the fourth sample test you can take the whole subsequence. | ```python
def subset_sum(nums, total):
dp = [False] * (total + 1)
dp[0] = True
n = len(nums)
for i in range(1, n + 1):
for j in range(total, -1, -1):
if j - nums[i - 1] >= 0:
dp[j] |= dp[j - nums[i - 1]]
return dp[total]
def solve(tc):
n, m = map(int, input().split())
v = list(map(int, input().split()))
def sum_vector(vec):
return sum(vec)
sum_val = sum_vector(v)
ans = False
v.sort()
if v[0] == 0:
print("yes")
return
subset_sum(v, m * (sum_val // m))
for i in range(1, (sum_val // m) + 1):
if i * m < sum_val:
ans |= dp[i * m]
if ans:
print("yes")
else:
print("no")
def main():
t = 1
# t = int(input())
for i in range(1, t + 1):
solve(i)
if __name__ == "__main__":
main()
``` | -1 | |
522 | A | Reposts | PROGRAMMING | 1,200 | [
"*special",
"dfs and similar",
"dp",
"graphs",
"trees"
] | null | null | One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user. | Print a single integer — the maximum length of a repost chain. | [
"5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n",
"6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n",
"1\nSoMeStRaNgEgUe repos... | [
"6\n",
"2\n",
"2\n"
] | none | 500 | [
{
"input": "5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya",
"output": "6"
},
{
"input": "6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforc... | 1,674,908,553 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 31 | 102,400 | from collections import defaultdict, deque
n = int(input())
g = {}
for _ in range(n):
p = input().split()
g[p[2].lower()] = g.get(p[0].lower(),[]) + [p[0].lower()]
ans = -float('inf')
elem = g.keys()
dp = 0
que = deque(['polycarp'])
while que:
l = len(que)
for _ in range(l):
temp = que.popleft()
if temp in g:
for nb in g[temp]:
que.append(nb)
dp+=1
print(dp)
| Title: Reposts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user.
Output Specification:
Print a single integer — the maximum length of a repost chain.
Demo Input:
['5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n', '6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n', '1\nSoMeStRaNgEgUe reposted PoLyCaRp\n']
Demo Output:
['6\n', '2\n', '2\n']
Note:
none | ```python
from collections import defaultdict, deque
n = int(input())
g = {}
for _ in range(n):
p = input().split()
g[p[2].lower()] = g.get(p[0].lower(),[]) + [p[0].lower()]
ans = -float('inf')
elem = g.keys()
dp = 0
que = deque(['polycarp'])
while que:
l = len(que)
for _ in range(l):
temp = que.popleft()
if temp in g:
for nb in g[temp]:
que.append(nb)
dp+=1
print(dp)
``` | 0 | |
748 | B | Santa Claus and Keyboard Check | PROGRAMMING | 1,500 | [
"implementation",
"strings"
] | null | null | Santa Claus decided to disassemble his keyboard to clean it. After he returned all the keys back, he suddenly realized that some pairs of keys took each other's place! That is, Santa suspects that each key is either on its place, or on the place of another key, which is located exactly where the first key should be.
In order to make sure that he's right and restore the correct order of keys, Santa typed his favorite patter looking only to his keyboard.
You are given the Santa's favorite patter and the string he actually typed. Determine which pairs of keys could be mixed. Each key must occur in pairs at most once. | The input consists of only two strings *s* and *t* denoting the favorite Santa's patter and the resulting string. *s* and *t* are not empty and have the same length, which is at most 1000. Both strings consist only of lowercase English letters. | If Santa is wrong, and there is no way to divide some of keys into pairs and swap keys in each pair so that the keyboard will be fixed, print «-1» (without quotes).
Otherwise, the first line of output should contain the only integer *k* (*k*<=≥<=0) — the number of pairs of keys that should be swapped. The following *k* lines should contain two space-separated letters each, denoting the keys which should be swapped. All printed letters must be distinct.
If there are several possible answers, print any of them. You are free to choose the order of the pairs and the order of keys in a pair.
Each letter must occur at most once. Santa considers the keyboard to be fixed if he can print his favorite patter without mistakes. | [
"helloworld\nehoolwlroz\n",
"hastalavistababy\nhastalavistababy\n",
"merrychristmas\nchristmasmerry\n"
] | [
"3\nh e\nl o\nd z\n",
"0\n",
"-1\n"
] | none | 1,000 | [
{
"input": "helloworld\nehoolwlroz",
"output": "3\nh e\nl o\nd z"
},
{
"input": "hastalavistababy\nhastalavistababy",
"output": "0"
},
{
"input": "merrychristmas\nchristmasmerry",
"output": "-1"
},
{
"input": "kusyvdgccw\nkusyvdgccw",
"output": "0"
},
{
"input": "... | 1,603,609,883 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 13 | 140 | 0 | #
# Created by Polusummator on 25.10.2020
# --------- Little PyPy Squad ---------
# Verdict:
#
orig = input()
s = input()
d = {}
z = True
for i in range(len(orig)):
# print(d.get(orig[i]), d.get(s[i]))
if orig[i] != s[i]:
if (d.get(orig[i]) is None and d.get(s[i])) is None:
d[orig[i]] = s[i]
d[s[i]] = orig[i]
elif d.get(orig[i]) == s[i] and d.get(s[i]) == orig[i]:
pass
else:
z = False
break
if not z:
print(-1)
else:
keys = list(d.keys())
print(len(keys) // 2)
b = set()
for i in keys:
if i not in b and d[i] not in b:
b.add(i)
b.add(d[i])
print(i, d[i])
| Title: Santa Claus and Keyboard Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus decided to disassemble his keyboard to clean it. After he returned all the keys back, he suddenly realized that some pairs of keys took each other's place! That is, Santa suspects that each key is either on its place, or on the place of another key, which is located exactly where the first key should be.
In order to make sure that he's right and restore the correct order of keys, Santa typed his favorite patter looking only to his keyboard.
You are given the Santa's favorite patter and the string he actually typed. Determine which pairs of keys could be mixed. Each key must occur in pairs at most once.
Input Specification:
The input consists of only two strings *s* and *t* denoting the favorite Santa's patter and the resulting string. *s* and *t* are not empty and have the same length, which is at most 1000. Both strings consist only of lowercase English letters.
Output Specification:
If Santa is wrong, and there is no way to divide some of keys into pairs and swap keys in each pair so that the keyboard will be fixed, print «-1» (without quotes).
Otherwise, the first line of output should contain the only integer *k* (*k*<=≥<=0) — the number of pairs of keys that should be swapped. The following *k* lines should contain two space-separated letters each, denoting the keys which should be swapped. All printed letters must be distinct.
If there are several possible answers, print any of them. You are free to choose the order of the pairs and the order of keys in a pair.
Each letter must occur at most once. Santa considers the keyboard to be fixed if he can print his favorite patter without mistakes.
Demo Input:
['helloworld\nehoolwlroz\n', 'hastalavistababy\nhastalavistababy\n', 'merrychristmas\nchristmasmerry\n']
Demo Output:
['3\nh e\nl o\nd z\n', '0\n', '-1\n']
Note:
none | ```python
#
# Created by Polusummator on 25.10.2020
# --------- Little PyPy Squad ---------
# Verdict:
#
orig = input()
s = input()
d = {}
z = True
for i in range(len(orig)):
# print(d.get(orig[i]), d.get(s[i]))
if orig[i] != s[i]:
if (d.get(orig[i]) is None and d.get(s[i])) is None:
d[orig[i]] = s[i]
d[s[i]] = orig[i]
elif d.get(orig[i]) == s[i] and d.get(s[i]) == orig[i]:
pass
else:
z = False
break
if not z:
print(-1)
else:
keys = list(d.keys())
print(len(keys) // 2)
b = set()
for i in keys:
if i not in b and d[i] not in b:
b.add(i)
b.add(d[i])
print(i, d[i])
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are $n$ banknote denominations on Mars: the value of $i$-th banknote is $a_i$. Natasha has an infinite number of banknotes of each denomination.
Martians have $k$ fingers on their hands, so they use a number system with base $k$. In addition, the Martians consider the digit $d$ (in the number system with base $k$) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base $k$ is $d$, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values $d$ Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change. | The first line contains two integers $n$ and $k$ ($1 \le n \le 100\,000$, $2 \le k \le 100\,000$) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — denominations of banknotes on Mars.
All numbers are given in decimal notation. | On the first line output the number of values $d$ for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation. | [
"2 8\n12 20\n",
"3 10\n10 20 30\n"
] | [
"2\n0 4 ",
"1\n0 "
] | Consider the first test case. It uses the octal number system.
If you take one banknote with the value of $12$, you will get $14_8$ in octal system. The last digit is $4_8$.
If you take one banknote with the value of $12$ and one banknote with the value of $20$, the total value will be $32$. In the octal system, it is $40_8$. The last digit is $0_8$.
If you take two banknotes with the value of $20$, the total value will be $40$, this is $50_8$ in the octal system. The last digit is $0_8$.
No other digits other than $0_8$ and $4_8$ can be obtained. Digits $0_8$ and $4_8$ could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | 0 | [
{
"input": "2 8\n12 20",
"output": "2\n0 4 "
},
{
"input": "3 10\n10 20 30",
"output": "1\n0 "
},
{
"input": "5 10\n20 16 4 16 2",
"output": "5\n0 2 4 6 8 "
},
{
"input": "10 5\n4 6 8 6 4 10 2 10 8 6",
"output": "5\n0 1 2 3 4 "
},
{
"input": "20 25\n15 10 5 20 10 ... | 1,579,386,636 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 18 | 202 | 10,547,200 | def hcfnaive(a,b):
if(b==0):
return a
else:
return hcfnaive(b,a%b)
n,k=map(int,input().split())
a=input().split()
ls=[]
for i in range(n):
ls.append(int(a[i])%k)
ls=list(set(ls))
if(len(ls)==1):
if(ls[0]==0):
print(1)
print(0)
else:
ans=[]
for i in range(k):
if(i%ls[0]==0):
ans.append(i)
print(len(ans))
print(*ans)
else:
gcd=hcfnaive(ls[0],ls[1])
for i in range(2,len(ls)):
gcd=hcfnaive(gcd,ls[i])
ans=[]
for i in range(k):
if(i%gcd==0):
ans.append(i)
print(len(ans))
print(*ans) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are $n$ banknote denominations on Mars: the value of $i$-th banknote is $a_i$. Natasha has an infinite number of banknotes of each denomination.
Martians have $k$ fingers on their hands, so they use a number system with base $k$. In addition, the Martians consider the digit $d$ (in the number system with base $k$) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base $k$ is $d$, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values $d$ Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le n \le 100\,000$, $2 \le k \le 100\,000$) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output Specification:
On the first line output the number of values $d$ for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Demo Input:
['2 8\n12 20\n', '3 10\n10 20 30\n']
Demo Output:
['2\n0 4 ', '1\n0 ']
Note:
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of $12$, you will get $14_8$ in octal system. The last digit is $4_8$.
If you take one banknote with the value of $12$ and one banknote with the value of $20$, the total value will be $32$. In the octal system, it is $40_8$. The last digit is $0_8$.
If you take two banknotes with the value of $20$, the total value will be $40$, this is $50_8$ in the octal system. The last digit is $0_8$.
No other digits other than $0_8$ and $4_8$ can be obtained. Digits $0_8$ and $4_8$ could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | ```python
def hcfnaive(a,b):
if(b==0):
return a
else:
return hcfnaive(b,a%b)
n,k=map(int,input().split())
a=input().split()
ls=[]
for i in range(n):
ls.append(int(a[i])%k)
ls=list(set(ls))
if(len(ls)==1):
if(ls[0]==0):
print(1)
print(0)
else:
ans=[]
for i in range(k):
if(i%ls[0]==0):
ans.append(i)
print(len(ans))
print(*ans)
else:
gcd=hcfnaive(ls[0],ls[1])
for i in range(2,len(ls)):
gcd=hcfnaive(gcd,ls[i])
ans=[]
for i in range(k):
if(i%gcd==0):
ans.append(i)
print(len(ans))
print(*ans)
``` | 0 | |
598 | D | Igor In the Museum | PROGRAMMING | 1,700 | [
"dfs and similar",
"graphs",
"shortest paths"
] | null | null | Igor is in the museum and he wants to see as many pictures as possible.
Museum can be represented as a rectangular field of *n*<=×<=*m* cells. Each cell is either empty or impassable. Empty cells are marked with '.', impassable cells are marked with '*'. Every two adjacent cells of different types (one empty and one impassable) are divided by a wall containing one picture.
At the beginning Igor is in some empty cell. At every moment he can move to any empty cell that share a side with the current one.
For several starting positions you should calculate the maximum number of pictures that Igor can see. Igor is able to see the picture only if he is in the cell adjacent to the wall with this picture. Igor have a lot of time, so he will examine every picture he can see. | First line of the input contains three integers *n*, *m* and *k* (3<=≤<=*n*,<=*m*<=≤<=1000,<=1<=≤<=*k*<=≤<=*min*(*n*·*m*,<=100<=000)) — the museum dimensions and the number of starting positions to process.
Each of the next *n* lines contains *m* symbols '.', '*' — the description of the museum. It is guaranteed that all border cells are impassable, so Igor can't go out from the museum.
Each of the last *k* lines contains two integers *x* and *y* (1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*y*<=≤<=*m*) — the row and the column of one of Igor's starting positions respectively. Rows are numbered from top to bottom, columns — from left to right. It is guaranteed that all starting positions are empty cells. | Print *k* integers — the maximum number of pictures, that Igor can see if he starts in corresponding position. | [
"5 6 3\n******\n*..*.*\n******\n*....*\n******\n2 2\n2 5\n4 3\n",
"4 4 1\n****\n*..*\n*.**\n****\n3 2\n"
] | [
"6\n4\n10\n",
"8\n"
] | none | 0 | [
{
"input": "5 6 3\n******\n*..*.*\n******\n*....*\n******\n2 2\n2 5\n4 3",
"output": "6\n4\n10"
},
{
"input": "4 4 1\n****\n*..*\n*.**\n****\n3 2",
"output": "8"
},
{
"input": "3 3 1\n***\n*.*\n***\n2 2",
"output": "4"
},
{
"input": "5 5 10\n*****\n*...*\n*..**\n*.***\n*****\... | 1,655,730,584 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | import sys, collections, numpy
m,n,k = map(int, sys.stdin.readline().split())
grid = []
for i in range(m):
s = sys.stdin.readline()
grid.append(list(s))
counts = {} # counts[k]: number of pictures in component k
def bfs(r, c, cId):
Q = collections.deque()
Q.append((r,c))
grid[r][c] = cId
picture = 0
while Q:
r,c = Q.pop()
for nr, nc in (r-1, c), (r+1,c), (r, c-1), (r, c+1):
if grid[nr][nc] == '*':
picture += 1
continue
elif grid[nr][nc] == '.':
grid[nr][nc] = cId
Q.appendleft((nr,nc))
counts[cId] = picture
compNum = 0
for _ in range(k):
r,c = map(int, sys.stdin.readline().split())
r -= 1
c -= 1
if grid[r][c] == '.':
bfs(r,c, compNum)
compNum += 1
compId = grid[r][c]
print(counts[compId])
| Title: Igor In the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor is in the museum and he wants to see as many pictures as possible.
Museum can be represented as a rectangular field of *n*<=×<=*m* cells. Each cell is either empty or impassable. Empty cells are marked with '.', impassable cells are marked with '*'. Every two adjacent cells of different types (one empty and one impassable) are divided by a wall containing one picture.
At the beginning Igor is in some empty cell. At every moment he can move to any empty cell that share a side with the current one.
For several starting positions you should calculate the maximum number of pictures that Igor can see. Igor is able to see the picture only if he is in the cell adjacent to the wall with this picture. Igor have a lot of time, so he will examine every picture he can see.
Input Specification:
First line of the input contains three integers *n*, *m* and *k* (3<=≤<=*n*,<=*m*<=≤<=1000,<=1<=≤<=*k*<=≤<=*min*(*n*·*m*,<=100<=000)) — the museum dimensions and the number of starting positions to process.
Each of the next *n* lines contains *m* symbols '.', '*' — the description of the museum. It is guaranteed that all border cells are impassable, so Igor can't go out from the museum.
Each of the last *k* lines contains two integers *x* and *y* (1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*y*<=≤<=*m*) — the row and the column of one of Igor's starting positions respectively. Rows are numbered from top to bottom, columns — from left to right. It is guaranteed that all starting positions are empty cells.
Output Specification:
Print *k* integers — the maximum number of pictures, that Igor can see if he starts in corresponding position.
Demo Input:
['5 6 3\n******\n*..*.*\n******\n*....*\n******\n2 2\n2 5\n4 3\n', '4 4 1\n****\n*..*\n*.**\n****\n3 2\n']
Demo Output:
['6\n4\n10\n', '8\n']
Note:
none | ```python
import sys, collections, numpy
m,n,k = map(int, sys.stdin.readline().split())
grid = []
for i in range(m):
s = sys.stdin.readline()
grid.append(list(s))
counts = {} # counts[k]: number of pictures in component k
def bfs(r, c, cId):
Q = collections.deque()
Q.append((r,c))
grid[r][c] = cId
picture = 0
while Q:
r,c = Q.pop()
for nr, nc in (r-1, c), (r+1,c), (r, c-1), (r, c+1):
if grid[nr][nc] == '*':
picture += 1
continue
elif grid[nr][nc] == '.':
grid[nr][nc] = cId
Q.appendleft((nr,nc))
counts[cId] = picture
compNum = 0
for _ in range(k):
r,c = map(int, sys.stdin.readline().split())
r -= 1
c -= 1
if grid[r][c] == '.':
bfs(r,c, compNum)
compNum += 1
compId = grid[r][c]
print(counts[compId])
``` | -1 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input":... | 1,686,766,590 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | def l_string(string):
if len(string)<=2:
return string
count=len(string)-2
s=string[0]+str(count)+string[-1]
return s
t=int(input())
for i in range(t):
string=input()
a_string=l_string(string)
print(a_string) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
def l_string(string):
if len(string)<=2:
return string
count=len(string)-2
s=string[0]+str(count)+string[-1]
return s
t=int(input())
for i in range(t):
string=input()
a_string=l_string(string)
print(a_string)
``` | 0 |
716 | A | Crazy Computer | PROGRAMMING | 800 | [
"implementation"
] | null | null | ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything. | The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word. | Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*. | [
"6 5\n1 3 8 14 19 20\n",
"6 1\n1 3 5 7 9 10\n"
] | [
"3",
"2"
] | The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | 500 | [
{
"input": "6 5\n1 3 8 14 19 20",
"output": "3"
},
{
"input": "6 1\n1 3 5 7 9 10",
"output": "2"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "5 5\n1 7 12 13 14",
"output": "4"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "2"
},
{
... | 1,677,455,539 | 2,147,483,647 | PyPy 3 | OK | TESTS | 81 | 202 | 10,854,400 | n, c = map(int, input().split())
lst = list(map(int, input().split()))
curr = 0
for i in range(len(lst)):
if i == 0:
curr += 1
elif abs(lst[i-1] - lst[i]) > c:
curr = 1
else:
curr += 1
print(curr) | Title: Crazy Computer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
Output Specification:
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
Demo Input:
['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n']
Demo Output:
['3', '2']
Note:
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | ```python
n, c = map(int, input().split())
lst = list(map(int, input().split()))
curr = 0
for i in range(len(lst)):
if i == 0:
curr += 1
elif abs(lst[i-1] - lst[i]) > c:
curr = 1
else:
curr += 1
print(curr)
``` | 3 | |
1,006 | C | Three Parts of the Array | PROGRAMMING | 1,200 | [
"binary search",
"data structures",
"two pointers"
] | null | null | You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible. | The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$. | Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$). | [
"5\n1 3 1 1 4\n",
"5\n1 3 2 1 4\n",
"3\n4 1 2\n"
] | [
"5\n",
"4\n",
"0\n"
] | In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$. | 0 | [
{
"input": "5\n1 3 1 1 4",
"output": "5"
},
{
"input": "5\n1 3 2 1 4",
"output": "4"
},
{
"input": "3\n4 1 2",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "5\n1 3 5 4 5",
"output": ... | 1,609,162,418 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 186 | 2,150,400 | n=int(input())
l1=list(map(int,input().split()))
sum1=0
for i in range (n):
index=i
sum1+=l1[i]
l2=l1[i+1:]
if(sum1>sum(l2)):
break
sum1-=l1[index]
sum2=0
for i in range (n):
sum2+=l2[len(l2)-1-i]
if(sum2==sum1):
break
if(sum2>sum1):
print(0)
else:
print(sum1) | Title: Three Parts of the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$.
Output Specification:
Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
Demo Input:
['5\n1 3 1 1 4\n', '5\n1 3 2 1 4\n', '3\n4 1 2\n']
Demo Output:
['5\n', '4\n', '0\n']
Note:
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$. | ```python
n=int(input())
l1=list(map(int,input().split()))
sum1=0
for i in range (n):
index=i
sum1+=l1[i]
l2=l1[i+1:]
if(sum1>sum(l2)):
break
sum1-=l1[index]
sum2=0
for i in range (n):
sum2+=l2[len(l2)-1-i]
if(sum2==sum1):
break
if(sum2>sum1):
print(0)
else:
print(sum1)
``` | -1 | |
985 | B | Switches and Lamps | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on. | Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch. | [
"4 5\n10101\n01000\n00111\n10000\n",
"4 5\n10100\n01000\n00110\n00101\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "4 5\n10101\n01000\n00111\n10000",
"output": "YES"
},
{
"input": "4 5\n10100\n01000\n00110\n00101",
"output": "NO"
},
{
"input": "1 5\n11111",
"output": "NO"
},
{
"input": "10 1\n1\n0\n0\n0\n0\n0\n0\n0\n0\n1",
"output": "YES"
},
{
"input": "1 1\n1",
... | 1,527,137,382 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 62 | 0 | a=list(map(int,input().split()))
n=a[0]
m=a[1]
a=[]
b=[0 for i in range(m)]
for i in range(n):
c=list(map(int,list(input())))
a.append(c)
for j in range(m):
b[j]+=a[i][j]
for i in range(m):
c=[1 for j in range(m) if b[j]-a[i][j] > 0 ]
if(len(c)==m):
print("YES")
break
else:
print("NO")
| Title: Switches and Lamps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on.
Output Specification:
Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch.
Demo Input:
['4 5\n10101\n01000\n00111\n10000\n', '4 5\n10100\n01000\n00110\n00101\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
a=list(map(int,input().split()))
n=a[0]
m=a[1]
a=[]
b=[0 for i in range(m)]
for i in range(n):
c=list(map(int,list(input())))
a.append(c)
for j in range(m):
b[j]+=a[i][j]
for i in range(m):
c=[1 for j in range(m) if b[j]-a[i][j] > 0 ]
if(len(c)==m):
print("YES")
break
else:
print("NO")
``` | -1 | |
203 | C | Photographer | PROGRAMMING | 1,400 | [
"greedy",
"sortings"
] | null | null | Valera's lifelong ambition was to be a photographer, so he bought a new camera. Every day he got more and more clients asking for photos, and one day Valera needed a program that would determine the maximum number of people he can serve.
The camera's memory is *d* megabytes. Valera's camera can take photos of high and low quality. One low quality photo takes *a* megabytes of memory, one high quality photo take *b* megabytes of memory. For unknown reasons, each client asks him to make several low quality photos and several high quality photos. More formally, the *i*-th client asks to make *x**i* low quality photos and *y**i* high quality photos.
Valera wants to serve as many clients per day as possible, provided that they will be pleased with his work. To please the *i*-th client, Valera needs to give him everything he wants, that is, to make *x**i* low quality photos and *y**i* high quality photos. To make one low quality photo, the camera must have at least *a* megabytes of free memory space. Similarly, to make one high quality photo, the camera must have at least *b* megabytes of free memory space. Initially the camera's memory is empty. Valera also does not delete photos from the camera so that the camera's memory gradually fills up.
