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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
a, b=input().split(' ')
print(int(a)*int(b))
```
Yes
| 98,600 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
n, m = map(int, input().split())
print(n * m)
```
Yes
| 98,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
print(eval(input().replace(' ','*')))
```
Yes
| 98,602 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
s = input().split()
print(int(s[0]) * int(s[1]))
```
Yes
| 98,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
from sys import stdin
trim = lambda s: s[:-1] if s[-1] == "\n" else s
T = int(stdin.readline())
for i in range(0, T):
n = int(stdin.readline())
s = trim(stdin.readline())
possible = True
for j in range(0, n//2 + 1):
if abs(ord(s[j])-ord(s[-(j+1)])) not in [0, 2]:
possible = False
if possible:
print("YES")
else:
print("NO")
```
| 98,604 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
import sys
import os
def solve(s):
n = len(s)
for i in range(n // 2):
a = ord(s[i])
b = ord(s[n - 1 - i])
diff = abs(a - b)
if diff != 0 and abs(diff) != 2:
return 'NO'
return 'YES'
def main():
t = int(input())
for i in range(t):
_ = int(input())
s = input()
print(solve(s))
if __name__ == '__main__':
main()
```
| 98,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
def is_pal(s):
for i in range(len(s)//2):
dif = abs(ord(s[i]) - ord(s[-i-1]))
if dif != 0 and dif != 2 :
return False
return True
t = int(input())
for i in range(t):
n = int(input())
s = input()
buf = []
print("YES" if is_pal(s) else "NO")
```
| 98,606 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
t = int(input())
while (t):
n = int(input())
s = input()
ok = 0
for i in range (n):
v = abs(ord(s[i]) - ord(s[n-i-1]))
if (v != 0 and v != 2):
ok = 1
if (ok):print("NO")
else: print("YES")
t -= 1
```
| 98,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
t = int(input())
alphabet = "abcdefghijklmnopqrstuvwxyz"
for i in range(t):
n = int(input())
string = input()
palindrome = True
i = 0
j = len(string) - 1
while(i <= j):
if(string[i] != string[j]):
esq = alphabet.index(string[i])
dir = alphabet.index(string[j])
if(esq + 1 != dir -1 and esq + 1 != dir + 1 and esq - 1 != dir - 1 and esq -1 != dir + 1):
palindrome = False
i+= 1
j-= 1
if(palindrome):
print("YES")
else:
print("NO")
```
| 98,608 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
import sys
import math
from collections import deque
def scan():
return list(map(int, sys.stdin.readline().strip().split()))
def print_primes_till_n(n):
i, j, flag = 0, 0, 0
a = []
c = 0
for i in range(1, n + 1, 1):
if i == 1 or i == 0:
continue
flag = 1
for j in range(2, ((i // 2) + 1), 1):
if i % j == 0:
flag = 0
break
if flag == 1:
a.append(i)
c += 1
return a, c
def is_square(n):
b = math.sqrt(n)
if n == int(b * b):
return True
return False
def solution():
for _ in range(int(input())):
n = int(input())
s = input()
s = [i for i in s]
c = 0
for i in range(n // 2):
if abs(ord(s[n-i-1])-ord(s[i])) == 2 or abs(ord(s[n-i-1])-ord(s[i])) == 0:
c += 1
else:
c = -1
break
if c == n//2:
print('YES')
else:
print('NO')
if __name__ == '__main__':
solution()
```
| 98,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
def can_be_transformed(word):
for i in range(len(word)//2):
if abs(ord(word[i])-ord(word[-i-1]))>2 or abs(ord(word[i])-ord(word[-i-1])) == 1:
return False
return True
def main():
answers = []
n=int(input())
for i in range(0,n):
m=int(input())
s=input()
if can_be_transformed(s):
answers.append('YES')
else:
answers.append('NO')
for i in range(0,n):
print(answers[i])
if __name__ == '__main__':
main()
```
| 98,610 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Tags: implementation, strings
Correct Solution:
```
def search(s,n):
for i in range(n//2):
x=abs(ord(s[i])-ord(s[n-i-1]))
if x!=0 and x!=2:
return(False)
return(True)
t=int(input())
for _ in range(t):
n=int(input())
s=input()
if search(s,n):
print("YES")
else:
print("NO")
```
| 98,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
t = int(input())
def get_str(s):
next = ''
prev = ''
if s == 'a':
next = chr(ord(s) + 1)
elif s == 'z':
prev = chr(ord(s) - 1)
else:
next = chr(ord(s) + 1)
prev = chr(ord(s) - 1)
return next + prev
def canPalindrome(n, s):
options = []
for i in range(n // 2):
a = get_str(s[i])
b = get_str(s[(-1 * i) -1])
l = len(options) // 2
options.insert(l,a)
options.insert(l + 1 ,b)
for i in range(n // 2):
if not len(''.join(set(options[i]).intersection(set(options[(-1 * i) - 1])))) > 0:
return False
return True
res = []
for i in range(t):
n = int(input())
s = input()
res.append(canPalindrome(n, s))
for i in res:
if i == True:
print('YES')
else:
print('NO')
```
Yes
| 98,612 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
import math
from sys import stdin
n_round = int(stdin.readline())
for _ in range(n_round):
n_letter = int(stdin.readline())
string = list(stdin.readline().strip())
numbers = [ord(x) for x in string]
#a-97 z-122
result = True
for _ in range(math.floor(n_letter/2)):
if not (numbers[_] == numbers[-(_+1)] or abs(numbers[_]-numbers[-(_+1)])==2):
result = False
if result == True:
print('YES')
else:
print('NO')
```
Yes
| 98,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
def canbe(c, s):
for i in range(int(c/2), c):
if (abs(ord(s[i]) - ord(s[c-i-1]))>2) or (abs(ord(s[i]) - ord(s[c-i-1]))==1):
return False
return True
t = int(input())
buf = []
for i in range(t):
c = int(input())
s = input()
if canbe(c, s):
buf.append('YES')
else:
buf.append('NO')
for el in buf:
print(el)
```
Yes
| 98,614 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
n = int(input())
tt = [-2, 0, 2]
for jj in range(n):
i, inp = input(), input()
need = 1
for j in range(len(inp)):
if (ord(inp[j]) - ord(inp[len(inp) - j - 1])) not in tt:
need = 0
if need:
print("YES")
else:
print("NO")
```
Yes
| 98,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
from __future__ import division, print_function
from collections import *
from math import *
from itertools import *
from time import time
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
'''
Notes:
n = number of balls
k = attract power
ni = ball's coords
'''
def main():
n = int(input())
s = input()
if s == s[::-1]:
print('YES')
return
opposite = sorted(s, reverse = True)
for i in range(n // 2):
first = ord(s[i])
second = ord(opposite[i])
if first - 1 == second or first == second or first + 1 == second:
continue
else:
print('NO')
return
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
t = int(input())
while(t):
main()
t -= 1
```
No
| 98,616 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
T=int(input())
for i in range(T):
n=int(input())
s=input()
l=len(s)-1
for i in range(len(s)):
if s[i]==s[l]:
l=l-1
if i==(len(s)-1):
print ("YES")
break
continue
if abs((ord(s[i])-ord(s[l])))!=2 and (s[i]!='a' or s[i]!='z') and (s[l]!='a' or s[l]!='z'):
print ("NO")
break
if (s[i]=='a' and (s[l] in ['c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'])):
print ("NO")
break
if (s[l]=='a' and (s[i] in ['c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'])):
print ("NO")
break
if (s[i]=='z' and (s[l] in ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x'])):
print ("NO")
break
if (s[l]=='z' and (s[i] in ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x'])):
print ("N0")
break
else:
if i==(len(s)-1):
print ("YES")
break
l=l-1
```
No
| 98,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
i=input
t=int(i())
while t:
t-=1
i();a=[*map(ord,i())]
print("YNEOS"[1-all((x-y)%26in{0,2,24}for x,y in zip(a,a[::-1]))::2])
```
No
| 98,618 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters. n is even.