Calculate the maximum number of clients Valera can successfully serve and print the numbers of these clients. | The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*d*<=≤<=109) — the number of clients and the camera memory size, correspondingly. The second line contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=104) — the size of one low quality photo and of one high quality photo, correspondingly.
Next *n* lines describe the clients. The *i*-th line contains two integers *x**i* and *y**i* (0<=≤<=*x**i*,<=*y**i*<=≤<=105) — the number of low quality photos and high quality photos the *i*-th client wants, correspondingly.
All numbers on all lines are separated by single spaces. | On the first line print the answer to the problem — the maximum number of clients that Valera can successfully serve. Print on the second line the numbers of the client in any order. All numbers must be distinct. If there are multiple answers, print any of them. The clients are numbered starting with 1 in the order in which they are defined in the input data. | [
"3 10\n2 3\n1 4\n2 1\n1 0\n",
"3 6\n6 6\n1 1\n1 0\n1 0\n"
] | [
"2\n3 2 ",
"1\n2 "
] | none | 1,500 | [
{
"input": "3 10\n2 3\n1 4\n2 1\n1 0",
"output": "2\n3 2 "
},
{
"input": "3 6\n6 6\n1 1\n1 0\n1 0",
"output": "1\n2 "
},
{
"input": "4 5\n6 8\n1 2\n3 0\n10 2\n0 4",
"output": "0"
},
{
"input": "4 10\n6 6\n1 2\n2 2\n0 0\n0 0",
"output": "2\n3 4 "
},
{
"input": "10 ... | 1,446,284,366 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 16 | 2,000 | 15,462,400 | class Pair:
def __init__(self,sumOfNumber,index):
self.sumOfNumber=sumOfNumber
self.index=index
def __lt__(self, other):
return self.sumOfNumber < other.sumOfNumber
n,d=map(int,input().split())
a,b=map(int,input().split())
s=[None]*n
for i in range(n):
x,y=map(int,input().split())
cur=Pair(0,0)
cur.sumOfNumber=(a*x)+(b*y)
cur.index=i+1
s[i]=cur
s.sort()
cur=0
while cur<n and d-s[cur].sumOfNumber >= 0:
d-=s[cur].sumOfNumber
cur+=1
print(cur)
for i in range(cur):
print(s[i].index,end=" ")
| Title: Photographer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera's lifelong ambition was to be a photographer, so he bought a new camera. Every day he got more and more clients asking for photos, and one day Valera needed a program that would determine the maximum number of people he can serve.
The camera's memory is *d* megabytes. Valera's camera can take photos of high and low quality. One low quality photo takes *a* megabytes of memory, one high quality photo take *b* megabytes of memory. For unknown reasons, each client asks him to make several low quality photos and several high quality photos. More formally, the *i*-th client asks to make *x**i* low quality photos and *y**i* high quality photos.
Valera wants to serve as many clients per day as possible, provided that they will be pleased with his work. To please the *i*-th client, Valera needs to give him everything he wants, that is, to make *x**i* low quality photos and *y**i* high quality photos. To make one low quality photo, the camera must have at least *a* megabytes of free memory space. Similarly, to make one high quality photo, the camera must have at least *b* megabytes of free memory space. Initially the camera's memory is empty. Valera also does not delete photos from the camera so that the camera's memory gradually fills up.
Calculate the maximum number of clients Valera can successfully serve and print the numbers of these clients.
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*d*<=≤<=109) — the number of clients and the camera memory size, correspondingly. The second line contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=104) — the size of one low quality photo and of one high quality photo, correspondingly.
Next *n* lines describe the clients. The *i*-th line contains two integers *x**i* and *y**i* (0<=≤<=*x**i*,<=*y**i*<=≤<=105) — the number of low quality photos and high quality photos the *i*-th client wants, correspondingly.
All numbers on all lines are separated by single spaces.
Output Specification:
On the first line print the answer to the problem — the maximum number of clients that Valera can successfully serve. Print on the second line the numbers of the client in any order. All numbers must be distinct. If there are multiple answers, print any of them. The clients are numbered starting with 1 in the order in which they are defined in the input data.
Demo Input:
['3 10\n2 3\n1 4\n2 1\n1 0\n', '3 6\n6 6\n1 1\n1 0\n1 0\n']
Demo Output:
['2\n3 2 ', '1\n2 ']
Note:
none | ```python
class Pair:
def __init__(self,sumOfNumber,index):
self.sumOfNumber=sumOfNumber
self.index=index
def __lt__(self, other):
return self.sumOfNumber < other.sumOfNumber
n,d=map(int,input().split())
a,b=map(int,input().split())
s=[None]*n
for i in range(n):
x,y=map(int,input().split())
cur=Pair(0,0)
cur.sumOfNumber=(a*x)+(b*y)
cur.index=i+1
s[i]=cur
s.sort()
cur=0
while cur<n and d-s[cur].sumOfNumber >= 0:
d-=s[cur].sumOfNumber
cur+=1
print(cur)
for i in range(cur):
print(s[i].index,end=" ")
``` | 0 | |
628 | A | Tennis Tournament | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | A tennis tournament with *n* participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.
The tournament takes place in the following way (below, *m* is the number of the participants of the current round):
- let *k* be the maximal power of the number 2 such that *k*<=≤<=*m*, - *k* participants compete in the current round and a half of them passes to the next round, the other *m*<=-<=*k* participants pass to the next round directly, - when only one participant remains, the tournament finishes.
Each match requires *b* bottles of water for each participant and one bottle for the judge. Besides *p* towels are given to each participant for the whole tournament.
Find the number of bottles and towels needed for the tournament.
Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose). | The only line contains three integers *n*,<=*b*,<=*p* (1<=≤<=*n*,<=*b*,<=*p*<=≤<=500) — the number of participants and the parameters described in the problem statement. | Print two integers *x* and *y* — the number of bottles and towels need for the tournament. | [
"5 2 3\n",
"8 2 4\n"
] | [
"20 15\n",
"35 32\n"
] | In the first example will be three rounds:
1. in the first round will be two matches and for each match 5 bottles of water are needed (two for each of the participants and one for the judge), 1. in the second round will be only one match, so we need another 5 bottles of water, 1. in the third round will also be only one match, so we need another 5 bottles of water.
So in total we need 20 bottles of water.
In the second example no participant will move on to some round directly. | 0 | [
{
"input": "5 2 3",
"output": "20 15"
},
{
"input": "8 2 4",
"output": "35 32"
},
{
"input": "10 1 500",
"output": "27 5000"
},
{
"input": "20 500 1",
"output": "19019 20"
},
{
"input": "100 123 99",
"output": "24453 9900"
},
{
"input": "500 1 1",
... | 1,455,904,238 | 2,147,483,647 | PyPy 3 | OK | TESTS | 38 | 108 | 0 | n,b,p = map(int,input().split())
print((n-1)*(2*b+1),n*p) | Title: Tennis Tournament
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tennis tournament with *n* participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.
The tournament takes place in the following way (below, *m* is the number of the participants of the current round):
- let *k* be the maximal power of the number 2 such that *k*<=≤<=*m*, - *k* participants compete in the current round and a half of them passes to the next round, the other *m*<=-<=*k* participants pass to the next round directly, - when only one participant remains, the tournament finishes.
Each match requires *b* bottles of water for each participant and one bottle for the judge. Besides *p* towels are given to each participant for the whole tournament.
Find the number of bottles and towels needed for the tournament.
Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose).
Input Specification:
The only line contains three integers *n*,<=*b*,<=*p* (1<=≤<=*n*,<=*b*,<=*p*<=≤<=500) — the number of participants and the parameters described in the problem statement.
Output Specification:
Print two integers *x* and *y* — the number of bottles and towels need for the tournament.
Demo Input:
['5 2 3\n', '8 2 4\n']
Demo Output:
['20 15\n', '35 32\n']
Note:
In the first example will be three rounds:
1. in the first round will be two matches and for each match 5 bottles of water are needed (two for each of the participants and one for the judge), 1. in the second round will be only one match, so we need another 5 bottles of water, 1. in the third round will also be only one match, so we need another 5 bottles of water.
So in total we need 20 bottles of water.
In the second example no participant will move on to some round directly. | ```python
n,b,p = map(int,input().split())
print((n-1)*(2*b+1),n*p)
``` | 3 | |
570 | A | Elections | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate.
The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.
At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.
Determine who will win the elections. | The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively.
Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*.
It is guaranteed that the total number of people in all the cities does not exceed 109. | Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one. | [
"3 3\n1 2 3\n2 3 1\n1 2 1\n",
"3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n"
] | [
"2",
"1"
] | Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.
Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index. | 500 | [
{
"input": "3 3\n1 2 3\n2 3 1\n1 2 1",
"output": "2"
},
{
"input": "3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7",
"output": "1"
},
{
"input": "1 3\n5\n3\n2",
"output": "1"
},
{
"input": "3 1\n1 2 3",
"output": "3"
},
{
"input": "3 1\n100 100 100",
"output": "1"
},
{... | 1,640,247,833 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 31 | 0 | n,m=[int(x) for x in input().split()]
a=[0]*n
mmax,ppos=0,-1
for i in range(m):
b=[int(x) for x in input().split()]
a[b.index(max(b))]+=1
if a[b.index(max(b))]>mmax:
mmax=a[b.index(max(b))]
ppos=b.index(max(b))
print(ppos+1) | Title: Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate.
The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.
At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.
Determine who will win the elections.
Input Specification:
The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively.
Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*.
It is guaranteed that the total number of people in all the cities does not exceed 109.
Output Specification:
Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.
Demo Input:
['3 3\n1 2 3\n2 3 1\n1 2 1\n', '3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n']
Demo Output:
['2', '1']
Note:
Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.
Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index. | ```python
n,m=[int(x) for x in input().split()]
a=[0]*n
mmax,ppos=0,-1
for i in range(m):
b=[int(x) for x in input().split()]
a[b.index(max(b))]+=1
if a[b.index(max(b))]>mmax:
mmax=a[b.index(max(b))]
ppos=b.index(max(b))
print(ppos+1)
``` | 0 | |
405 | A | Gravity Flip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch! | The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column. | Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch. | [
"4\n3 2 1 2\n",
"3\n2 3 8\n"
] | [
"1 2 2 3 \n",
"2 3 8 \n"
] | The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | 500 | [
{
"input": "4\n3 2 1 2",
"output": "1 2 2 3 "
},
{
"input": "3\n2 3 8",
"output": "2 3 8 "
},
{
"input": "5\n2 1 2 1 2",
"output": "1 1 2 2 2 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n4 3",
"output": "3 4 "
},
{
"input": "6\n100 40 60 20... | 1,699,454,502 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | n=int(input())
a=input().split()
for i in range (n) :
a[i]=int(a[i])
a.sort();
for i in range (n) :
print(a[i],end=' ')
| Title: Gravity Flip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch!
Input Specification:
The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column.
Output Specification:
Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch.
Demo Input:
['4\n3 2 1 2\n', '3\n2 3 8\n']
Demo Output:
['1 2 2 3 \n', '2 3 8 \n']
Note:
The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | ```python
n=int(input())
a=input().split()
for i in range (n) :
a[i]=int(a[i])
a.sort();
for i in range (n) :
print(a[i],end=' ')
``` | 3 | |
553 | A | Kyoya and Colored Balls | PROGRAMMING | 1,500 | [
"combinatorics",
"dp",
"math"
] | null | null | Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen. | The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors.
Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000).
The total number of balls doesn't exceed 1000. | A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007. | [
"3\n2\n2\n1\n",
"4\n1\n2\n3\n4\n"
] | [
"3\n",
"1680\n"
] | In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are: | 250 | [
{
"input": "3\n2\n2\n1",
"output": "3"
},
{
"input": "4\n1\n2\n3\n4",
"output": "1680"
},
{
"input": "10\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "12520708"
},
{
"input": "5\n10\n10\n10\n10\n10",
"output": "425711769"
},
{
"input": "11\n291\n3... | 1,670,374,283 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 46 | 0 | from math import comb
dec=10**9 + 7
k =int(input())
color_counts = [0] * k
for i in range(0, k):
color_counts[i] = int(input())
result = 1
curr = color_counts[0]
for i in range(1,k):
result = result * comb(curr + color_counts[i] - 1 , color_counts[i] - 1) % dec
curr = curr + color_counts[i]
print(result) | Title: Kyoya and Colored Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen.
Input Specification:
The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors.
Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000).
The total number of balls doesn't exceed 1000.
Output Specification:
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007.
Demo Input:
['3\n2\n2\n1\n', '4\n1\n2\n3\n4\n']
Demo Output:
['3\n', '1680\n']
Note:
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are: | ```python
from math import comb
dec=10**9 + 7
k =int(input())
color_counts = [0] * k
for i in range(0, k):
color_counts[i] = int(input())
result = 1
curr = color_counts[0]
for i in range(1,k):
result = result * comb(curr + color_counts[i] - 1 , color_counts[i] - 1) % dec
curr = curr + color_counts[i]
print(result)
``` | 3 | |
768 | A | Oath of the Night's Watch | PROGRAMMING | 900 | [
"constructive algorithms",
"sortings"
] | null | null | "Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support? | First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards. | Output a single integer representing the number of stewards which Jon will feed. | [
"2\n1 5\n",
"3\n1 2 5\n"
] | [
"0",
"1"
] | In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | 500 | [
{
"input": "2\n1 5",
"output": "0"
},
{
"input": "3\n1 2 5",
"output": "1"
},
{
"input": "4\n1 2 3 4",
"output": "2"
},
{
"input": "8\n7 8 9 4 5 6 1 2",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n100",
"output": "0"
},
... | 1,659,488,186 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 13 | 109 | 8,396,800 | # ridiculous
n = int(input())
a = sorted(list(map(int,input().split())))
print(n - a.count(a[0]) - a.count(a[-1]))
# bruh, wtf, runtime error
# @ isn't looking at your submission it cheating, what?
# You won't be able to see the tests before the contest ends, but if you've done the problem in practice you can see tests by clicking on the number of your ...
| Title: Oath of the Night's Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input Specification:
First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards.
Output Specification:
Output a single integer representing the number of stewards which Jon will feed.
Demo Input:
['2\n1 5\n', '3\n1 2 5\n']
Demo Output:
['0', '1']
Note:
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | ```python
# ridiculous
n = int(input())
a = sorted(list(map(int,input().split())))
print(n - a.count(a[0]) - a.count(a[-1]))
# bruh, wtf, runtime error
# @ isn't looking at your submission it cheating, what?
# You won't be able to see the tests before the contest ends, but if you've done the problem in practice you can see tests by clicking on the number of your ...
``` | 0 | |
812 | B | Sagheer, the Hausmeister | PROGRAMMING | 1,600 | [
"bitmasks",
"brute force",
"dp"
] | null | null | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of *n* floors with stairs at the left and the right sides. Each floor has *m* rooms on the same line with a corridor that connects the left and right stairs passing by all the rooms. In other words, the building can be represented as a rectangle with *n* rows and *m*<=+<=2 columns, where the first and the last columns represent the stairs, and the *m* columns in the middle represent rooms.
Sagheer is standing at the ground floor at the left stairs. He wants to turn all the lights off in such a way that he will not go upstairs until all lights in the floor he is standing at are off. Of course, Sagheer must visit a room to turn the light there off. It takes one minute for Sagheer to go to the next floor using stairs or to move from the current room/stairs to a neighboring room/stairs on the same floor. It takes no time for him to switch the light off in the room he is currently standing in. Help Sagheer find the minimum total time to turn off all the lights.
Note that Sagheer does not have to go back to his starting position, and he does not have to visit rooms where the light is already switched off. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=15 and 1<=≤<=*m*<=≤<=100) — the number of floors and the number of rooms in each floor, respectively.
The next *n* lines contains the building description. Each line contains a binary string of length *m*<=+<=2 representing a floor (the left stairs, then *m* rooms, then the right stairs) where 0 indicates that the light is off and 1 indicates that the light is on. The floors are listed from top to bottom, so that the last line represents the ground floor.
The first and last characters of each string represent the left and the right stairs, respectively, so they are always 0. | Print a single integer — the minimum total time needed to turn off all the lights. | [
"2 2\n0010\n0100\n",
"3 4\n001000\n000010\n000010\n",
"4 3\n01110\n01110\n01110\n01110\n"
] | [
"5\n",
"12\n",
"18\n"
] | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to the second room in the last floor.
In the third example, he will walk through the whole corridor alternating between the left and right stairs at each floor. | 1,000 | [
{
"input": "2 2\n0010\n0100",
"output": "5"
},
{
"input": "3 4\n001000\n000010\n000010",
"output": "12"
},
{
"input": "4 3\n01110\n01110\n01110\n01110",
"output": "18"
},
{
"input": "3 2\n0000\n0100\n0100",
"output": "4"
},
{
"input": "1 89\n0000000000000000000000... | 1,579,170,756 | 2,147,483,647 | PyPy 3 | OK | TESTS | 72 | 155 | 1,228,800 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
n,m=map(int,input().strip().split())
a=[]
L=[]
R=[]
t={}
for i in range(n):
b=input()
a.append(b)
L.append(9000000)
R.append(0)
t[i]=0
a.reverse()
for i in range(n):
for j in range(m+2):
if a[i][j]=='1':
L[i]=min(L[i],j)
R[i]=max(R[i],j)
t[i]=1
L1=0
R1=9000000
ans=0
for i in range(n):
if t[i]==0:
L1+=1
R1+=1
continue
ans=min(L1+R[i],R1+m+1-L[i])
L2=min(L1+R[i]*2+1,R1+m+2)
R2=min(R1+(m+1-L[i])*2+1,m+2+L1)
L1=L2
R1=R2
print(ans)
| Title: Sagheer, the Hausmeister
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of *n* floors with stairs at the left and the right sides. Each floor has *m* rooms on the same line with a corridor that connects the left and right stairs passing by all the rooms. In other words, the building can be represented as a rectangle with *n* rows and *m*<=+<=2 columns, where the first and the last columns represent the stairs, and the *m* columns in the middle represent rooms.
Sagheer is standing at the ground floor at the left stairs. He wants to turn all the lights off in such a way that he will not go upstairs until all lights in the floor he is standing at are off. Of course, Sagheer must visit a room to turn the light there off. It takes one minute for Sagheer to go to the next floor using stairs or to move from the current room/stairs to a neighboring room/stairs on the same floor. It takes no time for him to switch the light off in the room he is currently standing in. Help Sagheer find the minimum total time to turn off all the lights.
Note that Sagheer does not have to go back to his starting position, and he does not have to visit rooms where the light is already switched off.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=15 and 1<=≤<=*m*<=≤<=100) — the number of floors and the number of rooms in each floor, respectively.
The next *n* lines contains the building description. Each line contains a binary string of length *m*<=+<=2 representing a floor (the left stairs, then *m* rooms, then the right stairs) where 0 indicates that the light is off and 1 indicates that the light is on. The floors are listed from top to bottom, so that the last line represents the ground floor.
The first and last characters of each string represent the left and the right stairs, respectively, so they are always 0.
Output Specification:
Print a single integer — the minimum total time needed to turn off all the lights.
Demo Input:
['2 2\n0010\n0100\n', '3 4\n001000\n000010\n000010\n', '4 3\n01110\n01110\n01110\n01110\n']
Demo Output:
['5\n', '12\n', '18\n']
Note:
In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to the second room in the last floor.
In the third example, he will walk through the whole corridor alternating between the left and right stairs at each floor. | ```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n,m=map(int,input().strip().split())
a=[]
L=[]
R=[]
t={}
for i in range(n):
b=input()
a.append(b)
L.append(9000000)
R.append(0)
t[i]=0
a.reverse()
for i in range(n):
for j in range(m+2):
if a[i][j]=='1':
L[i]=min(L[i],j)
R[i]=max(R[i],j)
t[i]=1
L1=0
R1=9000000
ans=0
for i in range(n):
if t[i]==0:
L1+=1
R1+=1
continue
ans=min(L1+R[i],R1+m+1-L[i])
L2=min(L1+R[i]*2+1,R1+m+2)
R2=min(R1+(m+1-L[i])*2+1,m+2+L1)
L1=L2
R1=R2
print(ans)
``` | 3 | |
978 | B | File Name | PROGRAMMING | 800 | [
"greedy",
"strings"
] | null | null | You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.
Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".
You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii". | The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name.
The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name. | Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0. | [
"6\nxxxiii\n",
"5\nxxoxx\n",
"10\nxxxxxxxxxx\n"
] | [
"1\n",
"0\n",
"8\n"
] | In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters. | 0 | [
{
"input": "6\nxxxiii",
"output": "1"
},
{
"input": "5\nxxoxx",
"output": "0"
},
{
"input": "10\nxxxxxxxxxx",
"output": "8"
},
{
"input": "100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "98"
},
{
... | 1,699,200,292 | 2,147,483,647 | Python 3 | OK | TESTS | 13 | 31 | 0 | n = int(input())
file_name = input()
count_x,removals = 0,0
for i in range(n):
if file_name[i] == 'x':
count_x += 1
if count_x >= 3:removals += 1
else:count_x = 0
print(removals) | Title: File Name
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.
Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".
You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii".
Input Specification:
The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name.
The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name.
Output Specification:
Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.
Demo Input:
['6\nxxxiii\n', '5\nxxoxx\n', '10\nxxxxxxxxxx\n']
Demo Output:
['1\n', '0\n', '8\n']
Note:
In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters. | ```python
n = int(input())
file_name = input()
count_x,removals = 0,0
for i in range(n):
if file_name[i] == 'x':
count_x += 1
if count_x >= 3:removals += 1
else:count_x = 0
print(removals)
``` | 3 | |
355 | A | Vasya and Digital Root | PROGRAMMING | 1,100 | [
"constructive algorithms",
"implementation"
] | null | null | Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to:
1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=≥<=10.
For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=≤<=101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist. | The first line contains two integers *k* and *d* (1<=≤<=*k*<=≤<=1000; 0<=≤<=*d*<=≤<=9). | In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes. | [
"4 4\n",
"5 1\n",
"1 0\n"
] | [
"5881\n",
"36172\n",
"0\n"
] | For the first test sample *dr*(5881) = *dr*(22) = 4.
For the second test sample *dr*(36172) = *dr*(19) = *dr*(10) = 1. | 500 | [
{
"input": "4 4",
"output": "5881"
},
{
"input": "5 1",
"output": "36172"
},
{
"input": "1 0",
"output": "0"
},
{
"input": "8 7",
"output": "49722154"
},
{
"input": "487 0",
"output": "No solution"
},
{
"input": "1000 5",
"output": "854193955406789... | 1,587,392,158 | 2,147,483,647 | PyPy 3 | OK | TESTS | 20 | 140 | 0 | k,d = input().split()
k,d = int(k), int(d)
if k==1:
print(d)
elif d!=0:
s = '1'*(k-1)
m = int(s)
m = 9*m + d
print(m)
else:
print("No solution")
| Title: Vasya and Digital Root
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has recently found out what a digital root of a number is and he decided to share his knowledge with you.
Let's assume that *S*(*n*) is the sum of digits of number *n*, for example, *S*(4098)<==<=4<=+<=0<=+<=9<=+<=8<==<=21. Then the digital root of number *n* equals to:
1. *dr*(*n*)<==<=*S*(*n*), if *S*(*n*)<=<<=10; 1. *dr*(*n*)<==<=*dr*(<=*S*(*n*)<=), if *S*(*n*)<=≥<=10.
For example, *dr*(4098)<=<==<=<=*dr*(21)<=<==<=<=3.
Vasya is afraid of large numbers, so the numbers he works with are at most 101000. For all such numbers, he has proved that *dr*(*n*)<=<==<=<=*S*(<=*S*(<=*S*(<=*S*(*n*)<=)<=)<=) (*n*<=≤<=101000).
Now Vasya wants to quickly find numbers with the given digital root. The problem is, he hasn't learned how to do that and he asked you to help him. You task is, given numbers *k* and *d*, find the number consisting of exactly *k* digits (the leading zeroes are not allowed), with digital root equal to *d*, or else state that such number does not exist.
Input Specification:
The first line contains two integers *k* and *d* (1<=≤<=*k*<=≤<=1000; 0<=≤<=*d*<=≤<=9).
Output Specification:
In a single line print either any number that meets the requirements (without the leading zeroes) or "No solution" (without the quotes), if the corresponding number does not exist.
The chosen number must consist of exactly *k* digits. We assume that number 0 doesn't contain any leading zeroes.
Demo Input:
['4 4\n', '5 1\n', '1 0\n']
Demo Output:
['5881\n', '36172\n', '0\n']
Note:
For the first test sample *dr*(5881) = *dr*(22) = 4.
For the second test sample *dr*(36172) = *dr*(19) = *dr*(10) = 1. | ```python
k,d = input().split()
k,d = int(k), int(d)
if k==1:
print(d)
elif d!=0:
s = '1'*(k-1)
m = int(s)
m = 9*m + d
print(m)
else:
print("No solution")
``` | 3 | |
687 | A | NP-Hard Problem | PROGRAMMING | 1,500 | [
"dfs and similar",
"graphs"
] | null | null | Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.
Suppose the graph *G* is given. Subset *A* of its vertices is called a vertex cover of this graph, if for each edge *uv* there is at least one endpoint of it in this set, i.e. or (or both).
Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.
They have agreed to give you their graph and you need to find two disjoint subsets of its vertices *A* and *B*, such that both *A* and *B* are vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself). | The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — the number of vertices and the number of edges in the prize graph, respectively.
Each of the next *m* lines contains a pair of integers *u**i* and *v**i* (1<=<=≤<=<=*u**i*,<=<=*v**i*<=<=≤<=<=*n*), denoting an undirected edge between *u**i* and *v**i*. It's guaranteed the graph won't contain any self-loops or multiple edges. | If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).
If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer *k* denoting the number of vertices in that vertex cover, and the second line contains *k* integers — the indices of vertices. Note that because of *m*<=≥<=1, vertex cover cannot be empty. | [
"4 2\n1 2\n2 3\n",
"3 3\n1 2\n2 3\n1 3\n"
] | [
"1\n2 \n2\n1 3 \n",
"-1\n"
] | In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to Pari and keep vertex number 4 for yourself (or give it someone, if you wish).
In the second sample, there is no way to satisfy both Pari and Arya. | 500 | [
{
"input": "4 2\n1 2\n2 3",
"output": "1\n2 \n2\n1 3 "
},
{
"input": "3 3\n1 2\n2 3\n1 3",
"output": "-1"
},
{
"input": "5 7\n3 2\n5 4\n3 4\n1 3\n1 5\n1 4\n2 5",
"output": "-1"
},
{
"input": "10 11\n4 10\n8 10\n2 3\n2 4\n7 1\n8 5\n2 8\n7 2\n1 2\n2 9\n6 8",
"output": "-1"
... | 1,657,204,646 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 14 | 467 | 20,582,400 | from re import I
import sys
input = sys.stdin.readline
from math import *
def mrd(): return [int(x) for x in input().split()]
def rd(): return int(input())
MAXN = 2 * 10**5 + 5
INF = 10**16 * 2
mod = 10**9 + 7
#----------------------------------------------------------------------------------#
def solve():
n,m = mrd()
adj = [[] for _ in range(n + 1)]
for _ in range(m):
u,v = mrd()
adj[u].append(v)
adj[v].append(u)
col = [0 for _ in range(n + 1)]
def chk(u,c):
col[u] = c
for v in adj[u]:
if col[u] == col[v]:
return False
if not col[v] and not chk(v,3 - c):
return False
return True
for i in range(1,n + 1):
if not col[i]:
if not chk(i,1):
print(-1)
return
def wr(c):
ans = [i for i in range(1,n + 1) if col[i] == c]
print(len(ans))
for x in ans:
print(x,end=" ")
print("")
wr(1)
wr(2)
if __name__ == "__main__":
solve()
| Title: NP-Hard Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.
Suppose the graph *G* is given. Subset *A* of its vertices is called a vertex cover of this graph, if for each edge *uv* there is at least one endpoint of it in this set, i.e. or (or both).
Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.
They have agreed to give you their graph and you need to find two disjoint subsets of its vertices *A* and *B*, such that both *A* and *B* are vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).
Input Specification:
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100<=000, 1<=≤<=*m*<=≤<=100<=000) — the number of vertices and the number of edges in the prize graph, respectively.
Each of the next *m* lines contains a pair of integers *u**i* and *v**i* (1<=<=≤<=<=*u**i*,<=<=*v**i*<=<=≤<=<=*n*), denoting an undirected edge between *u**i* and *v**i*. It's guaranteed the graph won't contain any self-loops or multiple edges.
Output Specification:
If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).
If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer *k* denoting the number of vertices in that vertex cover, and the second line contains *k* integers — the indices of vertices. Note that because of *m*<=≥<=1, vertex cover cannot be empty.
Demo Input:
['4 2\n1 2\n2 3\n', '3 3\n1 2\n2 3\n1 3\n']
Demo Output:
['1\n2 \n2\n1 3 \n', '-1\n']
Note:
In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to Pari and keep vertex number 4 for yourself (or give it someone, if you wish).
In the second sample, there is no way to satisfy both Pari and Arya. | ```python
from re import I
import sys
input = sys.stdin.readline
from math import *
def mrd(): return [int(x) for x in input().split()]
def rd(): return int(input())
MAXN = 2 * 10**5 + 5
INF = 10**16 * 2
mod = 10**9 + 7
#----------------------------------------------------------------------------------#
def solve():
n,m = mrd()
adj = [[] for _ in range(n + 1)]
for _ in range(m):
u,v = mrd()
adj[u].append(v)
adj[v].append(u)
col = [0 for _ in range(n + 1)]
def chk(u,c):
col[u] = c
for v in adj[u]:
if col[u] == col[v]:
return False
if not col[v] and not chk(v,3 - c):
return False
return True
for i in range(1,n + 1):
if not col[i]:
if not chk(i,1):
print(-1)
return
def wr(c):
ans = [i for i in range(1,n + 1) if col[i] == c]
print(len(ans))
for x in ans:
print(x,end=" ")
print("")
wr(1)
wr(2)
if __name__ == "__main__":
solve()
``` | -1 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,678,493,057 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 46 | 0 | inp = input()
inp = inp.split()
for i in range(3):
inp[i] = int(inp[i])
D1 = inp[0]/inp[2]
if D1 > int(D1):
D1 = int(D1)+1
else:
D1 = int(D1)
D2 = inp[1]/inp[2]
if D2 > int(D2):
D2 = int(D2)+1
else:
D2 = int(D2)
num_b = D1*D2
print(num_b)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
inp = input()
inp = inp.split()
for i in range(3):
inp[i] = int(inp[i])
D1 = inp[0]/inp[2]
if D1 > int(D1):
D1 = int(D1)+1
else:
D1 = int(D1)
D2 = inp[1]/inp[2]
if D2 > int(D2):
D2 = int(D2)+1
else:
D2 = int(D2)
num_b = D1*D2
print(num_b)
``` | 3.977 |
492 | B | Vanya and Lanterns | PROGRAMMING | 1,200 | [
"binary search",
"implementation",
"math",
"sortings"
] | null | null | Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street? | The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street. | Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9. | [
"7 15\n15 5 3 7 9 14 0\n",
"2 5\n2 5\n"
] | [
"2.5000000000\n",
"2.0000000000\n"
] | Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit. | 1,000 | [
{
"input": "7 15\n15 5 3 7 9 14 0",
"output": "2.5000000000"
},
{
"input": "2 5\n2 5",
"output": "2.0000000000"
},
{
"input": "46 615683844\n431749087 271781274 274974690 324606253 480870261 401650581 13285442 478090364 266585394 425024433 588791449 492057200 391293435 563090494 317950 1... | 1,693,639,838 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | n, l = map(int, input().split())
lanterns = list(map(int, input().split()))
lanterns.sort()
max_gap = 0
for i in range(1, n):
max_gap = max(max_gap, lanterns[i] - lanterns[i - 1])
radius = max(max(lanterns[0], l - lanterns[-1]), max_gap / 2)
print("{:.10f}".format(radius))
| Title: Vanya and Lanterns
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?
Input Specification:
The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
Output Specification:
Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.
Demo Input:
['7 15\n15 5 3 7 9 14 0\n', '2 5\n2 5\n']
Demo Output:
['2.5000000000\n', '2.0000000000\n']
Note:
Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit. | ```python
n, l = map(int, input().split())
lanterns = list(map(int, input().split()))
lanterns.sort()
max_gap = 0
for i in range(1, n):
max_gap = max(max_gap, lanterns[i] - lanterns[i - 1])
radius = max(max(lanterns[0], l - lanterns[-1]), max_gap / 2)
print("{:.10f}".format(radius))
``` | 3 | |
762 | A | k-th divisor | PROGRAMMING | 1,400 | [
"math",
"number theory"
] | null | null | You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109). | If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*. | [
"4 2\n",
"5 3\n",
"12 5\n"
] | [
"2\n",
"-1\n",
"6\n"
] | In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | 0 | [
{
"input": "4 2",
"output": "2"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "12 5",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "866421317361600 26880",
"output": "866421317361600"
},
{
"input": "866421317361600 26881",
"ou... | 1,679,140,462 | 2,147,483,647 | PyPy 3 | OK | TESTS | 96 | 935 | 4,608,000 | n, k = map(int, input().split())
div = set()
for i in range(1, int(n**0.5)+1):
if n%i == 0:
div.add(i)
div.add(n//i)
if len(div) >= k:
div = sorted(div)
print(div[k-1])
else:
print("-1")
| Title: k-th divisor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109).
Output Specification:
If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*.
Demo Input:
['4 2\n', '5 3\n', '12 5\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | ```python
n, k = map(int, input().split())
div = set()
for i in range(1, int(n**0.5)+1):
if n%i == 0:
div.add(i)
div.add(n//i)
if len(div) >= k:
div = sorted(div)
print(div[k-1])
else:
print("-1")
``` | 3 | |
597 | B | Restaurant | PROGRAMMING | 1,600 | [
"dp",
"greedy",
"sortings"
] | null | null | A restaurant received *n* orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the *i*-th order is characterized by two time values — the start time *l**i* and the finish time *r**i* (*l**i*<=≤<=*r**i*).
Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?
No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both. | The first line contains integer number *n* (1<=≤<=*n*<=≤<=5·105) — number of orders. The following *n* lines contain integer values *l**i* and *r**i* each (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109). | Print the maximal number of orders that can be accepted. | [
"2\n7 11\n4 7\n",
"5\n1 2\n2 3\n3 4\n4 5\n5 6\n",
"6\n4 8\n1 5\n4 7\n2 5\n1 3\n6 8\n"
] | [
"1\n",
"3\n",
"2\n"
] | none | 1,000 | [
{
"input": "2\n7 11\n4 7",
"output": "1"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5\n5 6",
"output": "3"
},
{
"input": "6\n4 8\n1 5\n4 7\n2 5\n1 3\n6 8",
"output": "2"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "2\n4 6\n4 8",
"output": "1"
},
{
"inp... | 1,600,145,559 | 2,147,483,647 | Python 3 | OK | TESTS | 70 | 2,324 | 40,857,600 | n=int(input())
lis=[]
for i in range(n):
s,f=map(int,input().split())
lis.append([f,s])
lis.sort()
#now we get sorted wrt b and sorted with a aequentially
count=1
#count is already 1
maxl=lis[0][0]
#taking b as max ,strating with b
for i in range(1,len(lis)):
if lis[i][1]>maxl: #checking for a
count=count+1
maxl=lis[i][0] #updating b
print(count) | Title: Restaurant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A restaurant received *n* orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the *i*-th order is characterized by two time values — the start time *l**i* and the finish time *r**i* (*l**i*<=≤<=*r**i*).
Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?
No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=5·105) — number of orders. The following *n* lines contain integer values *l**i* and *r**i* each (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109).
Output Specification:
Print the maximal number of orders that can be accepted.
Demo Input:
['2\n7 11\n4 7\n', '5\n1 2\n2 3\n3 4\n4 5\n5 6\n', '6\n4 8\n1 5\n4 7\n2 5\n1 3\n6 8\n']
Demo Output:
['1\n', '3\n', '2\n']
Note:
none | ```python
n=int(input())
lis=[]
for i in range(n):
s,f=map(int,input().split())
lis.append([f,s])
lis.sort()
#now we get sorted wrt b and sorted with a aequentially
count=1
#count is already 1
maxl=lis[0][0]
#taking b as max ,strating with b
for i in range(1,len(lis)):
if lis[i][1]>maxl: #checking for a
count=count+1
maxl=lis[i][0] #updating b
print(count)
``` | 3 | |
835 | A | Key races | PROGRAMMING | 800 | [
"math"
] | null | null | Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game. | The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant. | If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship". | [
"5 1 2 1 2\n",
"3 3 1 1 1\n",
"4 5 3 1 5\n"
] | [
"First\n",
"Second\n",
"Friendship\n"
] | In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | 500 | [
{
"input": "5 1 2 1 2",
"output": "First"
},
{
"input": "3 3 1 1 1",
"output": "Second"
},
{
"input": "4 5 3 1 5",
"output": "Friendship"
},
{
"input": "1000 1000 1000 1000 1000",
"output": "Friendship"
},
{
"input": "1 1 1 1 1",
"output": "Friendship"
},
... | 1,571,737,924 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 124 | 0 | a,b,c,d,e=map(int,input().split())
sum1=sum2=0
sum1=(2*d)+(b*a)
sum2=(c*a)+(2*e)
if(sum1<sum2):
print("First")
elif sum1==sum2:
print("Friendship")
else:
print("Second")
| Title: Key races
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game.
Input Specification:
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
Output Specification:
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
Demo Input:
['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n']
Demo Output:
['First\n', 'Second\n', 'Friendship\n']
Note:
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | ```python
a,b,c,d,e=map(int,input().split())
sum1=sum2=0
sum1=(2*d)+(b*a)
sum2=(c*a)+(2*e)
if(sum1<sum2):
print("First")
elif sum1==sum2:
print("Friendship")
else:
print("Second")
``` | 3 | |
591 | A | Wizards' Duel | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight. | The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively. | Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if . | [
"100\n50\n50\n",
"199\n60\n40\n"
] | [
"50\n",
"119.4\n"
] | In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor. | 500 | [
{
"input": "100\n50\n50",
"output": "50"
},
{
"input": "199\n60\n40",
"output": "119.4"
},
{
"input": "1\n1\n1",
"output": "0.5"
},
{
"input": "1\n1\n500",
"output": "0.001996007984"
},
{
"input": "1\n500\n1",
"output": "0.998003992"
},
{
"input": "1\n... | 1,540,764,926 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 124 | 0 | l=int(input())
p=int(input())
q=int(input())
t=1+q/p
x=l/t
print(x) | Title: Wizards' Duel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
Input Specification:
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
Output Specification:
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['100\n50\n50\n', '199\n60\n40\n']
Demo Output:
['50\n', '119.4\n']
Note:
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor. | ```python
l=int(input())
p=int(input())
q=int(input())
t=1+q/p
x=l/t
print(x)
``` | 3 | |
66 | B | Petya and Countryside | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | B. Petya and Countryside | 2 | 256 | Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1<=×<=*n* in size, when viewed from above. This rectangle is divided into *n* equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1<=×<=5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000. | Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section. | [
"1\n2\n",
"5\n1 2 1 2 1\n",
"8\n1 2 1 1 1 3 3 4\n"
] | [
"1\n",
"3\n",
"6\n"
] | none | 1,000 | [
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 2 1 2 1",
"output": "3"
},
{
"input": "8\n1 2 1 1 1 3 3 4",
"output": "6"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "10"
},
{
"input... | 1,587,633,640 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 218 | 307,200 | n=int(input())
a=list(map(int,input().split()))
ans=1
low=set()
high=set()
for i in range(len(a)-1):
t=a[i]
j=i-1
k=i+1
p=1
while j!=-1 and a[j]<t:
j-=1
p+=1
while k!=n and a[k]<t:
k+=1
p+=1
ans=max(ans,p)
print(ans)
| Title: Petya and Countryside
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1<=×<=*n* in size, when viewed from above. This rectangle is divided into *n* equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1<=×<=5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
Output Specification:
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
Demo Input:
['1\n2\n', '5\n1 2 1 2 1\n', '8\n1 2 1 1 1 3 3 4\n']
Demo Output:
['1\n', '3\n', '6\n']
Note:
none | ```python
n=int(input())
a=list(map(int,input().split()))
ans=1
low=set()
high=set()
for i in range(len(a)-1):
t=a[i]
j=i-1
k=i+1
p=1
while j!=-1 and a[j]<t:
j-=1
p+=1
while k!=n and a[k]<t:
k+=1
p+=1
ans=max(ans,p)
print(ans)
``` | 0 |
716 | B | Complete the Word | PROGRAMMING | 1,300 | [
"greedy",
"two pointers"
] | null | null | ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him? | The first and only line of the input contains a single string *s* (1<=≤<=|*s*|<=≤<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember. | If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them. | [
"ABC??FGHIJK???OPQR?TUVWXY?\n",
"WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n",
"??????????????????????????\n",
"AABCDEFGHIJKLMNOPQRSTUVW??M\n"
] | [
"ABCDEFGHIJKLMNOPQRZTUVWXYS",
"-1",
"MNBVCXZLKJHGFDSAQPWOEIRUYT",
"-1"
] | In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is - 1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer. | 1,000 | [
{
"input": "ABC??FGHIJK???OPQR?TUVWXY?",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO",
"output": "-1"
},
{
"input": "??????????????????????????",
"output": "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
},
{
"input": "AABCDEFGHIJKLMNO... | 1,663,786,007 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 31 | 0 | def st(i) :
chk = [chr(i+65) for i in range(26)]; chk.sort()
alph = [chr(i+65) for i in range(26)]; ss = list(s[i:i+27]); j= 0
for x in ss :
if x in alph: alph.remove(x)
for x in range(26) :
if ss[x] == '?' : ss[x] = alph[j]; j+=1
ss.sort()
if ss == chk : return ''.join(ss)
else : return 0
s=input(); n=len(s)
for i in range(n-25) :
n=st(i)
if n : print(n); break
else : print(-1)
| Title: Complete the Word
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 26, no such substring exists and thus it is not nice.
Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?
Input Specification:
The first and only line of the input contains a single string *s* (1<=≤<=|*s*|<=≤<=50<=000), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.
Output Specification:
If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print <=-<=1 in the only line.
Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.
If there are multiple solutions, you may print any of them.
Demo Input:
['ABC??FGHIJK???OPQR?TUVWXY?\n', 'WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO\n', '??????????????????????????\n', 'AABCDEFGHIJKLMNOPQRSTUVW??M\n']
Demo Output:
['ABCDEFGHIJKLMNOPQRZTUVWXYS', '-1', 'MNBVCXZLKJHGFDSAQPWOEIRUYT', '-1']
Note:
In the first sample case, ABCDEFGHIJKLMNOPQRZTUVWXYS is a valid answer beacuse it contains a substring of length 26 (the whole string in this case) which contains all the letters of the English alphabet exactly once. Note that there are many possible solutions, such as ABCDEFGHIJKLMNOPQRSTUVWXYZ or ABCEDFGHIJKLMNOPQRZTUVWXYS.
In the second sample case, there are no missing letters. In addition, the given string does not have a substring of length 26 that contains all the letters of the alphabet, so the answer is - 1.
In the third sample case, any string of length 26 that contains all letters of the English alphabet fits as an answer. | ```python
def st(i) :
chk = [chr(i+65) for i in range(26)]; chk.sort()
alph = [chr(i+65) for i in range(26)]; ss = list(s[i:i+27]); j= 0
for x in ss :
if x in alph: alph.remove(x)
for x in range(26) :
if ss[x] == '?' : ss[x] = alph[j]; j+=1
ss.sort()
if ss == chk : return ''.join(ss)
else : return 0
s=input(); n=len(s)
for i in range(n-25) :
n=st(i)
if n : print(n); break
else : print(-1)
``` | 0 | |
981 | A | Antipalindrome | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into? | The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only. | If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique. | [
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] | [
"3\n",
"5\n",
"0\n"
] | "mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | 500 | [
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow"... | 1,633,626,055 | 2,147,483,647 | Python 3 | OK | TESTS | 133 | 77 | 6,758,400 | s = input()
print((len(s) - (s == s[::-1])) * (len(set(s)) > 1)) | Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | ```python
s = input()
print((len(s) - (s == s[::-1])) * (len(set(s)) > 1))
``` | 3 | |
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": "... | 1,587,874,995 | 2,147,483,647 | PyPy 3 | OK | TESTS | 102 | 155 | 0 | def xr(f,s):
return int(f)^int(s)
def nn(v):
f,s=v
n=len(f)
for x in range(n):
yield(xr(f[x],s[x]))
v=[input() for _ in range(2)]
for x in nn(v):
print(x,end='') | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
def xr(f,s):
return int(f)^int(s)
def nn(v):
f,s=v
n=len(f)
for x in range(n):
yield(xr(f[x],s[x]))
v=[input() for _ in range(2)]
for x in nn(v):
print(x,end='')
``` | 3.96125 |
572 | A | Arrays | PROGRAMMING | 900 | [
"sortings"
] | null | null | You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array. | The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*. | Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes). | [
"3 3\n2 1\n1 2 3\n3 4 5\n",
"3 3\n3 3\n1 2 3\n3 4 5\n",
"5 2\n3 1\n1 1 1 1 1\n2 2\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 500 | [
{
"input": "3 3\n2 1\n1 2 3\n3 4 5",
"output": "YES"
},
{
"input": "3 3\n3 3\n1 2 3\n3 4 5",
"output": "NO"
},
{
"input": "5 2\n3 1\n1 1 1 1 1\n2 2",
"output": "YES"
},
{
"input": "3 5\n1 1\n5 5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n1 1\n1\n1",
"ou... | 1,554,495,580 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 93 | 0 | m=input().split()
x=input().split()
a=int(x[0])
b=int(x[1])
emon=input().split()
emon.sort()
emon1=[]
mishu=input().split()
mishu.sort()
mishu1=[]
for i in range(a):
c=int(emon[i])
emon1.append(c)
for i in range(b):
c=int(mishu[i])
mishu1.append(c)
a=0
m=max(emon1)
n=min(mishu1)
if (n>m):
print("YES")
else :
print("NO") | Title: Arrays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.
Input Specification:
The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.
Output Specification:
Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).
Demo Input:
['3 3\n2 1\n1 2 3\n3 4 5\n', '3 3\n3 3\n1 2 3\n3 4 5\n', '5 2\n3 1\n1 1 1 1 1\n2 2\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
m=input().split()
x=input().split()
a=int(x[0])
b=int(x[1])
emon=input().split()
emon.sort()
emon1=[]
mishu=input().split()
mishu.sort()
mishu1=[]
for i in range(a):
c=int(emon[i])
emon1.append(c)
for i in range(b):
c=int(mishu[i])
mishu1.append(c)
a=0
m=max(emon1)
n=min(mishu1)
if (n>m):
print("YES")
else :
print("NO")
``` | 0 | |
414 | A | Mashmokh and Numbers | PROGRAMMING | 1,500 | [
"constructive algorithms",
"number theory"
] | null | null | It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
In this game Mashmokh writes sequence of *n* distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers *x* and *y* from the board, he gets *gcd*(*x*,<=*y*) points. At the beginning of the game Bimokh has zero points.
Mashmokh wants to win in the game. For this reason he wants his boss to get exactly *k* points in total. But the guy doesn't know how choose the initial sequence in the right way.