For each position i (1 ≤ i ≤ n) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters 'a' and 'z' have only one of these options). Letter in every position must be changed exactly once.
For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.
That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' → 'd', 'o' → 'p', 'd' → 'e', 'e' → 'd', 'f' → 'e', 'o' → 'p', 'r' → 'q', 'c' → 'b', 'e' → 'f', 's' → 't').
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings "abba" and "zz" are palindromes and strings "abca" and "zy" are not.
Your goal is to check if it's possible to make string s a palindrome by applying the aforementioned changes to every position. Print "YES" if string s can be transformed to a palindrome and "NO" otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input
The first line contains a single integer T (1 ≤ T ≤ 50) — the number of strings in a testcase.
Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th string. The first line of the pair contains a single integer n (2 ≤ n ≤ 100, n is even) — the length of the corresponding string. The second line of the pair contains a string s, consisting of n lowercase Latin letters.
Output
Print T lines. The i-th line should contain the answer to the i-th string of the input. Print "YES" if it's possible to make the i-th string a palindrome by applying the aforementioned changes to every position. Print "NO" otherwise.
Example
Input
5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml
Output
YES
NO
YES
NO
NO
Note
The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.
The third string can be changed to "beeb" which is a palindrome.
The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".
Submitted Solution:
```
def check(n, s):
for i in range(n//2):
c1, c2 = ord(s[i]), ord(s[n-i-1])
if abs(c1 - c2) > 2:
return False
return True
T = int(input())
for _ in range(T):
n = int(input())
s = input()
if check(n, s):
print("YES")
else:
print("NO")
```
No
| 98,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Since astronauts from BubbleCup XI mission finished their mission on the Moon and are big fans of famous singer, they decided to spend some fun time before returning to the Earth and hence created a so called "Moonwalk challenge" game.
Teams of astronauts are given the map of craters on the Moon and direct bidirectional paths from some craters to others that are safe for "Moonwalking". Each of those direct paths is colored in one color and there is unique path between each two craters. Goal of the game is to find two craters such that given array of colors appears most times as continuous subarray on the path between those two craters (overlapping appearances should be counted).
To help your favorite team win, you should make a program that, given the map, answers the queries of the following type: For two craters and array of colors answer how many times given array appears as continuous subarray on the path from the first crater to the second.
Colors are represented as lowercase English alphabet letters.
Input
In the first line, integer N (2 ≤ N ≤ 10^5) — number of craters on the Moon. Craters are numerated with numbers 1 to N.
In next N-1 lines, three values u, v, L (1 ≤ u, v ≤ N, L ∈ \\{a, ..., z\}) — denoting that there is a direct path with color L between craters u and v.
Next line contains integer Q (1 ≤ Q ≤ 10^5) — number of queries.
Next Q lines contain three values u, v (1 ≤ u, v ≤ N) and S (|S| ≤ 100), where u and v are the two cratersfor which you should find how many times array of colors S (represented as string) appears on the path from u to v.
Output
For each query output one number that represents number of occurrences of array S on the path from u to v.
Example
Input
6
2 3 g
3 4 n
5 3 o
6 1 n
1 2 d
7
1 6 n
6 4 dg
6 4 n
2 5 og
1 2 d
6 5 go
2 3 g
Output
1
1
2
0
1
1
1
Submitted Solution:
```
print(0)
```
No
| 98,620 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our brave travelers reached an island where pirates had buried treasure. However as the ship was about to moor, the captain found out that some rat ate a piece of the treasure map.
The treasure map can be represented as a rectangle n × m in size. Each cell stands for an islands' square (the square's side length equals to a mile). Some cells stand for the sea and they are impenetrable. All other cells are penetrable (i.e. available) and some of them contain local sights. For example, the large tree on the hills or the cave in the rocks.