Please, help him. Find *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* such that his boss will score exactly *k* points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109. | The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*<=≤<=105; 0<=≤<=*k*<=≤<=108). | If such sequence doesn't exist output -1 otherwise output *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | [
"5 2\n",
"5 3",
"7 2\n"
] | [
"1 2 3 4 5\n",
"2 4 3 7 1",
"-1\n"
] | *gcd*(*x*, *y*) is greatest common divisor of *x* and *y*. | 500 | [
{
"input": "5 2",
"output": "1 2 3 4 5"
},
{
"input": "5 3",
"output": "2 4 5 6 7"
},
{
"input": "7 2",
"output": "-1"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "2 0",
"output": "-1"
},
{
"input": "1 10",
"output": "-1"
},
{
"input"... | 1,420,654,222 | 2,147,483,647 | Python 3 | OK | TESTS | 84 | 343 | 102,400 | # 414A
from sys import stdin
__author__ = 'artyom'
n, k = list(map(int, stdin.readline().strip().split()))
if n == 1:
if k == 0:
print(1)
else:
print(-1)
exit()
pairs = n // 2
if k < pairs:
print(-1)
exit()
x, y = 1, 2
if k > pairs:
x = k - pairs + 1
y = x * 2
print(x, end=' ')
print(y, end=' ')
if n > 2:
for i in range(y + 1, y + n - 2):
print(i, end=' ')
print(y + n - 2) | Title: Mashmokh and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
In this game Mashmokh writes sequence of *n* distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers *x* and *y* from the board, he gets *gcd*(*x*,<=*y*) points. At the beginning of the game Bimokh has zero points.
Mashmokh wants to win in the game. For this reason he wants his boss to get exactly *k* points in total. But the guy doesn't know how choose the initial sequence in the right way.
Please, help him. Find *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* such that his boss will score exactly *k* points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109.
Input Specification:
The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*<=≤<=105; 0<=≤<=*k*<=≤<=108).
Output Specification:
If such sequence doesn't exist output -1 otherwise output *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Demo Input:
['5 2\n', '5 3', '7 2\n']
Demo Output:
['1 2 3 4 5\n', '2 4 3 7 1', '-1\n']
Note:
*gcd*(*x*, *y*) is greatest common divisor of *x* and *y*. | ```python
# 414A
from sys import stdin
__author__ = 'artyom'
n, k = list(map(int, stdin.readline().strip().split()))
if n == 1:
if k == 0:
print(1)
else:
print(-1)
exit()
pairs = n // 2
if k < pairs:
print(-1)
exit()
x, y = 1, 2
if k > pairs:
x = k - pairs + 1
y = x * 2
print(x, end=' ')
print(y, end=' ')
if n > 2:
for i in range(y + 1, y + n - 2):
print(i, end=' ')
print(y + n - 2)
``` | 3 | |
510 | A | Fox And Snake | PROGRAMMING | 800 | [
"implementation"
] | null | null | Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern. | The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number. | Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces. | [
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] | [
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] | none | 500 | [
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#... | 1,685,625,967 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 28 | 62 | 0 | r,c = map(int, input().split())
for i in range(1, r+1):
if i % 2 != 0:
print(c * "#")
else:
if (i / 2) % 2 == 1:
print((c-1) * "." + "#")
else:
print("#"+ (c-1) * ".")
| Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none | ```python
r,c = map(int, input().split())
for i in range(1, r+1):
if i % 2 != 0:
print(c * "#")
else:
if (i / 2) % 2 == 1:
print((c-1) * "." + "#")
else:
print("#"+ (c-1) * ".")
``` | 3 | |
697 | A | Pineapple Incident | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc.
Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time. | The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively. | Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output. | [
"3 10 4\n",
"3 10 3\n",
"3 8 51\n",
"3 8 52\n"
] | [
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] | In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.
In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52. | 500 | [
{
"input": "3 10 4",
"output": "NO"
},
{
"input": "3 10 3",
"output": "YES"
},
{
"input": "3 8 51",
"output": "YES"
},
{
"input": "3 8 52",
"output": "YES"
},
{
"input": "456947336 740144 45",
"output": "NO"
},
{
"input": "33 232603 599417964",
"ou... | 1,519,400,228 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 62 | 5,632,000 | t,s,x = map(int,input().split())
if (x == t+1) or (x<t) :print('NO')
elif ((x - t) % s) or ((x - t - 1) % s):print('YES')
else:print('NO')
| Title: Pineapple Incident
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc.
Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time.
Input Specification:
The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.
Output Specification:
Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output.
Demo Input:
['3 10 4\n', '3 10 3\n', '3 8 51\n', '3 8 52\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.
In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52. | ```python
t,s,x = map(int,input().split())
if (x == t+1) or (x<t) :print('NO')
elif ((x - t) % s) or ((x - t - 1) % s):print('YES')
else:print('NO')
``` | 0 | |
298 | B | Sail | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation"
] | null | null | The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)? | The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north). | If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes). | [
"5 0 0 1 1\nSESNW\n",
"10 5 3 3 6\nNENSWESNEE\n"
] | [
"4\n",
"-1\n"
] | In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination. | 500 | [
{
"input": "5 0 0 1 1\nSESNW",
"output": "4"
},
{
"input": "10 5 3 3 6\nNENSWESNEE",
"output": "-1"
},
{
"input": "19 -172106364 -468680119 -172106365 -468680119\nSSEEESSSESESWSEESSS",
"output": "13"
},
{
"input": "39 -1000000000 -1000000000 -999999997 -1000000000\nENEENWSWSS... | 1,594,208,356 | 2,147,483,647 | PyPy 3 | OK | TESTS | 43 | 310 | 22,630,400 | t,sx,sy,ex,ey=map(int,input().split())
s=input()
vert,hori="",""
if ey>sy:
vert='N'
else:
vert='S'
if ex>sx:
hori='E'
else:
hori='W'
x,y=abs(ex-sx),abs(ey-sy)
cnt=0
for i in range(0,len(s)):
if s[i]==vert:
y-=1
if s[i]==hori:
x-=1
if x<=0 and y<=0:
print(i+1)
break
else:
print(-1)
| Title: Sail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)?
Input Specification:
The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
Output Specification:
If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).
Demo Input:
['5 0 0 1 1\nSESNW\n', '10 5 3 3 6\nNENSWESNEE\n']
Demo Output:
['4\n', '-1\n']
Note:
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination. | ```python
t,sx,sy,ex,ey=map(int,input().split())
s=input()
vert,hori="",""
if ey>sy:
vert='N'
else:
vert='S'
if ex>sx:
hori='E'
else:
hori='W'
x,y=abs(ex-sx),abs(ey-sy)
cnt=0
for i in range(0,len(s)):
if s[i]==vert:
y-=1
if s[i]==hori:
x-=1
if x<=0 and y<=0:
print(i+1)
break
else:
print(-1)
``` | 3 | |
34 | A | Reconnaissance 2 | PROGRAMMING | 800 | [
"implementation"
] | A. Reconnaissance 2 | 2 | 256 | *n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit. | The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction. | Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle. | [
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] | [
"5 1\n",
"1 2\n"
] | none | 500 | [
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output"... | 1,629,267,387 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 154 | 6,963,200 | n = int(input())
soldiers = list(map(int, input().split()))
soldiers.append(soldiers[0])
m = float('inf')
i1, i2 = 0, 1
for i in range(1,n+1):
x = abs(soldiers[i]-soldiers[i-1])
if x<m:
m = x
i1, i2 = i-1, i
if i2==n:
i2 = 0
print(i1+1, i2+1)
| Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none | ```python
n = int(input())
soldiers = list(map(int, input().split()))
soldiers.append(soldiers[0])
m = float('inf')
i1, i2 = 0, 1
for i in range(1,n+1):
x = abs(soldiers[i]-soldiers[i-1])
if x<m:
m = x
i1, i2 = i-1, i
if i2==n:
i2 = 0
print(i1+1, i2+1)
``` | 3.94853 |
108 | A | Palindromic Times | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | A. Palindromic Times | 2 | 256 | Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him. | The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits. | Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time. | [
"12:21\n",
"23:59\n"
] | [
"13:31\n",
"00:00\n"
] | none | 500 | [
{
"input": "12:21",
"output": "13:31"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "15:51",
"output": "20:02"
},
{
"input": "10:44",
"output": "11:11"
},
{
"input": "04:02",
"output": "04:40"
},
{
"input": "02:11",
"output": "02:20"
},
... | 1,647,223,278 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 124 | 0 | a = input()
b = list(map(int, a.split(":")))
while True:
b[1] += 1
if b[1] > 59:
b[1] = 0
b[0] += 1
if b[0] > 23:
b[1] = 0
b[0] = 0
if str(b[0] + b[1]) == str(b[0] + b[1]):
print(str(b[0]) + ":" + str(b[1]))
break | Title: Palindromic Times
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
Input Specification:
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
Output Specification:
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
Demo Input:
['12:21\n', '23:59\n']
Demo Output:
['13:31\n', '00:00\n']
Note:
none | ```python
a = input()
b = list(map(int, a.split(":")))
while True:
b[1] += 1
if b[1] > 59:
b[1] = 0
b[0] += 1
if b[0] > 23:
b[1] = 0
b[0] = 0
if str(b[0] + b[1]) == str(b[0] + b[1]):
print(str(b[0]) + ":" + str(b[1]))
break
``` | 0 |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
... | 1,687,166,221 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 46 | 0 | k, r = [int(x) for x in input().split()]
a = 1
b = k
while True:
if k%10==0:
print(a)
break
elif (k-r)%10==0:
print(a)
break
else:
k+=b
a+=1 | Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
k, r = [int(x) for x in input().split()]
a = 1
b = k
while True:
if k%10==0:
print(a)
break
elif (k-r)%10==0:
print(a)
break
else:
k+=b
a+=1
``` | 3 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,680,076,803 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | a = input().split()
lenth = int(a[0])
width = int(a[1])
flagstone = int(a[2])
if flagstone == 1:
print(lenth * width)
elif flagstone ** 2 >= lenth * width:
print(1)
else:
baghi = lenth % flagstone
if baghi != 0:
lenth += (flagstone - baghi)
baghi = width % flagstone
if baghi != 0:
width += (flagstone - baghi)
number = (lenth * width) // flagstone ** 2
print(number)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
a = input().split()
lenth = int(a[0])
width = int(a[1])
flagstone = int(a[2])
if flagstone == 1:
print(lenth * width)
elif flagstone ** 2 >= lenth * width:
print(1)
else:
baghi = lenth % flagstone
if baghi != 0:
lenth += (flagstone - baghi)
baghi = width % flagstone
if baghi != 0:
width += (flagstone - baghi)
number = (lenth * width) // flagstone ** 2
print(number)
``` | 0 |
875 | B | Sorting the Coins | PROGRAMMING | 1,500 | [
"dsu",
"implementation",
"sortings",
"two pointers"
] | null | null | Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right; 1. If he sees that the *i*-th coin is still in circulation, and (*i*<=+<=1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (*i*<=+<=1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts *n* coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for *n* times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=300<=000) — number of coins that Sasha puts behind Dima.
Second line contains *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position *p*1, then coin located at position *p*2 and so on. Coins are numbered from left to right. | Print *n*<=+<=1 numbers *a*0,<=*a*1,<=...,<=*a**n*, where *a*0 is a hardness of ordering at the beginning, *a*1 is a hardness of ordering after the first replacement and so on. | [
"4\n1 3 4 2\n",
"8\n6 8 3 4 7 2 1 5\n"
] | [
"1 2 3 2 1\n",
"1 2 2 3 4 3 4 5 1\n"
] | Let's denote as O coin out of circulation, and as X — coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO → OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO → OXOX → OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX → OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | 1,000 | [
{
"input": "4\n1 3 4 2",
"output": "1 2 3 2 1"
},
{
"input": "8\n6 8 3 4 7 2 1 5",
"output": "1 2 2 3 4 3 4 5 1"
},
{
"input": "1\n1",
"output": "1 1"
},
{
"input": "11\n10 8 9 4 6 3 5 1 11 7 2",
"output": "1 2 3 4 5 6 7 8 9 6 2 1"
},
{
"input": "11\n10 8 9 4 3 5 ... | 1,677,194,656 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | n=int(input())
pyt= list(map(int, input().split(maxsplit = n)))
hod=[1]
arr=[0 for i in range(n)]
nul=n
ed=0
pyt.pop()
for i in pyt:
arr[i-1]=1
ed+=1
if i==nul:
arr.reverse()
nul=n-arr.index(0)
arr.reverse()
hod.append(ed-(n-nul-1))
hod.append(1)
print(hod)
| Title: Sorting the Coins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.
For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:
1. He looks through all the coins from left to right; 1. If he sees that the *i*-th coin is still in circulation, and (*i*<=+<=1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (*i*<=+<=1)-th.
Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.
Today Sasha invited Dima and proposed him a game. First he puts *n* coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for *n* times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.
The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=300<=000) — number of coins that Sasha puts behind Dima.
Second line contains *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position *p*1, then coin located at position *p*2 and so on. Coins are numbered from left to right.
Output Specification:
Print *n*<=+<=1 numbers *a*0,<=*a*1,<=...,<=*a**n*, where *a*0 is a hardness of ordering at the beginning, *a*1 is a hardness of ordering after the first replacement and so on.
Demo Input:
['4\n1 3 4 2\n', '8\n6 8 3 4 7 2 1 5\n']
Demo Output:
['1 2 3 2 1\n', '1 2 2 3 4 3 4 5 1\n']
Note:
Let's denote as O coin out of circulation, and as X — coin is circulation.
At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.
After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.
XOOO → OOOX
After replacement of the third coin, Dima's actions look this way:
XOXO → OXOX → OOXX
After replacement of the fourth coin, Dima's actions look this way:
XOXX → OXXX
Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges. | ```python
n=int(input())
pyt= list(map(int, input().split(maxsplit = n)))
hod=[1]
arr=[0 for i in range(n)]
nul=n
ed=0
pyt.pop()
for i in pyt:
arr[i-1]=1
ed+=1
if i==nul:
arr.reverse()
nul=n-arr.index(0)
arr.reverse()
hod.append(ed-(n-nul-1))
hod.append(1)
print(hod)
``` | 0 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,694,449,918 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 62 | 0 | a = input()
# Initialize pointers for both strings
i = 0 # Pointer for "hello"
for char in a:
if char == 'h' and i == 0:
i += 1
elif char == 'e' and i == 1:
i += 1
elif char == 'l' and i == 2:
i += 1
elif char == 'l' and i == 3:
i += 1
elif char == 'o' and i == 4:
i += 1
# Check if all characters in "hello" have been found in order
if i == 5:
print("YES")
else:
print("NO")
| Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
a = input()
# Initialize pointers for both strings
i = 0 # Pointer for "hello"
for char in a:
if char == 'h' and i == 0:
i += 1
elif char == 'e' and i == 1:
i += 1
elif char == 'l' and i == 2:
i += 1
elif char == 'l' and i == 3:
i += 1
elif char == 'o' and i == 4:
i += 1
# Check if all characters in "hello" have been found in order
if i == 5:
print("YES")
else:
print("NO")
``` | 3.969 |
456 | A | Laptops | PROGRAMMING | 1,100 | [
"sortings"
] | null | null | One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct. | If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes). | [
"2\n1 2\n2 1\n"
] | [
"Happy Alex\n"
] | none | 500 | [
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
... | 1,663,664,368 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n = int(input())
s = [[int(x) for x in input().split()] for _ in range(n)]
s.sort(reverse=True)
for i in s[1:]:
if i[1] > s[0][1]:
print('Happy Alex')
break
s[0][1] = i[1]
else:
print('Poor Alex') | Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none | ```python
n = int(input())
s = [[int(x) for x in input().split()] for _ in range(n)]
s.sort(reverse=True)
for i in s[1:]:
if i[1] > s[0][1]:
print('Happy Alex')
break
s[0][1] = i[1]
else:
print('Poor Alex')
``` | -1 | |
883 | F | Lost in Transliteration | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | There are some ambiguities when one writes Berland names with the letters of the Latin alphabet.
For example, the Berland sound u can be written in the Latin alphabet as "u", and can be written as "oo". For this reason, two words "ulyana" and "oolyana" denote the same name.
The second ambiguity is about the Berland sound h: one can use both "h" and "kh" to write it. For example, the words "mihail" and "mikhail" denote the same name.
There are *n* users registered on the Polycarp's website. Each of them indicated a name represented by the Latin letters. How many distinct names are there among them, if two ambiguities described above are taken into account?
Formally, we assume that two words denote the same name, if using the replacements "u" "oo" and "h" "kh", you can make the words equal. One can make replacements in both directions, in any of the two words an arbitrary number of times. A letter that resulted from the previous replacement can participate in the next replacements.
For example, the following pairs of words denote the same name:
- "koouper" and "kuooper". Making the replacements described above, you can make both words to be equal: "koouper" "kuuper" and "kuooper" "kuuper". - "khun" and "kkkhoon". With the replacements described above you can make both words to be equal: "khun" "khoon" and "kkkhoon" "kkhoon" "khoon".
For a given list of words, find the minimal number of groups where the words in each group denote the same name. | The first line contains integer number *n* (2<=≤<=*n*<=≤<=400) — number of the words in the list.
The following *n* lines contain words, one word per line. Each word consists of only lowercase Latin letters. The length of each word is between 1 and 20 letters inclusive. | Print the minimal number of groups where the words in each group denote the same name. | [
"10\nmihail\noolyana\nkooooper\nhoon\nulyana\nkoouper\nmikhail\nkhun\nkuooper\nkkkhoon\n",
"9\nhariton\nhkariton\nbuoi\nkkkhariton\nboooi\nbui\nkhariton\nboui\nboi\n",
"2\nalex\nalex\n"
] | [
"4\n",
"5\n",
"1\n"
] | There are four groups of words in the first example. Words in each group denote same name:
1. "mihail", "mikhail" 1. "oolyana", "ulyana" 1. "kooooper", "koouper" 1. "hoon", "khun", "kkkhoon"
There are five groups of words in the second example. Words in each group denote same name:
1. "hariton", "kkkhariton", "khariton" 1. "hkariton" 1. "buoi", "boooi", "boui" 1. "bui" 1. "boi"
In the third example the words are equal, so they denote the same name. | 0 | [
{
"input": "10\nmihail\noolyana\nkooooper\nhoon\nulyana\nkoouper\nmikhail\nkhun\nkuooper\nkkkhoon",
"output": "4"
},
{
"input": "9\nhariton\nhkariton\nbuoi\nkkkhariton\nboooi\nbui\nkhariton\nboui\nboi",
"output": "5"
},
{
"input": "2\nalex\nalex",
"output": "1"
},
{
"input": ... | 1,556,883,505 | 4,705 | PyPy 3 | RUNTIME_ERROR | TESTS | 3 | 171 | 1,024,000 | def k(s):
a = ''
for i in range(len(s)):
if s[i] == 'k' and s[i-1] != 'k':
temp = i
elif s[i] == 'k' and s[i-1] == 'k':
continue
elif s[i] == 'h' and s[i-1] == 'k':
a += s[i]
elif s[i] != 'h' and s[i-1] == 'k':
a += s[temp:i+1]
else:
a += s[i]
return a
def u(s):
a = ''
for i in s:
if i == 'u':
a += 'oo'
else:
a += i
return a
n = int(input())
s = set()
for i in range(n):
temp = input()
s.add(u(k(temp)))
print(len(s))
| Title: Lost in Transliteration
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are some ambiguities when one writes Berland names with the letters of the Latin alphabet.
For example, the Berland sound u can be written in the Latin alphabet as "u", and can be written as "oo". For this reason, two words "ulyana" and "oolyana" denote the same name.
The second ambiguity is about the Berland sound h: one can use both "h" and "kh" to write it. For example, the words "mihail" and "mikhail" denote the same name.
There are *n* users registered on the Polycarp's website. Each of them indicated a name represented by the Latin letters. How many distinct names are there among them, if two ambiguities described above are taken into account?
Formally, we assume that two words denote the same name, if using the replacements "u" "oo" and "h" "kh", you can make the words equal. One can make replacements in both directions, in any of the two words an arbitrary number of times. A letter that resulted from the previous replacement can participate in the next replacements.
For example, the following pairs of words denote the same name:
- "koouper" and "kuooper". Making the replacements described above, you can make both words to be equal: "koouper" "kuuper" and "kuooper" "kuuper". - "khun" and "kkkhoon". With the replacements described above you can make both words to be equal: "khun" "khoon" and "kkkhoon" "kkhoon" "khoon".
For a given list of words, find the minimal number of groups where the words in each group denote the same name.
Input Specification:
The first line contains integer number *n* (2<=≤<=*n*<=≤<=400) — number of the words in the list.
The following *n* lines contain words, one word per line. Each word consists of only lowercase Latin letters. The length of each word is between 1 and 20 letters inclusive.
Output Specification:
Print the minimal number of groups where the words in each group denote the same name.
Demo Input:
['10\nmihail\noolyana\nkooooper\nhoon\nulyana\nkoouper\nmikhail\nkhun\nkuooper\nkkkhoon\n', '9\nhariton\nhkariton\nbuoi\nkkkhariton\nboooi\nbui\nkhariton\nboui\nboi\n', '2\nalex\nalex\n']
Demo Output:
['4\n', '5\n', '1\n']
Note:
There are four groups of words in the first example. Words in each group denote same name:
1. "mihail", "mikhail" 1. "oolyana", "ulyana" 1. "kooooper", "koouper" 1. "hoon", "khun", "kkkhoon"
There are five groups of words in the second example. Words in each group denote same name:
1. "hariton", "kkkhariton", "khariton" 1. "hkariton" 1. "buoi", "boooi", "boui" 1. "bui" 1. "boi"
In the third example the words are equal, so they denote the same name. | ```python
def k(s):
a = ''
for i in range(len(s)):
if s[i] == 'k' and s[i-1] != 'k':
temp = i
elif s[i] == 'k' and s[i-1] == 'k':
continue
elif s[i] == 'h' and s[i-1] == 'k':
a += s[i]
elif s[i] != 'h' and s[i-1] == 'k':
a += s[temp:i+1]
else:
a += s[i]
return a
def u(s):
a = ''
for i in s:
if i == 'u':
a += 'oo'
else:
a += i
return a
n = int(input())
s = set()
for i in range(n):
temp = input()
s.add(u(k(temp)))
print(len(s))
``` | -1 | |
460 | A | Vasya and Socks | PROGRAMMING | 900 | [
"brute force",
"implementation",
"math"
] | null | null | Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks? | The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space. | Print a single integer — the answer to the problem. | [
"2 2\n",
"9 3\n"
] | [
"3\n",
"13\n"
] | In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | 500 | [
{
"input": "2 2",
"output": "3"
},
{
"input": "9 3",
"output": "13"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1 99",
"output": "1"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "10 2",
"outp... | 1,685,873,225 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 46 | 0 | tm = input().split()
n, m = int(tm[0]), int(tm[1])
it = n
p = 0
while it // m - p > 0:
l = p
p = it // m
it += it // m - l
print(it) | Title: Vasya and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
Input Specification:
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['2 2\n', '9 3\n']
Demo Output:
['3\n', '13\n']
Note:
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | ```python
tm = input().split()
n, m = int(tm[0]), int(tm[1])
it = n
p = 0
while it // m - p > 0:
l = p
p = it // m
it += it // m - l
print(it)
``` | 3 | |
460 | A | Vasya and Socks | PROGRAMMING | 900 | [
"brute force",
"implementation",
"math"
] | null | null | Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks? | The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space. | Print a single integer — the answer to the problem. | [
"2 2\n",
"9 3\n"
] | [
"3\n",
"13\n"
] | In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | 500 | [
{
"input": "2 2",
"output": "3"
},
{
"input": "9 3",
"output": "13"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1 99",
"output": "1"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "10 2",
"outp... | 1,655,841,908 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 46 | 0 | n, m = map(int, input().split())
count = 0
c=((((n//m)+n)//m)+n)
while c > 0:
c=c-1
count+=1
print(count) | Title: Vasya and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
Input Specification:
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['2 2\n', '9 3\n']
Demo Output:
['3\n', '13\n']
Note:
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day. | ```python
n, m = map(int, input().split())
count = 0
c=((((n//m)+n)//m)+n)
while c > 0:
c=c-1
count+=1
print(count)
``` | 0 | |
577 | A | Multiplication Table | PROGRAMMING | 1,000 | [
"implementation",
"number theory"
] | null | null | Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*. | The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table. | Print a single number: the number of times *x* occurs in the table. | [
"10 5\n",
"6 12\n",
"5 13\n"
] | [
"2\n",
"4\n",
"0\n"
] | A table for the second sample test is given below. The occurrences of number 12 are marked bold. | 500 | [
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 10000000... | 1,676,319,580 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 1,000 | 0 | n, x = map(int,input().split())
r = n
i = 1
j = 0
c = 1
if n * n >= x:
while r <= n * n:
if i == x:
j += 1
i += c
if i == r:
if i == x:
j += 1
c += 1
i = c
r = r + n
print(j) | Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold. | ```python
n, x = map(int,input().split())
r = n
i = 1
j = 0
c = 1
if n * n >= x:
while r <= n * n:
if i == x:
j += 1
i += c
if i == r:
if i == x:
j += 1
c += 1
i = c
r = r + n
print(j)
``` | 0 | |
835 | A | Key races | PROGRAMMING | 800 | [
"math"
] | null | null | Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game. | The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant. | If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship". | [
"5 1 2 1 2\n",
"3 3 1 1 1\n",
"4 5 3 1 5\n"
] | [
"First\n",
"Second\n",
"Friendship\n"
] | In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | 500 | [
{
"input": "5 1 2 1 2",
"output": "First"
},
{
"input": "3 3 1 1 1",
"output": "Second"
},
{
"input": "4 5 3 1 5",
"output": "Friendship"
},
{
"input": "1000 1000 1000 1000 1000",
"output": "Friendship"
},
{
"input": "1 1 1 1 1",
"output": "Friendship"
},
... | 1,653,536,231 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | s, v1, v2, t1, t2 = list(map(int, input().split()))
print(s,v1,v2,t1,t2)
def calc(s, t, v):
result = s * v + 2 * t
return result
first = calc(s, t1, v1)
second = calc(s, t2, v2)
if first > second:
print('Second')
elif second > first:
print('First')
else:
print('Friendship')
| Title: Key races
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game.