Besides, the map also has a set of k instructions. Each instruction is in the following form:
"Walk n miles in the y direction"
The possible directions are: north, south, east, and west. If you follow these instructions carefully (you should fulfill all of them, one by one) then you should reach exactly the place where treasures are buried.
Unfortunately the captain doesn't know the place where to start fulfilling the instructions — as that very piece of the map was lost. But the captain very well remembers that the place contained some local sight. Besides, the captain knows that the whole way goes through the island's penetrable squares.
The captain wants to know which sights are worth checking. He asks you to help him with that.
Input
The first line contains two integers n and m (3 ≤ n, m ≤ 1000).
Then follow n lines containing m integers each — the island map's description. "#" stands for the sea. It is guaranteed that all cells along the rectangle's perimeter are the sea. "." stands for a penetrable square without any sights and the sights are marked with uppercase Latin letters from "A" to "Z". Not all alphabet letters can be used. However, it is guaranteed that at least one of them is present on the map. All local sights are marked by different letters.
The next line contains number k (1 ≤ k ≤ 105), after which k lines follow. Each line describes an instruction. Each instruction possesses the form "dir len", where dir stands for the direction and len stands for the length of the way to walk. dir can take values "N", "S", "W" and "E" for North, South, West and East correspondingly. At that, north is to the top, South is to the bottom, west is to the left and east is to the right. len is an integer from 1 to 1000.
Output
Print all local sights that satisfy to the instructions as a string without any separators in the alphabetical order. If no sight fits, print "no solution" without the quotes.
Examples
Input
6 10
##########
#K#..#####
#.#..##.##
#..L.#...#
###D###A.#
##########
4
N 2
S 1
E 1
W 2
Output
AD
Input
3 4
####
#.A#
####
2
W 1
N 2
Output
no solution
Tags: brute force, implementation
Correct Solution:
```
#!/usr/bin/env python3
from sys import stdin
n, m = map(int, stdin.readline().rstrip().split())
island = []
pos = {}
for i in range(n):
island.append(stdin.readline().rstrip())
for j, c in enumerate(island[i]):
if c >= 'A' and c <= 'Z':
pos[c] = [i, j]
l_reach = [[-1 for j in range(m)] for i in range(n)]
r_reach = [[-1 for j in range(m)] for i in range(n)]
u_reach = [[-1 for j in range(m)] for i in range(n)]
d_reach = [[-1 for j in range(m)] for i in range(n)]
for i in range(1, n-1):
for j in range(1, m-1):
if island[i][j] != '#':
l_reach[i][j] = 1 + l_reach[i][j-1]
u_reach[i][j] = 1 + u_reach[i-1][j]
for i in range(n-2, 0, -1):
for j in range(m-2, 0, -1):
if island[i][j] != '#':
r_reach[i][j] = 1 + r_reach[i][j+1]
d_reach[i][j] = 1 + d_reach[i+1][j]
dir = [None] * 100
dir[ord('N')] = [-1, 0, u_reach]
dir[ord('W')] = [0, -1, l_reach]
dir[ord('S')] = [1, 0, d_reach]
dir[ord('E')] = [0, 1, r_reach]
for c in range(int(stdin.readline().rstrip())):
x, y, d = dir[ord(stdin.read(1))]
c = int(stdin.readline()[1:-1])
to_delete = []
for k, v in pos.items():
if c > d[v[0]][v[1]]:
to_delete.append(k)
else:
v[0] += c * x
v[1] += c * y
for k in to_delete:
del pos[k]
ans = ''.join(sorted(pos.keys()))
print(ans if ans else 'no solution')
```
| 98,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our brave travelers reached an island where pirates had buried treasure. However as the ship was about to moor, the captain found out that some rat ate a piece of the treasure map.
The treasure map can be represented as a rectangle n × m in size. Each cell stands for an islands' square (the square's side length equals to a mile). Some cells stand for the sea and they are impenetrable. All other cells are penetrable (i.e. available) and some of them contain local sights. For example, the large tree on the hills or the cave in the rocks.