Input Specification:
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
Output Specification:
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
Demo Input:
['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n']
Demo Output:
['First\n', 'Second\n', 'Friendship\n']
Note:
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | ```python
s, v1, v2, t1, t2 = list(map(int, input().split()))
print(s,v1,v2,t1,t2)
def calc(s, t, v):
result = s * v + 2 * t
return result
first = calc(s, t1, v1)
second = calc(s, t2, v2)
if first > second:
print('Second')
elif second > first:
print('First')
else:
print('Friendship')
``` | 0 | |
186 | A | Comparing Strings | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not. | The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length. | Print "YES", if the dwarves belong to the same race. Otherwise, print "NO". | [
"ab\nba\n",
"aa\nab\n"
] | [
"YES\n",
"NO\n"
] | - First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b". | 500 | [
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "aa\nab",
"output": "NO"
},
{
"input": "a\nza",
"output": "NO"
},
{
"input": "vvea\nvvae",
"output": "YES"
},
{
"input": "rtfabanpc\natfabrnpc",
"output": "YES"
},
{
"input": "mt\ntm",
"output": "Y... | 1,579,252,859 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 310 | 6,656,000 | S1, S2, Dif = input(), input() , []
if len(S1) != len(S2): print("NO"); exit()
for i in zip(S1 , S2):
if i[0] != i[1]:
Dif.append(sorted(i))
print("YES" if len(Dif)==2 and Dif[0]==Dif[1] else "NO")
# Show you deserve being the best to whom doesn't believe in you.
| Title: Comparing Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
Input Specification:
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
Output Specification:
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
Demo Input:
['ab\nba\n', 'aa\nab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b". | ```python
S1, S2, Dif = input(), input() , []
if len(S1) != len(S2): print("NO"); exit()
for i in zip(S1 , S2):
if i[0] != i[1]:
Dif.append(sorted(i))
print("YES" if len(Dif)==2 and Dif[0]==Dif[1] else "NO")
# Show you deserve being the best to whom doesn't believe in you.
``` | 3 | |
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",
... | 1,611,046,087 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 248 | 0 | n = int(input())
d = {}
for i in range(n):
x = input()
if x not in d:
d[x] = 1
else:
d[x] += 1
print(max(d, key = d.get)) | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none | ```python
n = int(input())
d = {}
for i in range(n):
x = input()
if x not in d:
d[x] = 1
else:
d[x] += 1
print(max(d, key = d.get))
``` | 3.938 |
916 | C | Jamie and Interesting Graph | PROGRAMMING | 1,600 | [
"constructive algorithms",
"graphs",
"shortest paths"
] | null | null | Jamie has recently found undirected weighted graphs with the following properties very interesting:
- The graph is connected and contains exactly *n* vertices and *m* edges. - All edge weights are integers and are in range [1,<=109] inclusive. - The length of shortest path from 1 to *n* is a prime number. - The sum of edges' weights in the minimum spanning tree (MST) of the graph is a prime number. - The graph contains no loops or multi-edges.
If you are not familiar with some terms from the statement you can find definitions of them in notes section.
Help Jamie construct any graph with given number of vertices and edges that is interesting! | First line of input contains 2 integers *n*, *m* — the required number of vertices and edges. | In the first line output 2 integers *sp*, *mstw* (1<=≤<=*sp*,<=*mstw*<=≤<=1014) — the length of the shortest path and the sum of edges' weights in the minimum spanning tree.
In the next *m* lines output the edges of the graph. In each line output 3 integers *u*, *v*, *w* (1<=≤<=*u*,<=*v*<=≤<=*n*,<=1<=≤<=*w*<=≤<=109) describing the edge connecting *u* and *v* and having weight *w*. | [
"4 4\n",
"5 4\n"
] | [
"7 7\n1 2 3\n2 3 2\n3 4 2\n2 4 4\n",
"7 13\n1 2 2\n1 3 4\n1 4 3\n4 5 4\n"
] | The graph of sample 1: <img class="tex-graphics" src="https://espresso.codeforces.com/42f9750de41b0d9a6b21e8615170113cfe19b0f2.png" style="max-width: 100.0%;max-height: 100.0%;"/> Shortest path sequence: {1, 2, 3, 4}. MST edges are marked with an asterisk (*).
Definition of terms used in the problem statement:
A shortest path in an undirected graph is a sequence of vertices (*v*<sub class="lower-index">1</sub>, *v*<sub class="lower-index">2</sub>, ... , *v*<sub class="lower-index">*k*</sub>) such that *v*<sub class="lower-index">*i*</sub> is adjacent to *v*<sub class="lower-index">*i* + 1</sub> 1 ≤ *i* < *k* and the sum of weight <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e7f62cfd5c2c3b81dc80aaf2f512898495354f03.png" style="max-width: 100.0%;max-height: 100.0%;"/> is minimized where *w*(*i*, *j*) is the edge weight between *i* and *j*. ([https://en.wikipedia.org/wiki/Shortest_path_problem](https://en.wikipedia.org/wiki/Shortest_path_problem))
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. ([https://en.wikipedia.org/wiki/Prime_number](https://en.wikipedia.org/wiki/Prime_number))
A minimum spanning tree (MST) is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. ([https://en.wikipedia.org/wiki/Minimum_spanning_tree](https://en.wikipedia.org/wiki/Minimum_spanning_tree))
[https://en.wikipedia.org/wiki/Multiple_edges](https://en.wikipedia.org/wiki/Multiple_edges) | 1,500 | [
{
"input": "4 4",
"output": "100003 100003\n1 2 100001\n2 3 1\n3 4 1\n1 3 1000000000"
},
{
"input": "5 4",
"output": "100003 100003\n1 2 100000\n2 3 1\n3 4 1\n4 5 1"
},
{
"input": "2 1",
"output": "100003 100003\n1 2 100003"
},
{
"input": "10 19",
"output": "100003 100003... | 1,528,189,682 | 2,147,483,647 | Python 3 | OK | TESTS | 79 | 841 | 7,270,400 | def doit(n, m):
if (n == 2):
print(2, 2)
print(1, 2, 2)
return
sp = 2
mstw = 100003
print(sp, mstw)
print(1, n, sp)
print(2, n, mstw - n + 3 - sp)
for i in range(3, n):
print(i, n, 1)
for i in range(2, n):
for j in range(1, i):
if (m == n - 1):
return
print(i, j, mstw)
m -= 1
n, m = input().split()
doit(int(n), int(m))
| Title: Jamie and Interesting Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jamie has recently found undirected weighted graphs with the following properties very interesting:
- The graph is connected and contains exactly *n* vertices and *m* edges. - All edge weights are integers and are in range [1,<=109] inclusive. - The length of shortest path from 1 to *n* is a prime number. - The sum of edges' weights in the minimum spanning tree (MST) of the graph is a prime number. - The graph contains no loops or multi-edges.
If you are not familiar with some terms from the statement you can find definitions of them in notes section.
Help Jamie construct any graph with given number of vertices and edges that is interesting!
Input Specification:
First line of input contains 2 integers *n*, *m* — the required number of vertices and edges.
Output Specification:
In the first line output 2 integers *sp*, *mstw* (1<=≤<=*sp*,<=*mstw*<=≤<=1014) — the length of the shortest path and the sum of edges' weights in the minimum spanning tree.
In the next *m* lines output the edges of the graph. In each line output 3 integers *u*, *v*, *w* (1<=≤<=*u*,<=*v*<=≤<=*n*,<=1<=≤<=*w*<=≤<=109) describing the edge connecting *u* and *v* and having weight *w*.
Demo Input:
['4 4\n', '5 4\n']
Demo Output:
['7 7\n1 2 3\n2 3 2\n3 4 2\n2 4 4\n', '7 13\n1 2 2\n1 3 4\n1 4 3\n4 5 4\n']
Note:
The graph of sample 1: <img class="tex-graphics" src="https://espresso.codeforces.com/42f9750de41b0d9a6b21e8615170113cfe19b0f2.png" style="max-width: 100.0%;max-height: 100.0%;"/> Shortest path sequence: {1, 2, 3, 4}. MST edges are marked with an asterisk (*).
Definition of terms used in the problem statement:
A shortest path in an undirected graph is a sequence of vertices (*v*<sub class="lower-index">1</sub>, *v*<sub class="lower-index">2</sub>, ... , *v*<sub class="lower-index">*k*</sub>) such that *v*<sub class="lower-index">*i*</sub> is adjacent to *v*<sub class="lower-index">*i* + 1</sub> 1 ≤ *i* < *k* and the sum of weight <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e7f62cfd5c2c3b81dc80aaf2f512898495354f03.png" style="max-width: 100.0%;max-height: 100.0%;"/> is minimized where *w*(*i*, *j*) is the edge weight between *i* and *j*. ([https://en.wikipedia.org/wiki/Shortest_path_problem](https://en.wikipedia.org/wiki/Shortest_path_problem))
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. ([https://en.wikipedia.org/wiki/Prime_number](https://en.wikipedia.org/wiki/Prime_number))
A minimum spanning tree (MST) is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. ([https://en.wikipedia.org/wiki/Minimum_spanning_tree](https://en.wikipedia.org/wiki/Minimum_spanning_tree))
[https://en.wikipedia.org/wiki/Multiple_edges](https://en.wikipedia.org/wiki/Multiple_edges) | ```python
def doit(n, m):
if (n == 2):
print(2, 2)
print(1, 2, 2)
return
sp = 2
mstw = 100003
print(sp, mstw)
print(1, n, sp)
print(2, n, mstw - n + 3 - sp)
for i in range(3, n):
print(i, n, 1)
for i in range(2, n):
for j in range(1, i):
if (m == n - 1):
return
print(i, j, mstw)
m -= 1
n, m = input().split()
doit(int(n), int(m))
``` | 3 | |
621 | B | Wet Shark and Bishops | PROGRAMMING | 1,300 | [
"combinatorics",
"implementation"
] | null | null | Today, Wet Shark is given *n* bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right.
Wet Shark thinks that two bishops attack each other if they share the same diagonal. Note, that this is the only criteria, so two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them. Now Wet Shark wants to count the number of pairs of bishops that attack each other. | The first line of the input contains *n* (1<=≤<=*n*<=≤<=200<=000) — the number of bishops.
Each of next *n* lines contains two space separated integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the number of row and the number of column where *i*-th bishop is positioned. It's guaranteed that no two bishops share the same position. | Output one integer — the number of pairs of bishops which attack each other. | [
"5\n1 1\n1 5\n3 3\n5 1\n5 5\n",
"3\n1 1\n2 3\n3 5\n"
] | [
"6\n",
"0\n"
] | In the first sample following pairs of bishops attack each other: (1, 3), (1, 5), (2, 3), (2, 4), (3, 4) and (3, 5). Pairs (1, 2), (1, 4), (2, 5) and (4, 5) do not attack each other because they do not share the same diagonal. | 1,000 | [
{
"input": "5\n1 1\n1 5\n3 3\n5 1\n5 5",
"output": "6"
},
{
"input": "3\n1 1\n2 3\n3 5",
"output": "0"
},
{
"input": "3\n859 96\n634 248\n808 72",
"output": "0"
},
{
"input": "3\n987 237\n891 429\n358 145",
"output": "0"
},
{
"input": "3\n411 81\n149 907\n611 114"... | 1,571,686,941 | 2,147,483,647 | PyPy 3 | OK | TESTS | 68 | 1,560 | 9,728,000 | size = 10
n = int(input())
cnt = 0
dp1, dp2 = {}, {}
for i in range(n):
i, j = map(int, input().split())
s, d = i + j, i - j
if not s in dp1.keys():
dp1[s] = 0
if not d in dp2.keys():
dp2[d] = 0
dp1[s] += 1
dp2[d] += 1
for v in dp1.values():
cnt += (v * (v - 1)) // 2
for v in dp2.values():
cnt += (v * (v - 1)) // 2
print(cnt)
| Title: Wet Shark and Bishops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right.
Wet Shark thinks that two bishops attack each other if they share the same diagonal. Note, that this is the only criteria, so two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them. Now Wet Shark wants to count the number of pairs of bishops that attack each other.
Input Specification:
The first line of the input contains *n* (1<=≤<=*n*<=≤<=200<=000) — the number of bishops.
Each of next *n* lines contains two space separated integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the number of row and the number of column where *i*-th bishop is positioned. It's guaranteed that no two bishops share the same position.
Output Specification:
Output one integer — the number of pairs of bishops which attack each other.
Demo Input:
['5\n1 1\n1 5\n3 3\n5 1\n5 5\n', '3\n1 1\n2 3\n3 5\n']
Demo Output:
['6\n', '0\n']
Note:
In the first sample following pairs of bishops attack each other: (1, 3), (1, 5), (2, 3), (2, 4), (3, 4) and (3, 5). Pairs (1, 2), (1, 4), (2, 5) and (4, 5) do not attack each other because they do not share the same diagonal. | ```python
size = 10
n = int(input())
cnt = 0
dp1, dp2 = {}, {}
for i in range(n):
i, j = map(int, input().split())
s, d = i + j, i - j
if not s in dp1.keys():
dp1[s] = 0
if not d in dp2.keys():
dp2[d] = 0
dp1[s] += 1
dp2[d] += 1
for v in dp1.values():
cnt += (v * (v - 1)) // 2
for v in dp2.values():
cnt += (v * (v - 1)) // 2
print(cnt)
``` | 3 | |
615 | A | Bulbs | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on. | The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs. | If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO". | [
"3 4\n2 1 4\n3 1 3 1\n1 2\n",
"3 3\n1 1\n1 2\n1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | 500 | [
{
"input": "3 4\n2 1 4\n3 1 3 1\n1 2",
"output": "YES"
},
{
"input": "3 3\n1 1\n1 2\n1 1",
"output": "NO"
},
{
"input": "3 4\n1 1\n1 2\n1 3",
"output": "NO"
},
{
"input": "1 5\n5 1 2 3 4 5",
"output": "YES"
},
{
"input": "1 5\n5 4 4 1 2 3",
"output": "NO"
},... | 1,665,471,594 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 31 | 0 | a,b=map(int,input().split())
z=[]
for i in range(a):
c=input().split()
for v in c:
z.append(v)
f=set(z)
if len(f)==b:
print('YES')
else:
print('NO') | Title: Bulbs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input Specification:
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
Output Specification:
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
Demo Input:
['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | ```python
a,b=map(int,input().split())
z=[]
for i in range(a):
c=input().split()
for v in c:
z.append(v)
f=set(z)
if len(f)==b:
print('YES')
else:
print('NO')
``` | 0 | |
585 | D | Lizard Era: Beginning | PROGRAMMING | 2,300 | [
"meet-in-the-middle"
] | null | null | In the game Lizard Era: Beginning the protagonist will travel with three companions: Lynn, Meliana and Worrigan. Overall the game has *n* mandatory quests. To perform each of them, you need to take exactly two companions.
The attitude of each of the companions to the hero is an integer. Initially, the attitude of each of them to the hero of neutral and equal to 0. As the hero completes quests, he makes actions that change the attitude of the companions, whom he took to perform this task, in positive or negative direction.
Tell us what companions the hero needs to choose to make their attitude equal after completing all the quests. If this can be done in several ways, choose the one in which the value of resulting attitude is greatest possible. | The first line contains positive integer *n* (1<=≤<=*n*<=≤<=25) — the number of important tasks.
Next *n* lines contain the descriptions of the tasks — the *i*-th line contains three integers *l**i*,<=*m**i*,<=*w**i* — the values by which the attitude of Lynn, Meliana and Worrigan respectively will change towards the hero if the hero takes them on the *i*-th task. All the numbers in the input are integers and do not exceed 107 in absolute value. | If there is no solution, print in the first line "Impossible".
Otherwise, print *n* lines, two characters is each line — in the *i*-th line print the first letters of the companions' names that hero should take to complete the *i*-th task ('L' for Lynn, 'M' for Meliana, 'W' for Worrigan). Print the letters in any order, if there are multiple solutions, print any of them. | [
"3\n1 0 0\n0 1 0\n0 0 1\n",
"7\n0 8 9\n5 9 -2\n6 -8 -7\n9 4 5\n-4 -9 9\n-4 5 2\n-6 8 -7\n",
"2\n1 0 0\n1 1 0\n"
] | [
"LM\nMW\nMW\n",
"LM\nMW\nLM\nLW\nMW\nLM\nLW\n",
"Impossible\n"
] | none | 1,750 | [
{
"input": "3\n1 0 0\n0 1 0\n0 0 1",
"output": "LM\nMW\nMW"
},
{
"input": "7\n0 8 9\n5 9 -2\n6 -8 -7\n9 4 5\n-4 -9 9\n-4 5 2\n-6 8 -7",
"output": "LM\nMW\nLM\nLW\nMW\nLM\nLW"
},
{
"input": "2\n1 0 0\n1 1 0",
"output": "Impossible"
},
{
"input": "25\n26668 10412 12658\n25216 1... | 1,585,368,873 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 261,529,600 | from itertools import combinations
def solve():
for I in range(N // 2 + 2):
X[I] = list(combinations(RX, I))
for I in range(N - N // 2):
Z[I] = list(combinations(RZ, I))
LX = []
LY = []
MX = []
MY = []
for I in range(len(X)):
for J in range(len(X[I])):
XL = XM = 0
for A in X[I][J]:
XL += L[A]
XM += M[A]
LX.append([XL, I, J])
MX.append([XM, I, J])
for I in range(len(Z)):
for J in range(len(Z[I])):
YL = YM = 0
for B in Z[I][J]:
YL += L[B]
YM += M[B]
LY.append([YL, I, J])
MY.append([YM, I, J])
LXY = []
MXY = []
PS = set()
MS = set()
for I in range(len(LX)):
for J in range(len(LY)):
if I | J:
P = LX[I][0] + LY[J][0]
LXY.append([P, I, J])
PS.add(P)
for I in range(len(MX)):
for J in range(len(MY)):
if I | J:
P = MX[I][0] + MY[J][0]
if P in PS:
MXY.append([P, I, J])
MS.add(P)
DS = ''
MAX = -10**10
for I in LXY:
for J in MXY:
if I[0] == J[0]:
S1 = set(X[LX[I[1]][1]][LX[I[1]][2]]+Z[LY[I[2]][1]][LY[I[2]][2]])
S2 = set(X[MX[J[1]][1]][MX[J[1]][2]]+Z[MY[J[2]][1]][MY[J[2]][2]])
S3 = (S1-S2).union(S2-S1)
S3S = 0
for U in S3:
S3S+=W[U]
if S3S == I[0]:
if S1.intersection(S2).union(S2.intersection(S3)).union(S3.intersection(S1)) == S:
D = ''
T = 0
for Q in range(N):
if Q in S1:
D+='L'
T+=L[Q]
if Q in S2:
D+='M'
T+=M[Q]
if Q in S3:
D+='W'
T+=W[Q]
D+='\n'
if T > MAX:
DS = D
MAX = T
if DS:
print(DS)
else:
print('Impossible')
N = int(input())
L = list(range(N))
M = list(range(N))
W = list(range(N))
S = set(L)
X = list(range(N//2+2))
Z = list(range(N-N//2))
RX = list(range(N//2+1))
RZ = list(range(N//2+1, N))
for I in range(N):
L[I], M[I], W[I] = map(int, input().split())
solve() | Title: Lizard Era: Beginning
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the game Lizard Era: Beginning the protagonist will travel with three companions: Lynn, Meliana and Worrigan. Overall the game has *n* mandatory quests. To perform each of them, you need to take exactly two companions.
The attitude of each of the companions to the hero is an integer. Initially, the attitude of each of them to the hero of neutral and equal to 0. As the hero completes quests, he makes actions that change the attitude of the companions, whom he took to perform this task, in positive or negative direction.
Tell us what companions the hero needs to choose to make their attitude equal after completing all the quests. If this can be done in several ways, choose the one in which the value of resulting attitude is greatest possible.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=25) — the number of important tasks.
Next *n* lines contain the descriptions of the tasks — the *i*-th line contains three integers *l**i*,<=*m**i*,<=*w**i* — the values by which the attitude of Lynn, Meliana and Worrigan respectively will change towards the hero if the hero takes them on the *i*-th task. All the numbers in the input are integers and do not exceed 107 in absolute value.
Output Specification:
If there is no solution, print in the first line "Impossible".
Otherwise, print *n* lines, two characters is each line — in the *i*-th line print the first letters of the companions' names that hero should take to complete the *i*-th task ('L' for Lynn, 'M' for Meliana, 'W' for Worrigan). Print the letters in any order, if there are multiple solutions, print any of them.
Demo Input:
['3\n1 0 0\n0 1 0\n0 0 1\n', '7\n0 8 9\n5 9 -2\n6 -8 -7\n9 4 5\n-4 -9 9\n-4 5 2\n-6 8 -7\n', '2\n1 0 0\n1 1 0\n']
Demo Output:
['LM\nMW\nMW\n', 'LM\nMW\nLM\nLW\nMW\nLM\nLW\n', 'Impossible\n']
Note:
none | ```python
from itertools import combinations
def solve():
for I in range(N // 2 + 2):
X[I] = list(combinations(RX, I))
for I in range(N - N // 2):
Z[I] = list(combinations(RZ, I))
LX = []
LY = []
MX = []
MY = []
for I in range(len(X)):
for J in range(len(X[I])):
XL = XM = 0
for A in X[I][J]:
XL += L[A]
XM += M[A]
LX.append([XL, I, J])
MX.append([XM, I, J])
for I in range(len(Z)):
for J in range(len(Z[I])):
YL = YM = 0
for B in Z[I][J]:
YL += L[B]
YM += M[B]
LY.append([YL, I, J])
MY.append([YM, I, J])
LXY = []
MXY = []
PS = set()
MS = set()
for I in range(len(LX)):
for J in range(len(LY)):
if I | J:
P = LX[I][0] + LY[J][0]
LXY.append([P, I, J])
PS.add(P)
for I in range(len(MX)):
for J in range(len(MY)):
if I | J:
P = MX[I][0] + MY[J][0]
if P in PS:
MXY.append([P, I, J])
MS.add(P)
DS = ''
MAX = -10**10
for I in LXY:
for J in MXY:
if I[0] == J[0]:
S1 = set(X[LX[I[1]][1]][LX[I[1]][2]]+Z[LY[I[2]][1]][LY[I[2]][2]])
S2 = set(X[MX[J[1]][1]][MX[J[1]][2]]+Z[MY[J[2]][1]][MY[J[2]][2]])
S3 = (S1-S2).union(S2-S1)
S3S = 0
for U in S3:
S3S+=W[U]
if S3S == I[0]:
if S1.intersection(S2).union(S2.intersection(S3)).union(S3.intersection(S1)) == S:
D = ''
T = 0
for Q in range(N):
if Q in S1:
D+='L'
T+=L[Q]
if Q in S2:
D+='M'
T+=M[Q]
if Q in S3:
D+='W'
T+=W[Q]
D+='\n'
if T > MAX:
DS = D
MAX = T
if DS:
print(DS)
else:
print('Impossible')
N = int(input())
L = list(range(N))
M = list(range(N))
W = list(range(N))
S = set(L)
X = list(range(N//2+2))
Z = list(range(N-N//2))
RX = list(range(N//2+1))
RZ = list(range(N//2+1, N))
for I in range(N):
L[I], M[I], W[I] = map(int, input().split())
solve()
``` | 0 | |
950 | A | Left-handers, Right-handers and Ambidexters | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively. | The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training. | Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players. | [
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] | [
"6\n",
"14\n",
"0\n"
] | In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand. | 500 | [
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
... | 1,579,615,942 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 109 | 0 | l,r,a=map(int,input().split())
if l==r:
print(2*l+(a//2)*2)
else:
if l>r:
l,r=r,l
if r-l>=a:
print((l+a)*2)
else:
print(2*r+(a//2)*2) | Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand. | ```python
l,r,a=map(int,input().split())
if l==r:
print(2*l+(a//2)*2)
else:
if l>r:
l,r=r,l
if r-l>=a:
print((l+a)*2)
else:
print(2*r+(a//2)*2)
``` | 0 | |
342 | A | Xenia and Divisors | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist. | The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3. | If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1. | [
"6\n1 1 1 2 2 2\n",
"6\n2 2 1 1 4 6\n"
] | [
"-1\n",
"1 2 4\n1 2 6\n"
] | none | 500 | [
{
"input": "6\n1 1 1 2 2 2",
"output": "-1"
},
{
"input": "6\n2 2 1 1 4 6",
"output": "1 2 4\n1 2 6"
},
{
"input": "3\n1 2 3",
"output": "-1"
},
{
"input": "3\n7 5 7",
"output": "-1"
},
{
"input": "3\n1 3 4",
"output": "-1"
},
{
"input": "3\n1 1 1",
... | 1,625,060,524 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | test_list = []
res = [sub for sub in test_list if all(ele % 3 == 0 for ele in sub)]
print("tuples : " + str(res))
| Title: Xenia and Divisors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held:
- *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*.
Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.
Help Xenia, find the required partition or else say that it doesn't exist.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7.
It is guaranteed that *n* is divisible by 3.
Output Specification:
If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.
If there is no solution, print -1.
Demo Input:
['6\n1 1 1 2 2 2\n', '6\n2 2 1 1 4 6\n']
Demo Output:
['-1\n', '1 2 4\n1 2 6\n']
Note:
none | ```python
test_list = []
res = [sub for sub in test_list if all(ele % 3 == 0 for ele in sub)]
print("tuples : " + str(res))
``` | 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",
... | 1,623,408,883 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 466 | 10,240,000 | from fractions import Fraction
n=input().split()
s=max(n)
p=6-int(s)+1
s=p/6
if type(s)=="int":
print(s)
else:
print(Fraction(s)) | 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
from fractions import Fraction
n=input().split()
s=max(n)
p=6-int(s)+1
s=p/6
if type(s)=="int":
print(s)
else:
print(Fraction(s))
``` | 0 |
819 | B | Mister B and PR Shifts | PROGRAMMING | 1,900 | [
"data structures",
"implementation",
"math"
] | null | null | Some time ago Mister B detected a strange signal from the space, which he started to study.
After some transformation the signal turned out to be a permutation *p* of length *n* or its cyclic shift. For the further investigation Mister B need some basis, that's why he decided to choose cyclic shift of this permutation which has the minimum possible deviation.
Let's define the deviation of a permutation *p* as .
Find a cyclic shift of permutation *p* with minimum possible deviation. If there are multiple solutions, print any of them.
Let's denote id *k* (0<=≤<=*k*<=<<=*n*) of a cyclic shift of permutation *p* as the number of right shifts needed to reach this shift, for example:
- *k*<==<=0: shift *p*1,<=*p*2,<=... *p**n*, - *k*<==<=1: shift *p**n*,<=*p*1,<=... *p**n*<=-<=1, - ..., - *k*<==<=*n*<=-<=1: shift *p*2,<=*p*3,<=... *p**n*,<=*p*1. | First line contains single integer *n* (2<=≤<=*n*<=≤<=106) — the length of the permutation.
The second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the elements of the permutation. It is guaranteed that all elements are distinct. | Print two integers: the minimum deviation of cyclic shifts of permutation *p* and the id of such shift. If there are multiple solutions, print any of them. | [
"3\n1 2 3\n",
"3\n2 3 1\n",
"3\n3 2 1\n"
] | [
"0 0\n",
"0 1\n",
"2 1\n"
] | In the first sample test the given permutation *p* is the identity permutation, that's why its deviation equals to 0, the shift id equals to 0 as well.
In the second sample test the deviation of *p* equals to 4, the deviation of the 1-st cyclic shift (1, 2, 3) equals to 0, the deviation of the 2-nd cyclic shift (3, 1, 2) equals to 4, the optimal is the 1-st cyclic shift.
In the third sample test the deviation of *p* equals to 4, the deviation of the 1-st cyclic shift (1, 3, 2) equals to 2, the deviation of the 2-nd cyclic shift (2, 1, 3) also equals to 2, so the optimal are both 1-st and 2-nd cyclic shifts. | 1,000 | [
{
"input": "3\n1 2 3",
"output": "0 0"
},
{
"input": "3\n2 3 1",
"output": "0 1"
},
{
"input": "3\n3 2 1",
"output": "2 1"
},
{
"input": "2\n1 2",
"output": "0 0"
},
{
"input": "2\n2 1",
"output": "0 1"
},
{
"input": "10\n10 1 9 2 8 3 7 4 6 5",
"ou... | 1,691,243,538 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1691243538.4798694")# 1691243538.479886 | Title: Mister B and PR Shifts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some time ago Mister B detected a strange signal from the space, which he started to study.
After some transformation the signal turned out to be a permutation *p* of length *n* or its cyclic shift. For the further investigation Mister B need some basis, that's why he decided to choose cyclic shift of this permutation which has the minimum possible deviation.
Let's define the deviation of a permutation *p* as .
Find a cyclic shift of permutation *p* with minimum possible deviation. If there are multiple solutions, print any of them.
Let's denote id *k* (0<=≤<=*k*<=<<=*n*) of a cyclic shift of permutation *p* as the number of right shifts needed to reach this shift, for example:
- *k*<==<=0: shift *p*1,<=*p*2,<=... *p**n*, - *k*<==<=1: shift *p**n*,<=*p*1,<=... *p**n*<=-<=1, - ..., - *k*<==<=*n*<=-<=1: shift *p*2,<=*p*3,<=... *p**n*,<=*p*1.
Input Specification:
First line contains single integer *n* (2<=≤<=*n*<=≤<=106) — the length of the permutation.
The second line contains *n* space-separated integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the elements of the permutation. It is guaranteed that all elements are distinct.
Output Specification:
Print two integers: the minimum deviation of cyclic shifts of permutation *p* and the id of such shift. If there are multiple solutions, print any of them.
Demo Input:
['3\n1 2 3\n', '3\n2 3 1\n', '3\n3 2 1\n']
Demo Output:
['0 0\n', '0 1\n', '2 1\n']
Note:
In the first sample test the given permutation *p* is the identity permutation, that's why its deviation equals to 0, the shift id equals to 0 as well.
In the second sample test the deviation of *p* equals to 4, the deviation of the 1-st cyclic shift (1, 2, 3) equals to 0, the deviation of the 2-nd cyclic shift (3, 1, 2) equals to 4, the optimal is the 1-st cyclic shift.
In the third sample test the deviation of *p* equals to 4, the deviation of the 1-st cyclic shift (1, 3, 2) equals to 2, the deviation of the 2-nd cyclic shift (2, 1, 3) also equals to 2, so the optimal are both 1-st and 2-nd cyclic shifts. | ```python
print("_RANDOM_GUESS_1691243538.4798694")# 1691243538.479886
``` | 0 | |
822 | A | I'm bored with life | PROGRAMMING | 800 | [
"implementation",
"math",
"number theory"
] | null | null | Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you? | The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12). | Print a single integer denoting the greatest common divisor of integers *A*! and *B*!. | [
"4 3\n"
] | [
"6\n"
] | Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | 500 | [
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"ou... | 1,665,206,048 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 1 | 1,000 | 15,155,200 | #By Anan instagram: anan.ssw
from math import factorial
A,B=map(int, input().split())
a,b=factorial(A),factorial(B)
if A==B:
print(A)
elif a<b:
i=1
ans=a
while ans>=a/2:
if b%ans==0 and a%ans==0:
break
ans-=1
print([1,ans][ans>=a/2])
else:
i=1
ans=b
while ans>=b/2:
if b%ans==0 and a%ans==0:
break
ans-=1
print([1,ans][ans>=b/2])
| Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | ```python
#By Anan instagram: anan.ssw
from math import factorial
A,B=map(int, input().split())
a,b=factorial(A),factorial(B)
if A==B:
print(A)
elif a<b:
i=1
ans=a
while ans>=a/2:
if b%ans==0 and a%ans==0:
break
ans-=1
print([1,ans][ans>=a/2])
else:
i=1
ans=b
while ans>=b/2:
if b%ans==0 and a%ans==0:
break
ans-=1
print([1,ans][ans>=b/2])
``` | 0 | |
501 | B | Misha and Changing Handles | PROGRAMMING | 1,100 | [
"data structures",
"dsu",
"strings"
] | null | null | Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that. | The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone. | In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description. | [
"5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n"
] | [
"3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n"
] | none | 500 | [
{
"input": "5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov",
"output": "3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123"
},
{
"input": "1\nMisha Vasya",
"output": "1\nMisha Vasya"
},
{
"input": "10\na b\nb c\nc d\nd... | 1,543,353,810 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 93 | 0 | def add_cconnection(new_change, data):
old, new = new_change.split(" ")
value = data.get(old)
if value:
del data[old]
data[new] = value
else:
data[new] = old
num = int(input())
data = {}
_input = input().split("\n")
print(_input)
for line in _input:
add_cconnection(line, data)
results = list(data.items())
results.reverse()
for new, old in results:
print(old, new)
| Title: Misha and Changing Handles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input Specification:
The first line contains integer *q* (1<=≤<=*q*<=≤<=1000), the number of handle change requests.
Next *q* lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
Output Specification:
In the first line output the integer *n* — the number of users that changed their handles at least once.
In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Demo Input:
['5\nMisha ILoveCodeforces\nVasya Petrov\nPetrov VasyaPetrov123\nILoveCodeforces MikeMirzayanov\nPetya Ivanov\n']
Demo Output:
['3\nPetya Ivanov\nMisha MikeMirzayanov\nVasya VasyaPetrov123\n']
Note:
none | ```python
def add_cconnection(new_change, data):
old, new = new_change.split(" ")
value = data.get(old)
if value:
del data[old]
data[new] = value
else:
data[new] = old
num = int(input())
data = {}
_input = input().split("\n")
print(_input)
for line in _input:
add_cconnection(line, data)
results = list(data.items())
results.reverse()
for new, old in results:
print(old, new)
``` | 0 | |
637 | B | Chat Order | PROGRAMMING | 1,200 | [
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null | Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus. | The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10. | Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom. | [
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] | [
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] | In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex | 1,000 | [
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"outpu... | 1,636,613,200 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 3,000 | 46,694,400 | n = int(input())
s = []
for i in range(n):
if len(s) == 0:
s.append(input())
else:
v = input()
if v in s:
s.remove(v)
s1 = s[:]
s = [v] + s1[:]
else:
s1 = s[:]
s = [v] + s1[:]
[print(x) for x in s] | Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex | ```python
n = int(input())
s = []
for i in range(n):
if len(s) == 0:
s.append(input())
else:
v = input()
if v in s:
s.remove(v)
s1 = s[:]
s = [v] + s1[:]
else:
s1 = s[:]
s = [v] + s1[:]
[print(x) for x in s]
``` | 0 | |
195 | B | After Training | PROGRAMMING | 1,300 | [
"data structures",
"implementation",
"math"
] | null | null | After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on. | The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly. | Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball. | [
"4 3\n",
"3 1\n"
] | [
"2\n1\n3\n2\n",
"1\n1\n1\n"
] | none | 1,000 | [
{
"input": "4 3",
"output": "2\n1\n3\n2"
},
{
"input": "3 1",
"output": "1\n1\n1"
},
{
"input": "10 3",
"output": "2\n1\n3\n2\n1\n3\n2\n1\n3\n2"
},
{
"input": "6 5",
"output": "3\n2\n4\n1\n5\n3"
},
{
"input": "2 6",
"output": "3\n4"
},
{
"input": "5 2"... | 1,648,913,288 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 530 | 0 | #----------------------------
# Matheus de Souza Oliveira |
# RA: 203407 |
#----------------------------
n, m = list(map(int, input().split()))
for index in range(0, n):
factor = m+index%m
isFactorEven = not(bool(factor%2))
if isFactorEven:
result = int((m-(index%m))/2)
print(result)
else:
result = int((factor+1)/2)
print(result)
| Title: After Training
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has *n* balls and *m* baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to *m*, correspondingly. The balls are numbered with numbers from 1 to *n*.
Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which is minimum, where *i* is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.
For every ball print the number of the basket where it will go according to Valeric's scheme.
Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.
Input Specification:
The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of balls and baskets, correspondingly.
Output Specification:
Print *n* numbers, one per line. The *i*-th line must contain the number of the basket for the *i*-th ball.
Demo Input:
['4 3\n', '3 1\n']
Demo Output:
['2\n1\n3\n2\n', '1\n1\n1\n']
Note:
none | ```python
#----------------------------
# Matheus de Souza Oliveira |
# RA: 203407 |
#----------------------------
n, m = list(map(int, input().split()))
for index in range(0, n):
factor = m+index%m
isFactorEven = not(bool(factor%2))
if isFactorEven:
result = int((m-(index%m))/2)
print(result)
else:
result = int((factor+1)/2)
print(result)
``` | 3 | |
304 | B | Calendar | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | Calendars in widespread use today include the Gregorian calendar, which is the de facto international standard, and is used almost everywhere in the world for civil purposes. The Gregorian reform modified the Julian calendar's scheme of leap years as follows:
Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100; the centurial years that are exactly divisible by 400 are still leap years. For example, the year 1900 is not a leap year; the year 2000 is a leap year.
In this problem, you have been given two dates and your task is to calculate how many days are between them. Note, that leap years have unusual number of days in February.
Look at the sample to understand what borders are included in the aswer. | The first two lines contain two dates, each date is in the format yyyy:mm:dd (1900<=≤<=*yyyy*<=≤<=2038 and yyyy:mm:dd is a legal date). | Print a single integer — the answer to the problem. | [
"1900:01:01\n2038:12:31\n",
"1996:03:09\n1991:11:12\n"
] | [
"50768\n",
"1579\n"
] | none | 1,000 | [
{
"input": "1900:01:01\n2038:12:31",
"output": "50768"
},
{
"input": "1996:03:09\n1991:11:12",
"output": "1579"
},
{
"input": "1999:12:31\n2000:02:29",
"output": "60"
},
{
"input": "1903:09:27\n1988:06:15",
"output": "30943"
},
{
"input": "1913:11:14\n1901:05:11",... | 1,541,059,773 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 404 | 1,126,400 | import datetime
d1=datetime.datetime.strptime(input(),"%Y:%m:%d")
d2=datetime.datetime.strptime(input(),"%Y:%m:%d")
print(abs((d2-d1).days)) | Title: Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Calendars in widespread use today include the Gregorian calendar, which is the de facto international standard, and is used almost everywhere in the world for civil purposes. The Gregorian reform modified the Julian calendar's scheme of leap years as follows:
Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100; the centurial years that are exactly divisible by 400 are still leap years. For example, the year 1900 is not a leap year; the year 2000 is a leap year.
In this problem, you have been given two dates and your task is to calculate how many days are between them. Note, that leap years have unusual number of days in February.
Look at the sample to understand what borders are included in the aswer.
Input Specification:
The first two lines contain two dates, each date is in the format yyyy:mm:dd (1900<=≤<=*yyyy*<=≤<=2038 and yyyy:mm:dd is a legal date).
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['1900:01:01\n2038:12:31\n', '1996:03:09\n1991:11:12\n']
Demo Output:
['50768\n', '1579\n']
Note:
none | ```python
import datetime
d1=datetime.datetime.strptime(input(),"%Y:%m:%d")
d2=datetime.datetime.strptime(input(),"%Y:%m:%d")
print(abs((d2-d1).days))
``` | 3 | |
723 | A | The New Year: Meeting Friends | PROGRAMMING | 800 | [
"implementation",
"math",
"sortings"
] | null | null | There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer. | The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively. | Print one integer — the minimum total distance the friends need to travel in order to meet together. | [
"7 1 4\n",
"30 20 10\n"
] | [
"6\n",
"20\n"
] | In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | 500 | [
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"... | 1,694,946,086 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | a,b,c=map(int,input().split())
s=(a+b+c)//3
print(abs(a-s)+abs(s-b)+abs(s-c)) | Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | ```python
a,b,c=map(int,input().split())
s=(a+b+c)//3
print(abs(a-s)+abs(s-b)+abs(s-c))
``` | 0 | |
31 | C | Schedule | PROGRAMMING | 1,700 | [
"implementation"
] | C. Schedule | 2 | 256 | At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that. | The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1). | Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order. | [
"3\n3 10\n20 30\n1 3\n",
"4\n3 10\n20 30\n1 3\n1 39\n",
"3\n1 5\n2 6\n3 7\n"
] | [
"3\n1 2 3 ",
"1\n4 ",
"0\n"
] | none | 1,500 | [
{
"input": "3\n3 10\n20 30\n1 3",
"output": "3\n1 2 3 "
},
{
"input": "4\n3 10\n20 30\n1 3\n1 39",
"output": "1\n4 "
},
{
"input": "3\n1 5\n2 6\n3 7",
"output": "0"
},
{
"input": "4\n1 5\n5 7\n6 9\n9 10",
"output": "2\n2 3 "
},
{
"input": "11\n717170 795210\n86642... | 1,643,679,527 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 204,800 | while True:
try:
n = int(input())
times = []
s = [0] * 5000
sum = 0
for i in range(n):
l, r = input().split()
l = int(l)
r = int(r)
times.append((l, r))
for i in range(n):
for j in range(i):
if times[j][0] < times[i][1] and times[i][0] < times[j][1]:
s[i] += 1
s[j] += 1
sum += 1
temp = []
for i in range(n):
if s[i] == sum:
temp.append(i + 1)
print(len(temp))
print(*temp)
except EOFError:
break
| Title: Schedule
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1).
Output Specification:
Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order.
Demo Input:
['3\n3 10\n20 30\n1 3\n', '4\n3 10\n20 30\n1 3\n1 39\n', '3\n1 5\n2 6\n3 7\n']
Demo Output:
['3\n1 2 3 ', '1\n4 ', '0\n']
Note:
none | ```python
while True:
try:
n = int(input())
times = []
s = [0] * 5000
sum = 0
for i in range(n):
l, r = input().split()
l = int(l)
r = int(r)
times.append((l, r))
for i in range(n):
for j in range(i):
if times[j][0] < times[i][1] and times[i][0] < times[j][1]:
s[i] += 1
s[j] += 1
sum += 1
temp = []
for i in range(n):
if s[i] == sum:
temp.append(i + 1)
print(len(temp))
print(*temp)
except EOFError:
break
``` | 0 |
614 | B | Gena's Code | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case. | The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000. | Print a single number without leading zeroes — the product of the number of tanks presented by each country. | [
"3\n5 10 1\n",
"4\n1 1 10 11\n",
"5\n0 3 1 100 1\n"
] | [
"50",
"110",
"0"
] | In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | 1,000 | [
{
"input": "3\n5 10 1",
"output": "50"
},
{
"input": "4\n1 1 10 11",
"output": "110"
},
{
"input": "5\n0 3 1 100 1",
"output": "0"
},
{
"input": "40\n10 100 10 1 10 10 100 10 10 100 10 100 100 10 1824868942 100 100 1 10 100 100 10 100 100 10 100 10 1 10 100 100 100 10 1 10 1 ... | 1,461,235,529 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 21 | 139 | 6,348,800 | from sys import stdin
def isBad(s):
if s[0] != '1':
return 1
for i in s[1:]:
if i != '0':
return 1
return 0
def getBad(a):
for i in a:
if isBad(i):
return i
return None
def getPowerOfTen(a):
res = 0
for i in a:
if not isBad(i) and not i == '0':
res += len(i) - 1
return res
n = int(stdin.readline().strip())
a = stdin.readline().strip().split(' ')
b = getBad(a)
# print('bad :', b)
if b is None:
print('1' + '0' * getPowerOfTen(a))
elif b == '0':
print('0')
else:
print(b + '0' * getPowerOfTen(a)) | Title: Gena's Code
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly *n* distinct countries in the world and the *i*-th country added *a**i* tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input Specification:
The first line of the input contains the number of countries *n* (1<=≤<=*n*<=≤<=100<=000). The second line contains *n* non-negative integers *a**i* without leading zeroes — the number of tanks of the *i*-th country.
It is guaranteed that the second line contains at least *n*<=-<=1 beautiful numbers and the total length of all these number's representations doesn't exceed 100<=000.