Besides, the map also has a set of k instructions. Each instruction is in the following form:
"Walk n miles in the y direction"
The possible directions are: north, south, east, and west. If you follow these instructions carefully (you should fulfill all of them, one by one) then you should reach exactly the place where treasures are buried.
Unfortunately the captain doesn't know the place where to start fulfilling the instructions — as that very piece of the map was lost. But the captain very well remembers that the place contained some local sight. Besides, the captain knows that the whole way goes through the island's penetrable squares.
The captain wants to know which sights are worth checking. He asks you to help him with that.
Input
The first line contains two integers n and m (3 ≤ n, m ≤ 1000).
Then follow n lines containing m integers each — the island map's description. "#" stands for the sea. It is guaranteed that all cells along the rectangle's perimeter are the sea. "." stands for a penetrable square without any sights and the sights are marked with uppercase Latin letters from "A" to "Z". Not all alphabet letters can be used. However, it is guaranteed that at least one of them is present on the map. All local sights are marked by different letters.
The next line contains number k (1 ≤ k ≤ 105), after which k lines follow. Each line describes an instruction. Each instruction possesses the form "dir len", where dir stands for the direction and len stands for the length of the way to walk. dir can take values "N", "S", "W" and "E" for North, South, West and East correspondingly. At that, north is to the top, South is to the bottom, west is to the left and east is to the right. len is an integer from 1 to 1000.
Output
Print all local sights that satisfy to the instructions as a string without any separators in the alphabetical order. If no sight fits, print "no solution" without the quotes.
Examples
Input
6 10
##########
#K#..#####
#.#..##.##
#..L.#...#
###D###A.#
##########
4
N 2
S 1
E 1
W 2
Output
AD
Input
3 4
####
#.A#
####
2
W 1
N 2
Output
no solution
Tags: brute force, implementation
Correct Solution:
```
from sys import stdin
n, m = map(int, stdin.readline().rstrip().split())
island = []
pos = {}
for i in range(n):
island.append(stdin.readline().rstrip())
for j, c in enumerate(island[i]):
if c >= 'A' and c <= 'Z':
pos[c] = [i, j]
l_reach = [[-1 for j in range(m)] for i in range(n)]
r_reach = [[-1 for j in range(m)] for i in range(n)]
u_reach = [[-1 for j in range(m)] for i in range(n)]
d_reach = [[-1 for j in range(m)] for i in range(n)]
for i in range(1, n-1):
for j in range(1, m-1):
if island[i][j] != '#':
l_reach[i][j] = 1 + l_reach[i][j-1]
u_reach[i][j] = 1 + u_reach[i-1][j]
for i in range(n-2, 0, -1):
for j in range(m-2, 0, -1):
if island[i][j] != '#':
r_reach[i][j] = 1 + r_reach[i][j+1]
d_reach[i][j] = 1 + d_reach[i+1][j]
dir = [None] * 100
dir[ord('N')] = [-1, 0, u_reach]
dir[ord('W')] = [0, -1, l_reach]
dir[ord('S')] = [1, 0, d_reach]
dir[ord('E')] = [0, 1, r_reach]
for c in range(int(stdin.readline().rstrip())):
x, y, d = dir[ord(stdin.read(1))]
c = int(stdin.readline()[1:-1])
to_delete = []
for k, v in pos.items():
if c > d[v[0]][v[1]]:
to_delete.append(k)
else:
v[0] += c * x
v[1] += c * y
for k in to_delete:
del pos[k]
ans = ''.join(sorted(pos.keys()))
print(ans if ans else 'no solution')
```
| 98,622 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our brave travelers reached an island where pirates had buried treasure. However as the ship was about to moor, the captain found out that some rat ate a piece of the treasure map.
The treasure map can be represented as a rectangle n × m in size. Each cell stands for an islands' square (the square's side length equals to a mile). Some cells stand for the sea and they are impenetrable. All other cells are penetrable (i.e. available) and some of them contain local sights. For example, the large tree on the hills or the cave in the rocks.