Output Specification:
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Demo Input:
['3\n5 10 1\n', '4\n1 1 10 11\n', '5\n0 3 1 100 1\n']
Demo Output:
['50', '110', '0']
Note:
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | ```python
from sys import stdin
def isBad(s):
if s[0] != '1':
return 1
for i in s[1:]:
if i != '0':
return 1
return 0
def getBad(a):
for i in a:
if isBad(i):
return i
return None
def getPowerOfTen(a):
res = 0
for i in a:
if not isBad(i) and not i == '0':
res += len(i) - 1
return res
n = int(stdin.readline().strip())
a = stdin.readline().strip().split(' ')
b = getBad(a)
# print('bad :', b)
if b is None:
print('1' + '0' * getPowerOfTen(a))
elif b == '0':
print('0')
else:
print(b + '0' * getPowerOfTen(a))
``` | 0 | |
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "... | 1,577,005,173 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 312 | 0 | n,m = list(map(int,input().split()))
l = list(map(int,input().split()))
l.sort()
money_earned = 0
for i in range(0,m):
if(l[i]<0):
money_earned-=l[i]
print(money_earned) | Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
n,m = list(map(int,input().split()))
l = list(map(int,input().split()))
l.sort()
money_earned = 0
for i in range(0,m):
if(l[i]<0):
money_earned-=l[i]
print(money_earned)
``` | 3.922 |
15 | A | Cottage Village | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | A. Cottage Village | 2 | 64 | A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house? | The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000). | Output the amount of possible positions of the new house. | [
"2 2\n0 4\n6 2\n",
"2 2\n0 4\n5 2\n",
"2 3\n0 4\n5 2\n"
] | [
"4\n",
"3\n",
"2\n"
] | It is possible for the *x*-coordinate of the new house to have non-integer value. | 0 | [
{
"input": "2 2\n0 4\n6 2",
"output": "4"
},
{
"input": "2 2\n0 4\n5 2",
"output": "3"
},
{
"input": "2 3\n0 4\n5 2",
"output": "2"
},
{
"input": "1 1\n1 1",
"output": "2"
},
{
"input": "1 2\n2 1",
"output": "2"
},
{
"input": "2 1\n2 1\n1 1",
"outp... | 1,690,443,779 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 35 | 154 | 2,355,200 | from collections import deque
lst = [w.rstrip() for w in open(0).readlines()]
n, t = map(int, lst[0].split())
data = [tuple(map(int, x.split())) for x in lst[1:]]
L, R = [], []
for x in data:
L.append(x[0] - x[1] / 2)
R.append(x[0] + x[1] / 2)
L.sort()
R.sort()
d = deque()
for a, b in zip(L, R):
if len(d) > 0 and d[-1] == a:
d[-1] = b
else:
d.append(a)
d.append(b)
d.popleft()
d = list(d)
r = 2
for a, b in zip(d[::2], d[1::2]):
if b-a > t:
r += 2
elif b-a == t:
r += 1
print(r) | Title: Cottage Village
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other.
The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village.
Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?
Input Specification:
The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000).
Output Specification:
Output the amount of possible positions of the new house.
Demo Input:
['2 2\n0 4\n6 2\n', '2 2\n0 4\n5 2\n', '2 3\n0 4\n5 2\n']
Demo Output:
['4\n', '3\n', '2\n']
Note:
It is possible for the *x*-coordinate of the new house to have non-integer value. | ```python
from collections import deque
lst = [w.rstrip() for w in open(0).readlines()]
n, t = map(int, lst[0].split())
data = [tuple(map(int, x.split())) for x in lst[1:]]
L, R = [], []
for x in data:
L.append(x[0] - x[1] / 2)
R.append(x[0] + x[1] / 2)
L.sort()
R.sort()
d = deque()
for a, b in zip(L, R):
if len(d) > 0 and d[-1] == a:
d[-1] = b
else:
d.append(a)
d.append(b)
d.popleft()
d = list(d)
r = 2
for a, b in zip(d[::2], d[1::2]):
if b-a > t:
r += 2
elif b-a == t:
r += 1
print(r)
``` | 3.943952 |
977 | A | Wrong Subtraction | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number. | The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly. | Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number. | [
"512 4\n",
"1000000000 9\n"
] | [
"50\n",
"1\n"
] | The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | 0 | [
{
"input": "512 4",
"output": "50"
},
{
"input": "1000000000 9",
"output": "1"
},
{
"input": "131203 11",
"output": "12"
},
{
"input": "999999999 50",
"output": "9999"
},
{
"input": "999999999 49",
"output": "99990"
},
{
"input": "131203 9",
"outpu... | 1,695,110,998 | 2,147,483,647 | Python 3 | OK | TESTS | 11 | 46 | 0 | n,k=map(int,input().split())
a=0#记录操作次数#
while a<k:
a+=1
if n%10==0:
n=n//10
else:
n-=1
print(n) | Title: Wrong Subtraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
Input Specification:
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly.
Output Specification:
Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
Demo Input:
['512 4\n', '1000000000 9\n']
Demo Output:
['50\n', '1\n']
Note:
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | ```python
n,k=map(int,input().split())
a=0#记录操作次数#
while a<k:
a+=1
if n%10==0:
n=n//10
else:
n-=1
print(n)
``` | 3 | |
318 | A | Even Odds | PROGRAMMING | 900 | [
"math"
] | null | null | Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*. | The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print the number that will stand at the position number *k* after Volodya's manipulations. | [
"10 3\n",
"7 7\n"
] | [
"5",
"6"
] | In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | 500 | [
{
"input": "10 3",
"output": "5"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "7 1",
"output": "1"
},
{
"input": "7 2",
"output": "3"
},
{
"input": "8 5",
"output": "2"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output... | 1,695,345,741 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 1,000 | 138,956,800 | n, k = [ int(n) for n in input().split()]
odds = [x for x in range(n+1) if x % 2 != 0]
evens = [x for x in range(n+1) if x % 2 == 0 and x != 0]
combined = odds + evens
if k != 0:
print(combined[k-1])
else:
print(combined[k])
| Title: Even Odds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
Input Specification:
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the number that will stand at the position number *k* after Volodya's manipulations.
Demo Input:
['10 3\n', '7 7\n']
Demo Output:
['5', '6']
Note:
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | ```python
n, k = [ int(n) for n in input().split()]
odds = [x for x in range(n+1) if x % 2 != 0]
evens = [x for x in range(n+1) if x % 2 == 0 and x != 0]
combined = odds + evens
if k != 0:
print(combined[k-1])
else:
print(combined[k])
``` | 0 | |
499 | A | Watching a movie | PROGRAMMING | 1,000 | [
"greedy",
"implementation"
] | null | null | You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments? | The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*. | Output a single number — the answer to the problem. | [
"2 3\n5 6\n10 12\n",
"1 1\n1 100000\n"
] | [
"6\n",
"100000\n"
] | In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie. | 500 | [
{
"input": "2 3\n5 6\n10 12",
"output": "6"
},
{
"input": "1 1\n1 100000",
"output": "100000"
},
{
"input": "10 1\n2156 3497\n4784 7775\n14575 31932\n33447 35902\n36426 47202\n48772 60522\n63982 68417\n78537 79445\n90081 90629\n94325 95728",
"output": "53974"
},
{
"input": "1... | 1,599,204,696 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 204,800 | a=list(map(int,input().split()))
n=int(a[0])
x=int(a[1])
l=[0]*n
r=[0]*n
for i in range(0,n):
l[i]=int(input())
r[i]=int(input())
j=0
for i in range(0,n):
if i==0:
j=j+((l[i]-1)%x)+abs(l[i]-r[i])+1
else:
j=j+((l[i]-r[i-1])%x)+abs(l[i]-r[i])
print(j)
| Title: Watching a movie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments?
Input Specification:
The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*.
Output Specification:
Output a single number — the answer to the problem.
Demo Input:
['2 3\n5 6\n10 12\n', '1 1\n1 100000\n']
Demo Output:
['6\n', '100000\n']
Note:
In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie. | ```python
a=list(map(int,input().split()))
n=int(a[0])
x=int(a[1])
l=[0]*n
r=[0]*n
for i in range(0,n):
l[i]=int(input())
r[i]=int(input())
j=0
for i in range(0,n):
if i==0:
j=j+((l[i]-1)%x)+abs(l[i]-r[i])+1
else:
j=j+((l[i]-r[i-1])%x)+abs(l[i]-r[i])
print(j)
``` | -1 | |
798 | A | Mike and palindrome | PROGRAMMING | 1,000 | [
"brute force",
"constructive algorithms",
"strings"
] | null | null | Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not. | The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15). | Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise. | [
"abccaa\n",
"abbcca\n",
"abcda\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "abccaa",
"output": "YES"
},
{
"input": "abbcca",
"output": "NO"
},
{
"input": "abcda",
"output": "YES"
},
{
"input": "kyw",
"output": "YES"
},
{
"input": "fccf",
"output": "NO"
},
{
"input": "mnlm",
"output": "YES"
},
{
"input":... | 1,637,337,480 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | s=input()
l=len(s)
m=0
for i in range(l):
if s[i]!=s[l-1-i]:
m+=1
continue
else:
m+=1
if m==1:
print('YES')
else:
print('NO') | Title: Mike and palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
Input Specification:
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
Output Specification:
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
Demo Input:
['abccaa\n', 'abbcca\n', 'abcda\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
s=input()
l=len(s)
m=0
for i in range(l):
if s[i]!=s[l-1-i]:
m+=1
continue
else:
m+=1
if m==1:
print('YES')
else:
print('NO')
``` | 0 | |
991 | C | Candies | PROGRAMMING | 1,500 | [
"binary search",
"implementation"
] | null | null | After passing a test, Vasya got himself a box of $n$ candies. He decided to eat an equal amount of candies each morning until there are no more candies. However, Petya also noticed the box and decided to get some candies for himself.
This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$, same for all days. After that, in the morning he eats $k$ candies from the box (if there are less than $k$ candies in the box, he eats them all), then in the evening Petya eats $10\%$ of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats $k$ candies again, and Petya — $10\%$ of the candies left in a box, and so on.
If the amount of candies in the box is not divisible by $10$, Petya rounds the amount he takes from the box down. For example, if there were $97$ candies in the box, Petya would eat only $9$ of them. In particular, if there are less than $10$ candies in a box, Petya won't eat any at all.
Your task is to find out the minimal amount of $k$ that can be chosen by Vasya so that he would eat at least half of the $n$ candies he initially got. Note that the number $k$ must be integer. | The first line contains a single integer $n$ ($1 \leq n \leq 10^{18}$) — the initial amount of candies in the box. | Output a single integer — the minimal amount of $k$ that would allow Vasya to eat at least half of candies he got. | [
"68\n"
] | [
"3\n"
] | In the sample, the amount of candies, with $k=3$, would change in the following way (Vasya eats first):
$68 \to 65 \to 59 \to 56 \to 51 \to 48 \to 44 \to 41 \\ \to 37 \to 34 \to 31 \to 28 \to 26 \to 23 \to 21 \to 18 \to 17 \to 14 \\ \to 13 \to 10 \to 9 \to 6 \to 6 \to 3 \to 3 \to 0$.
In total, Vasya would eat $39$ candies, while Petya — $29$. | 1,250 | [
{
"input": "68",
"output": "3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "42",
"output": "1"
},
{
"input": "43",
"output": "2"
},
{
"input": "756",
"output": "29"
},
{
"input": "999999972",
"output"... | 1,546,188,924 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 1 | 1,000 | 0 | n=int(input())
def w(k):
s=n
p=0
while s>0:
p+=min(s,k)
s=max(0,s-k)
s-=s//10
if p>=n/2:
return True
return False
l,r=0,n
while l<r:
m=(l+r)//2
if w(m):
r=m
else:
l=m+1
print(r) | Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After passing a test, Vasya got himself a box of $n$ candies. He decided to eat an equal amount of candies each morning until there are no more candies. However, Petya also noticed the box and decided to get some candies for himself.
This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$, same for all days. After that, in the morning he eats $k$ candies from the box (if there are less than $k$ candies in the box, he eats them all), then in the evening Petya eats $10\%$ of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats $k$ candies again, and Petya — $10\%$ of the candies left in a box, and so on.
If the amount of candies in the box is not divisible by $10$, Petya rounds the amount he takes from the box down. For example, if there were $97$ candies in the box, Petya would eat only $9$ of them. In particular, if there are less than $10$ candies in a box, Petya won't eat any at all.
Your task is to find out the minimal amount of $k$ that can be chosen by Vasya so that he would eat at least half of the $n$ candies he initially got. Note that the number $k$ must be integer.
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 10^{18}$) — the initial amount of candies in the box.
Output Specification:
Output a single integer — the minimal amount of $k$ that would allow Vasya to eat at least half of candies he got.
Demo Input:
['68\n']
Demo Output:
['3\n']
Note:
In the sample, the amount of candies, with $k=3$, would change in the following way (Vasya eats first):
$68 \to 65 \to 59 \to 56 \to 51 \to 48 \to 44 \to 41 \\ \to 37 \to 34 \to 31 \to 28 \to 26 \to 23 \to 21 \to 18 \to 17 \to 14 \\ \to 13 \to 10 \to 9 \to 6 \to 6 \to 3 \to 3 \to 0$.
In total, Vasya would eat $39$ candies, while Petya — $29$. | ```python
n=int(input())
def w(k):
s=n
p=0
while s>0:
p+=min(s,k)
s=max(0,s-k)
s-=s//10
if p>=n/2:
return True
return False
l,r=0,n
while l<r:
m=(l+r)//2
if w(m):
r=m
else:
l=m+1
print(r)
``` | 0 | |
987 | C | Three displays | PROGRAMMING | 1,400 | [
"brute force",
"dp",
"implementation"
] | null | null | It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are $n$ displays placed along a road, and the $i$-th of them can display a text with font size $s_i$ only. Maria Stepanovna wants to rent such three displays with indices $i < j < k$ that the font size increases if you move along the road in a particular direction. Namely, the condition $s_i < s_j < s_k$ should be held.
The rent cost is for the $i$-th display is $c_i$. Please determine the smallest cost Maria Stepanovna should pay. | The first line contains a single integer $n$ ($3 \le n \le 3\,000$) — the number of displays.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^9$) — the font sizes on the displays in the order they stand along the road.
The third line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^8$) — the rent costs for each display. | If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $i < j < k$ such that $s_i < s_j < s_k$. | [
"5\n2 4 5 4 10\n40 30 20 10 40\n",
"3\n100 101 100\n2 4 5\n",
"10\n1 2 3 4 5 6 7 8 9 10\n10 13 11 14 15 12 13 13 18 13\n"
] | [
"90\n",
"-1\n",
"33\n"
] | In the first example you can, for example, choose displays $1$, $4$ and $5$, because $s_1 < s_4 < s_5$ ($2 < 4 < 10$), and the rent cost is $40 + 10 + 40 = 90$.
In the second example you can't select a valid triple of indices, so the answer is -1. | 1,250 | [
{
"input": "5\n2 4 5 4 10\n40 30 20 10 40",
"output": "90"
},
{
"input": "3\n100 101 100\n2 4 5",
"output": "-1"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10\n10 13 11 14 15 12 13 13 18 13",
"output": "33"
},
{
"input": "3\n1 2 3\n100000000 100000000 100000000",
"output": "300... | 1,604,770,881 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 10 | 296 | 7,270,400 |
n = int(input())
s = [int(x) for x in input().strip().split()]
c = [int(x) for x in input().strip().split()]
from functools import lru_cache
@lru_cache(None)
def helper(i, last, nums):
if nums==3:
return 0
if i==len(s):
return float('inf')
ans = helper(i+1, last, nums)
if last==None:
ans = min(ans, helper(i+1, i, nums+1)+c[i])
else:
if s[i]>s[last]:
ans = min(ans, helper(i+1, i, nums+1)+c[i])
return ans
val = helper(0, None, 0)
print(val if val!=float('inf') else -1)
| Title: Three displays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are $n$ displays placed along a road, and the $i$-th of them can display a text with font size $s_i$ only. Maria Stepanovna wants to rent such three displays with indices $i < j < k$ that the font size increases if you move along the road in a particular direction. Namely, the condition $s_i < s_j < s_k$ should be held.
The rent cost is for the $i$-th display is $c_i$. Please determine the smallest cost Maria Stepanovna should pay.
Input Specification:
The first line contains a single integer $n$ ($3 \le n \le 3\,000$) — the number of displays.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^9$) — the font sizes on the displays in the order they stand along the road.
The third line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^8$) — the rent costs for each display.
Output Specification:
If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $i < j < k$ such that $s_i < s_j < s_k$.
Demo Input:
['5\n2 4 5 4 10\n40 30 20 10 40\n', '3\n100 101 100\n2 4 5\n', '10\n1 2 3 4 5 6 7 8 9 10\n10 13 11 14 15 12 13 13 18 13\n']
Demo Output:
['90\n', '-1\n', '33\n']
Note:
In the first example you can, for example, choose displays $1$, $4$ and $5$, because $s_1 < s_4 < s_5$ ($2 < 4 < 10$), and the rent cost is $40 + 10 + 40 = 90$.
In the second example you can't select a valid triple of indices, so the answer is -1. | ```python
n = int(input())
s = [int(x) for x in input().strip().split()]
c = [int(x) for x in input().strip().split()]
from functools import lru_cache
@lru_cache(None)
def helper(i, last, nums):
if nums==3:
return 0
if i==len(s):
return float('inf')
ans = helper(i+1, last, nums)
if last==None:
ans = min(ans, helper(i+1, i, nums+1)+c[i])
else:
if s[i]>s[last]:
ans = min(ans, helper(i+1, i, nums+1)+c[i])
return ans
val = helper(0, None, 0)
print(val if val!=float('inf') else -1)
``` | -1 | |
393 | A | Nineteen | PROGRAMMING | 0 | [] | null | null | Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string. | The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100. | Print a single integer — the maximum number of "nineteen"s that she can get in her string. | [
"nniinneetteeeenn\n",
"nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n",
"nineteenineteen\n"
] | [
"2",
"2",
"2"
] | none | 500 | [
{
"input": "nniinneetteeeenn",
"output": "2"
},
{
"input": "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii",
"output": "2"
},
{
"input": "nineteenineteen",
"output": "2"
},
{
"input": "nssemsnnsitjtihtthij",
"output": "0"
},
{
"input": "eehihnttehtherjsihihn... | 1,584,463,642 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 18 | 109 | 0 | x=input()
n=x.count('n')
e=x.count('e')
i=x.count('i')
t=x.count('t')
n1=(n-1)//2
e1=e//3
lim=min(n1,e1,i,t)
print(lim)
| Title: Nineteen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string.
Input Specification:
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
Output Specification:
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
Demo Input:
['nniinneetteeeenn\n', 'nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n', 'nineteenineteen\n']
Demo Output:
['2', '2', '2']
Note:
none | ```python
x=input()
n=x.count('n')
e=x.count('e')
i=x.count('i')
t=x.count('t')
n1=(n-1)//2
e1=e//3
lim=min(n1,e1,i,t)
print(lim)
``` | 0 | |
27 | A | Next Test | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | A. Next Test | 2 | 256 | «Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests. | Output the required default value for the next test index. | [
"3\n1 7 2\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "3\n3 4 1",
"output": "2"
},
{
"input": "4\n6 4 3 5",
"output": "1"
},
{
"input": "5\n3 2 1 7 4",
"output": "5"
},
{
"input": "6\n4 1 2 5 3 7",
"output": "6"
},
... | 1,659,519,501 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 30 | 124 | 1,740,800 | n = int(input())
tasks = list(map(int, input().split()))
index = -1
for i in range(1, 3001):
if not (i in tasks):
index = i
break
if index == -1:
tasks.sort()
index = tasks[n-1] + 1
print(index) | Title: Next Test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
Output Specification:
Output the required default value for the next test index.
Demo Input:
['3\n1 7 2\n']
Demo Output:
['3\n']
Note:
none | ```python
n = int(input())
tasks = list(map(int, input().split()))
index = -1
for i in range(1, 3001):
if not (i in tasks):
index = i
break
if index == -1:
tasks.sort()
index = tasks[n-1] + 1
print(index)
``` | 3.965758 |
514 | A | Chewbaсca and Number | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero. | The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca. | Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes. | [
"27\n",
"4545\n"
] | [
"22\n",
"4444\n"
] | none | 500 | [
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
... | 1,666,180,899 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 31 | 0 | n = input()
s = 0
for i in n:
if 9 - int(i) < int(i):
print(9 - int(i) ,end='')
else:
print(i ,end = '')
| Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none | ```python
n = input()
s = 0
for i in n:
if 9 - int(i) < int(i):
print(9 - int(i) ,end='')
else:
print(i ,end = '')
``` | 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",... | 1,560,206,228 | 2,147,483,647 | Python 3 | OK | TESTS | 93 | 124 | 0 | n = input().split()
a = int(n[0])
b = int(n[1])
if(a==b):
print(a)
else:
print('2')
| 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
n = input().split()
a = int(n[0])
b = int(n[1])
if(a==b):
print(a)
else:
print('2')
``` | 3 | |
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",
... | 1,596,530,873 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 140 | 20,172,800 | a,b = [int(x) for x in input().split(" ")]
time = 0
while a and b:
if a==1 and b==1:
break
if a<b:
a += 1
b -= 2
else:
b += 1
a -= 2
time+=1
print(time) | 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(" ")]
time = 0
while a and b:
if a==1 and b==1:
break
if a<b:
a += 1
b -= 2
else:
b += 1
a -= 2
time+=1
print(time)
``` | 3 | |
255 | C | Almost Arithmetical Progression | PROGRAMMING | 1,500 | [
"brute force",
"dp"
] | null | null | Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:
- *a*1<==<=*p*, where *p* is some integer; - *a**i*<==<=*a**i*<=-<=1<=+<=(<=-<=1)*i*<=+<=1·*q* (*i*<=><=1), where *q* is some integer.
Right now Gena has a piece of paper with sequence *b*, consisting of *n* integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.
Sequence *s*1,<=<=*s*2,<=<=...,<=<=*s**k* is a subsequence of sequence *b*1,<=<=*b*2,<=<=...,<=<=*b**n*, if there is such increasing sequence of indexes *i*1,<=*i*2,<=...,<=*i**k* (1<=<=≤<=<=*i*1<=<=<<=<=*i*2<=<=<<=... <=<=<<=<=*i**k*<=<=≤<=<=*n*), that *b**i**j*<=<==<=<=*s**j*. In other words, sequence *s* can be obtained from *b* by crossing out some elements. | The first line contains integer *n* (1<=≤<=*n*<=≤<=4000). The next line contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=106). | Print a single integer — the length of the required longest subsequence. | [
"2\n3 5\n",
"4\n10 20 10 30\n"
] | [
"2\n",
"3\n"
] | In the first test the sequence actually is the suitable subsequence.
In the second test the following subsequence fits: 10, 20, 10. | 1,500 | [
{
"input": "2\n3 5",
"output": "2"
},
{
"input": "4\n10 20 10 30",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "2"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "3"
},
{
"input": "2\n468 335",
... | 1,605,155,894 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 21 | 249 | 5,427,200 | #!/usr/bin/env python3
from sys import stdin
from bisect import bisect_left
def solve(tc):
n = int(stdin.readline().strip())
seq = list(map(int, stdin.readline().split()))
elems = []
idxcnt = 0
idx = dict()
for i in range(n):
if seq[i] not in idx:
idx[seq[i]] = idxcnt
idxcnt += 1
elems.append([])
elems[idx[seq[i]]].append(i)
ans = 1
m = len(elems)
for i in range(m):
for j in range(m):
if i == j:
continue
cur = -1
cnt = 0
leni, lenj = len(elems[i]), len(elems[j])
pi, pj = 0, 0
while True:
k = bisect_left(elems[i], cur, pi, leni)
if k == leni:
break
pi = k+1
cur = elems[i][k]
cnt += 1
k = bisect_left(elems[j], cur, pj, lenj)
if k == lenj:
break
pj = k+1
cur = elems[j][k]
cnt += 1
ans = max(ans, cnt)
print(ans)
tc = 1
solve(tc)
| Title: Almost Arithmetical Progression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:
- *a*1<==<=*p*, where *p* is some integer; - *a**i*<==<=*a**i*<=-<=1<=+<=(<=-<=1)*i*<=+<=1·*q* (*i*<=><=1), where *q* is some integer.
Right now Gena has a piece of paper with sequence *b*, consisting of *n* integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.
Sequence *s*1,<=<=*s*2,<=<=...,<=<=*s**k* is a subsequence of sequence *b*1,<=<=*b*2,<=<=...,<=<=*b**n*, if there is such increasing sequence of indexes *i*1,<=*i*2,<=...,<=*i**k* (1<=<=≤<=<=*i*1<=<=<<=<=*i*2<=<=<<=... <=<=<<=<=*i**k*<=<=≤<=<=*n*), that *b**i**j*<=<==<=<=*s**j*. In other words, sequence *s* can be obtained from *b* by crossing out some elements.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=4000). The next line contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Print a single integer — the length of the required longest subsequence.
Demo Input:
['2\n3 5\n', '4\n10 20 10 30\n']
Demo Output:
['2\n', '3\n']
Note:
In the first test the sequence actually is the suitable subsequence.
In the second test the following subsequence fits: 10, 20, 10. | ```python
#!/usr/bin/env python3
from sys import stdin
from bisect import bisect_left
def solve(tc):
n = int(stdin.readline().strip())
seq = list(map(int, stdin.readline().split()))
elems = []
idxcnt = 0
idx = dict()
for i in range(n):
if seq[i] not in idx:
idx[seq[i]] = idxcnt
idxcnt += 1
elems.append([])
elems[idx[seq[i]]].append(i)
ans = 1
m = len(elems)
for i in range(m):
for j in range(m):
if i == j:
continue
cur = -1
cnt = 0
leni, lenj = len(elems[i]), len(elems[j])
pi, pj = 0, 0
while True:
k = bisect_left(elems[i], cur, pi, leni)
if k == leni:
break
pi = k+1
cur = elems[i][k]
cnt += 1
k = bisect_left(elems[j], cur, pj, lenj)
if k == lenj:
break
pj = k+1
cur = elems[j][k]
cnt += 1
ans = max(ans, cnt)
print(ans)
tc = 1
solve(tc)
``` | 0 | |
353 | A | Domino | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | null | null | Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half. | Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1. | [
"2\n4 2\n6 4\n",
"1\n2 3\n",
"3\n1 4\n2 3\n4 4\n"
] | [
"0\n",
"-1\n",
"1\n"
] | In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8. | 500 | [
{
"input": "2\n4 2\n6 4",
"output": "0"
},
{
"input": "1\n2 3",
"output": "-1"
},
{
"input": "3\n1 4\n2 3\n4 4",
"output": "1"
},
{
"input": "5\n5 4\n5 4\n1 5\n5 5\n3 3",
"output": "1"
},
{
"input": "20\n1 3\n5 2\n5 2\n2 6\n2 4\n1 1\n1 3\n1 4\n2 6\n4 2\n5 6\n2 2\n... | 1,599,243,594 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 7 | 248 | 0 | n=int(input())
r=[list(map(int,input().split())) for _ in range(n)]
s1,s2=0,0
for i in range(n):
s1+=r[i][0]
s2+=r[i][1]
if s1%2==0 and s2%2==0:
print(0)
elif s1%2!=s2%2:
print(-1)
else:
print(1) | Title: Domino
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera has got *n* domino pieces in a row. Each piece consists of two halves — the upper one and the lower one. Each of the halves contains a number from 1 to 6. Valera loves even integers very much, so he wants the sum of the numbers on the upper halves and the sum of the numbers on the lower halves to be even.
To do that, Valera can rotate the dominoes by 180 degrees. After the rotation the upper and the lower halves swap places. This action takes one second. Help Valera find out the minimum time he must spend rotating dominoes to make his wish come true.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100), denoting the number of dominoes Valera has. Next *n* lines contain two space-separated integers *x**i*,<=*y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=6). Number *x**i* is initially written on the upper half of the *i*-th domino, *y**i* is initially written on the lower half.
Output Specification:
Print a single number — the minimum required number of seconds. If Valera can't do the task in any time, print <=-<=1.
Demo Input:
['2\n4 2\n6 4\n', '1\n2 3\n', '3\n1 4\n2 3\n4 4\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
In the first test case the sum of the numbers on the upper halves equals 10 and the sum of the numbers on the lower halves equals 6. Both numbers are even, so Valera doesn't required to do anything.
In the second sample Valera has only one piece of domino. It is written 3 on the one of its halves, therefore one of the sums will always be odd.
In the third case Valera can rotate the first piece, and after that the sum on the upper halves will be equal to 10, and the sum on the lower halves will be equal to 8. | ```python
n=int(input())
r=[list(map(int,input().split())) for _ in range(n)]
s1,s2=0,0
for i in range(n):
s1+=r[i][0]
s2+=r[i][1]
if s1%2==0 and s2%2==0:
print(0)
elif s1%2!=s2%2:
print(-1)
else:
print(1)
``` | 0 | |
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... | 1,604,319,746 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 218 | 0 | line_s = input().strip()
line_t = input().strip()
if line_s == line_t[-1::-1]:
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
line_s = input().strip()
line_t = input().strip()
if line_s == line_t[-1::-1]:
print('YES')
else:
print('NO')
``` | 3.9455 |
999 | E | Reachability from the Capital | PROGRAMMING | 2,000 | [
"dfs and similar",
"graphs",
"greedy"
] | null | null | There are $n$ cities and $m$ roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way. | The first line of input consists of three integers $n$, $m$ and $s$ ($1 \le n \le 5000, 0 \le m \le 5000, 1 \le s \le n$) — the number of cities, the number of roads and the index of the capital. Cities are indexed from $1$ to $n$.
The following $m$ lines contain roads: road $i$ is given as a pair of cities $u_i$, $v_i$ ($1 \le u_i, v_i \le n$, $u_i \ne v_i$). For each pair of cities $(u, v)$, there can be at most one road from $u$ to $v$. Roads in opposite directions between a pair of cities are allowed (i.e. from $u$ to $v$ and from $v$ to $u$). | Print one integer — the minimum number of extra roads needed to make all the cities reachable from city $s$. If all the cities are already reachable from $s$, print 0. | [
"9 9 1\n1 2\n1 3\n2 3\n1 5\n5 6\n6 1\n1 8\n9 8\n7 1\n",
"5 4 5\n1 2\n2 3\n3 4\n4 1\n"
] | [
"3\n",
"1\n"
] | The first example is illustrated by the following:
For example, you can add roads ($6, 4$), ($7, 9$), ($1, 7$) to make all the cities reachable from $s = 1$.
The second example is illustrated by the following:
In this example, you can add any one of the roads ($5, 1$), ($5, 2$), ($5, 3$), ($5, 4$) to make all the cities reachable from $s = 5$. | 0 | [
{
"input": "9 9 1\n1 2\n1 3\n2 3\n1 5\n5 6\n6 1\n1 8\n9 8\n7 1",
"output": "3"
},
{
"input": "5 4 5\n1 2\n2 3\n3 4\n4 1",
"output": "1"
},
{
"input": "5000 0 2956",
"output": "4999"
},
{
"input": "2 0 2",
"output": "1"
},
{
"input": "2 1 1\n1 2",
"output": "0"... | 1,608,714,662 | 1,802 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 514 | 11,468,800 | import sys
import math
import random
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
def FACT(n, mod):
s = 1
facts = [1]
for i in range(1,n+1):
s*=i
s%=mod
facts.append(s)
return facts[n]
def C(n, k, mod):
return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
n,m,s = MI()
d = [[] for i in range(n+1)]
d1 = [[] for i in range(n+1)]
for i in range(m):
u,v = MI()
d[u].append(v)
d1[v].append(u)
count = 0
v = [0 for i in range(n+1)]
def dfs(i, isD1 = False):
global v
global d
v[i] = 1
if isD1:
for x in d1[i]:
if v[x] == 0:
dfs(x, isD1)
else:
for x in d[i]:
if v[x] == 0:
dfs(x,isD1)
dfs(s)
for i in range(1,n+1):
if v[i] == 0:
count+=1
dfs(i, True)
print(count)
| Title: Reachability from the Capital
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ cities and $m$ roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input Specification:
The first line of input consists of three integers $n$, $m$ and $s$ ($1 \le n \le 5000, 0 \le m \le 5000, 1 \le s \le n$) — the number of cities, the number of roads and the index of the capital. Cities are indexed from $1$ to $n$.
The following $m$ lines contain roads: road $i$ is given as a pair of cities $u_i$, $v_i$ ($1 \le u_i, v_i \le n$, $u_i \ne v_i$). For each pair of cities $(u, v)$, there can be at most one road from $u$ to $v$. Roads in opposite directions between a pair of cities are allowed (i.e. from $u$ to $v$ and from $v$ to $u$).
Output Specification:
Print one integer — the minimum number of extra roads needed to make all the cities reachable from city $s$. If all the cities are already reachable from $s$, print 0.
Demo Input:
['9 9 1\n1 2\n1 3\n2 3\n1 5\n5 6\n6 1\n1 8\n9 8\n7 1\n', '5 4 5\n1 2\n2 3\n3 4\n4 1\n']
Demo Output:
['3\n', '1\n']
Note:
The first example is illustrated by the following:
For example, you can add roads ($6, 4$), ($7, 9$), ($1, 7$) to make all the cities reachable from $s = 1$.
The second example is illustrated by the following:
In this example, you can add any one of the roads ($5, 1$), ($5, 2$), ($5, 3$), ($5, 4$) to make all the cities reachable from $s = 5$. | ```python
import sys
import math
import random
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
def FACT(n, mod):
s = 1
facts = [1]
for i in range(1,n+1):
s*=i
s%=mod
facts.append(s)
return facts[n]
def C(n, k, mod):
return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
n,m,s = MI()
d = [[] for i in range(n+1)]
d1 = [[] for i in range(n+1)]
for i in range(m):
u,v = MI()
d[u].append(v)
d1[v].append(u)
count = 0
v = [0 for i in range(n+1)]
def dfs(i, isD1 = False):
global v
global d
v[i] = 1
if isD1:
for x in d1[i]:
if v[x] == 0:
dfs(x, isD1)
else:
for x in d[i]:
if v[x] == 0:
dfs(x,isD1)
dfs(s)
for i in range(1,n+1):
if v[i] == 0:
count+=1
dfs(i, True)
print(count)
``` | 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... | 1,608,894,784 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 278 | 5,120,000 | import fileinput
s = []
left = True
for line in fileinput.input():
s.append(line[:-1])
l = len(max(s, key=len))
for i in range(len(s)):
free = l - len(s[i])
sp = free//2
if free%2==0:
s[i] = '*' + sp*' ' + s[i] + sp*' ' + '*'
else:
if left:
s[i] = '*' + sp*' ' + s[i] + (sp+1)*' ' + '*'
else:
s[i] = '*' + (sp+1)*' ' + s[i] + sp*' ' + '*'
left = not left
print('*'*(l+2))
print('\n'.join(s))
print('*'*(l+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 fileinput
s = []
left = True
for line in fileinput.input():
s.append(line[:-1])
l = len(max(s, key=len))
for i in range(len(s)):
free = l - len(s[i])
sp = free//2
if free%2==0:
s[i] = '*' + sp*' ' + s[i] + sp*' ' + '*'
else:
if left:
s[i] = '*' + sp*' ' + s[i] + (sp+1)*' ' + '*'
else:
s[i] = '*' + (sp+1)*' ' + s[i] + sp*' ' + '*'
left = not left
print('*'*(l+2))
print('\n'.join(s))
print('*'*(l+2))
``` | 3.822853 |
777 | C | Alyona and Spreadsheet | PROGRAMMING | 1,600 | [
"binary search",
"data structures",
"dp",
"greedy",
"implementation",
"two pointers"
] | null | null | During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.
Now she has a table filled with integers. The table consists of *n* rows and *m* columns. By *a**i*,<=*j* we will denote the integer located at the *i*-th row and the *j*-th column. We say that the table is sorted in non-decreasing order in the column *j* if *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from 1 to *n*<=-<=1.
Teacher gave Alyona *k* tasks. For each of the tasks two integers *l* and *r* are given and Alyona has to answer the following question: if one keeps the rows from *l* to *r* inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such *j* that *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from *l* to *r*<=-<=1 inclusive.
Alyona is too small to deal with this task and asks you to help! | The first line of the input contains two positive integers *n* and *m* (1<=≤<=*n*·*m*<=≤<=100<=000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.
Each of the following *n* lines contains *m* integers. The *j*-th integers in the *i* of these lines stands for *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109).
The next line of the input contains an integer *k* (1<=≤<=*k*<=≤<=100<=000) — the number of task that teacher gave to Alyona.
The *i*-th of the next *k* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). | Print "Yes" to the *i*-th line of the output if the table consisting of rows from *l**i* to *r**i* inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No". | [
"5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5\n"
] | [
"Yes\nNo\nYes\nYes\nYes\nNo\n"
] | In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3. | 1,500 | [
{
"input": "5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5",
"output": "Yes\nNo\nYes\nYes\nYes\nNo"
},
{
"input": "1 1\n1\n1\n1 1",
"output": "Yes"
},
{
"input": "10 1\n523130301\n127101624\n15573616\n703140639\n628818570\n957494759\n161270109\n38686565... | 1,686,596,592 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 966 | 23,654,400 | import sys
input = lambda: sys.stdin.readline().rstrip()
N,M = map(int, input().split())
S = []
for _ in range(N):
S.append(list(map(int, input().split())))
A = []
for i in range(int(input())):
l,r = map(int, input().split())
A.append((l-1,r-1,i))
k = len(A)
A.sort()
#print(A)
ans = [0]*len(A)
for j in range(M):
pre=0
l=0
for i in range(N):
if S[i][j]<S[i-1][j]:
pre=i
while l<k and A[l][0]<pre:
l+=1
while l<k and A[l][1]<=i:
ans[A[l][2]]=1
l+=1
#print(l,i-pre+1)
for i in range(k):
if ans[i]:
print('Yes')
else:
print('No')
| Title: Alyona and Spreadsheet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.
Now she has a table filled with integers. The table consists of *n* rows and *m* columns. By *a**i*,<=*j* we will denote the integer located at the *i*-th row and the *j*-th column. We say that the table is sorted in non-decreasing order in the column *j* if *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from 1 to *n*<=-<=1.
Teacher gave Alyona *k* tasks. For each of the tasks two integers *l* and *r* are given and Alyona has to answer the following question: if one keeps the rows from *l* to *r* inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such *j* that *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from *l* to *r*<=-<=1 inclusive.
Alyona is too small to deal with this task and asks you to help!
Input Specification:
The first line of the input contains two positive integers *n* and *m* (1<=≤<=*n*·*m*<=≤<=100<=000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.
Each of the following *n* lines contains *m* integers. The *j*-th integers in the *i* of these lines stands for *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109).
The next line of the input contains an integer *k* (1<=≤<=*k*<=≤<=100<=000) — the number of task that teacher gave to Alyona.
The *i*-th of the next *k* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print "Yes" to the *i*-th line of the output if the table consisting of rows from *l**i* to *r**i* inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No".
Demo Input:
['5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5\n']
Demo Output:
['Yes\nNo\nYes\nYes\nYes\nNo\n']
Note:
In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3. | ```python
import sys
input = lambda: sys.stdin.readline().rstrip()
N,M = map(int, input().split())
S = []
for _ in range(N):
S.append(list(map(int, input().split())))
A = []
for i in range(int(input())):
l,r = map(int, input().split())
A.append((l-1,r-1,i))
k = len(A)
A.sort()
#print(A)
ans = [0]*len(A)
for j in range(M):
pre=0
l=0
for i in range(N):
if S[i][j]<S[i-1][j]:
pre=i
while l<k and A[l][0]<pre:
l+=1
while l<k and A[l][1]<=i:
ans[A[l][2]]=1
l+=1
#print(l,i-pre+1)
for i in range(k):
if ans[i]:
print('Yes')
else:
print('No')
``` | 0 | |
124 | A | The number of positions | PROGRAMMING | 1,000 | [
"math"
] | null | null | Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy. | The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100). | Print the single number — the number of the sought positions. | [
"3 1 1\n",
"5 2 3\n"
] | [
"2\n",
"3\n"
] | The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | 500 | [
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 ... | 1,673,434,261 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 92 | 0 | l=list(map(int,input().split()))
if l[1]==0:print(1)
elif l[2]==0 :print(l[0])
else: print(l[1]+1) | Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | ```python
l=list(map(int,input().split()))
if l[1]==0:print(1)
elif l[2]==0 :print(l[0])
else: print(l[1]+1)
``` | 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... | 1,686,211,714 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | import math
import os
from collections import Counter
n, m = map(int, input().split())
arr = list(map(int, input().split()))
maxN = 0
ans = 0
for i in range(n):
a = arr[i] % m
b = (arr[i] - a) / m
maxN = max(b, maxN)
if maxN == b :
ans = i+1
print(ans)
# for _ in range(int(input())):
# grid = [list(map(int, input().split())) for _ in range(3)]
# result = [[1] * 3 for _ in range(3)]
# n, s, r = map(int, input().split())
# arr = list(map(int, input().split()))
# n = input()
# n = int(n) | 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
import math
import os
from collections import Counter
n, m = map(int, input().split())
arr = list(map(int, input().split()))
maxN = 0
ans = 0
for i in range(n):
a = arr[i] % m
b = (arr[i] - a) / m
maxN = max(b, maxN)
if maxN == b :
ans = i+1
print(ans)
# for _ in range(int(input())):
# grid = [list(map(int, input().split())) for _ in range(3)]
# result = [[1] * 3 for _ in range(3)]
# n, s, r = map(int, input().split())
# arr = list(map(int, input().split()))
# n = input()
# n = int(n)
``` | 0 | |
510 | A | Fox And Snake | PROGRAMMING | 800 | [
"implementation"
] | null | null | Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern. | The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number. | Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces. | [
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] | [
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] | none | 500 | [
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#... | 1,680,323,600 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 46 | 0 | n,m=input().split()
n = int(n)
m = int(m)
arah = 'kanan'
for n in range(n):
if n%2==0:
print('#'*m)
else:
if arah=='kiri':
arah='kanan'
print('#','.'*(m-1),sep='')
else:
arah='kiri'
print('.'*(m-1),'#',sep='') | Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none | ```python
n,m=input().split()
n = int(n)
m = int(m)
arah = 'kanan'
for n in range(n):
if n%2==0:
print('#'*m)
else:
if arah=='kiri':
arah='kanan'
print('#','.'*(m-1),sep='')
else:
arah='kiri'
print('.'*(m-1),'#',sep='')
``` | 3 | |
543 | A | Writing Code | PROGRAMMING | 1,800 | [
"dp"
] | null | null | Programmers working on a large project have just received a task to write exactly *m* lines of code. There are *n* programmers working on a project, the *i*-th of them makes exactly *a**i* bugs in every line of code that he writes.
Let's call a sequence of non-negative integers *v*1,<=*v*2,<=...,<=*v**n* a plan, if *v*1<=+<=*v*2<=+<=...<=+<=*v**n*<==<=*m*. The programmers follow the plan like that: in the beginning the first programmer writes the first *v*1 lines of the given task, then the second programmer writes *v*2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most *b* bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer *mod*. | The first line contains four integers *n*, *m*, *b*, *mod* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*b*<=≤<=500; 1<=≤<=*mod*<=≤<=109<=+<=7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=500) — the number of bugs per line for each programmer. | Print a single integer — the answer to the problem modulo *mod*. | [
"3 3 3 100\n1 1 1\n",
"3 6 5 1000000007\n1 2 3\n",
"3 5 6 11\n1 2 1\n"
] | [
"10\n",
"0\n",
"0\n"
] | none | 500 | [
{
"input": "3 3 3 100\n1 1 1",
"output": "10"
},
{
"input": "3 6 5 1000000007\n1 2 3",
"output": "0"
},
{
"input": "3 5 6 11\n1 2 1",
"output": "0"
},
{
"input": "2 3 3 1000\n1 2",
"output": "1"
},
{
"input": "3 10 10 150691913\n8 7 10",
"output": "0"
},
{... | 1,680,342,675 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 3,000 | 14,745,600 | n,m,b,mod = map(int, input().split())
a = list(map(int, input().split()))
d = []
for i in range(n+1):
d.append([])
for j in range(m+1):
d[i].append([])
for k in range(b+1):
d[i][j].append(0)
if i == 0 and j == 0 and k == 0:
d[i][j][k] = 1
if i > 0 and j > 0 and k >= a[i-1]:
d[i][j][k] = (d[i-1][j][k]%mod + d[i][j-1][k-a[i-1]])%mod
elif i > 0:
d[i][j][k] = d[i-1][j][k]%mod
print(sum(d[-1][-1])%mod)
| Title: Writing Code
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmers working on a large project have just received a task to write exactly *m* lines of code. There are *n* programmers working on a project, the *i*-th of them makes exactly *a**i* bugs in every line of code that he writes.
Let's call a sequence of non-negative integers *v*1,<=*v*2,<=...,<=*v**n* a plan, if *v*1<=+<=*v*2<=+<=...<=+<=*v**n*<==<=*m*. The programmers follow the plan like that: in the beginning the first programmer writes the first *v*1 lines of the given task, then the second programmer writes *v*2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most *b* bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer *mod*.
Input Specification:
The first line contains four integers *n*, *m*, *b*, *mod* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*b*<=≤<=500; 1<=≤<=*mod*<=≤<=109<=+<=7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=500) — the number of bugs per line for each programmer.
Output Specification:
Print a single integer — the answer to the problem modulo *mod*.
Demo Input:
['3 3 3 100\n1 1 1\n', '3 6 5 1000000007\n1 2 3\n', '3 5 6 11\n1 2 1\n']
Demo Output:
['10\n', '0\n', '0\n']
Note:
none | ```python
n,m,b,mod = map(int, input().split())
a = list(map(int, input().split()))
d = []
for i in range(n+1):
d.append([])
for j in range(m+1):
d[i].append([])
for k in range(b+1):
d[i][j].append(0)
if i == 0 and j == 0 and k == 0:
d[i][j][k] = 1
if i > 0 and j > 0 and k >= a[i-1]:
d[i][j][k] = (d[i-1][j][k]%mod + d[i][j-1][k-a[i-1]])%mod
elif i > 0:
d[i][j][k] = d[i-1][j][k]%mod
print(sum(d[-1][-1])%mod)
``` | 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... | 1,651,160,208 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | cc,sc=0,0
s=input()
for i in range(len(s)):
if(ord(s[i])>=65 and ord(s[i])<=90):
cc+=1
elif(ord(s[i])>=97 and ord(s[i])<=122):
sc+=1
if(sc<cc):
print(s.upper())
elif(sc>cc):
print(s.lower())
elif(sc==cc):
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
cc,sc=0,0
s=input()
for i in range(len(s)):
if(ord(s[i])>=65 and ord(s[i])<=90):
cc+=1
elif(ord(s[i])>=97 and ord(s[i])<=122):
sc+=1
if(sc<cc):
print(s.upper())
elif(sc>cc):
print(s.lower())
elif(sc==cc):
print(s.lower())
``` | 3.977 |
378 | A | Playing with Dice | PROGRAMMING | 800 | [
"brute force"
] | null | null | Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins? | The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly. | Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly. | [
"2 5\n",
"2 4\n"
] | [
"3 0 3\n",
"2 1 3\n"
] | The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|. | 500 | [
{
"input": "2 5",
"output": "3 0 3"
},
{
"input": "2 4",
"output": "2 1 3"
},
{
"input": "5 3",
"output": "2 1 3"
},
{
"input": "1 6",
"output": "3 0 3"
},
{
"input": "5 1",
"output": "3 1 2"
},
{
"input": "6 3",
"output": "2 0 4"
},
{
"inp... | 1,664,205,140 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 38 | 62 | 0 | m=list(map(int,input().split()))
result_a_win=result_draw=result_b_win=0
for i in range(1,7):
if abs(m[0]-i)<abs(m[1]-i):
result_a_win += 1
if abs(m[0]-i)==abs(m[1]-i):
result_draw += 1
if abs(m[0]-i)>abs(m[1]-i):
result_b_win += 1
print(result_a_win,result_draw,result_b_win) | Title: Playing with Dice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.
The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?
Input Specification:
The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.
Output Specification:
Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.
Demo Input:
['2 5\n', '2 4\n']
Demo Output:
['3 0 3\n', '2 1 3\n']
Note:
The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.
You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|. | ```python
m=list(map(int,input().split()))
result_a_win=result_draw=result_b_win=0
for i in range(1,7):
if abs(m[0]-i)<abs(m[1]-i):
result_a_win += 1
if abs(m[0]-i)==abs(m[1]-i):
result_draw += 1
if abs(m[0]-i)>abs(m[1]-i):
result_b_win += 1
print(result_a_win,result_draw,result_b_win)
``` | 3 |
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