Besides, the map also has a set of k instructions. Each instruction is in the following form:
"Walk n miles in the y direction"
The possible directions are: north, south, east, and west. If you follow these instructions carefully (you should fulfill all of them, one by one) then you should reach exactly the place where treasures are buried.
Unfortunately the captain doesn't know the place where to start fulfilling the instructions — as that very piece of the map was lost. But the captain very well remembers that the place contained some local sight. Besides, the captain knows that the whole way goes through the island's penetrable squares.
The captain wants to know which sights are worth checking. He asks you to help him with that.
Input
The first line contains two integers n and m (3 ≤ n, m ≤ 1000).
Then follow n lines containing m integers each — the island map's description. "#" stands for the sea. It is guaranteed that all cells along the rectangle's perimeter are the sea. "." stands for a penetrable square without any sights and the sights are marked with uppercase Latin letters from "A" to "Z". Not all alphabet letters can be used. However, it is guaranteed that at least one of them is present on the map. All local sights are marked by different letters.
The next line contains number k (1 ≤ k ≤ 105), after which k lines follow. Each line describes an instruction. Each instruction possesses the form "dir len", where dir stands for the direction and len stands for the length of the way to walk. dir can take values "N", "S", "W" and "E" for North, South, West and East correspondingly. At that, north is to the top, South is to the bottom, west is to the left and east is to the right. len is an integer from 1 to 1000.
Output
Print all local sights that satisfy to the instructions as a string without any separators in the alphabetical order. If no sight fits, print "no solution" without the quotes.
Examples
Input
6 10
##########
#K#..#####
#.#..##.##
#..L.#...#
###D###A.#
##########
4
N 2
S 1
E 1
W 2
Output
AD
Input
3 4
####
#.A#
####
2
W 1
N 2
Output
no solution
Submitted Solution:
```
import sys
from math import gcd,sqrt,ceil,log2
from collections import defaultdict,Counter,deque
from bisect import bisect_left,bisect_right
import math
import heapq
from itertools import permutations
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_sum(bit,i):
s = 0
i+=1
while i>0:
s+=bit[i]
i-=i&(-i)
return s
def update(bit,n,i,v):
i+=1
while i<=n:
bit[i]+=v
i+=i&(-i)
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def primeFactors(n):
sa = set()
sa.add(n)
while n % 2 == 0:
sa.add(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
sa.add(i)
n = n // i
# sa.add(n)
return sa
def seive(n):
pri = [True]*(n+1)
p = 2
while p*p<=n:
if pri[p] == True:
for i in range(p*p,n+1,p):
pri[i] = False
p+=1
return pri
def check_prim(n):
if n<0:
return False
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
return False
return True
def getZarr(string, z):
n = len(string)
# [L,R] make a window which matches
# with prefix of s
l, r, k = 0, 0, 0
for i in range(1, n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
def search(text, pattern):
# Create concatenated string "P$T"
concat = pattern + "$" + text
l = len(concat)
z = [0] * l
getZarr(concat, z)
ha = []
for i in range(l):
if z[i] == len(pattern):
ha.append(i - len(pattern) - 1)
return ha
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# n = int(input())
# l = list(map(int,input().split()))
#
# hash = defaultdict(list)
# la = []
#
# for i in range(n):
# la.append([l[i],i+1])
#
# la.sort(key = lambda x: (x[0],-x[1]))
# ans = []
# r = n
# flag = 0
# lo = []
# ha = [i for i in range(n,0,-1)]
# yo = []
# for a,b in la:
#
# if a == 1:
# ans.append([r,b])
# # hash[(1,1)].append([b,r])
# lo.append((r,b))
# ha.pop(0)
# yo.append([r,b])
# r-=1
#
# elif a == 2:
# # print(yo,lo)
# # print(hash[1,1])
# if lo == []:
# flag = 1
# break
# c,d = lo.pop(0)
# yo.pop(0)
# if b>=d:
# flag = 1
# break
# ans.append([c,b])
# yo.append([c,b])
#
#
#
# elif a == 3:
#
# if yo == []:
# flag = 1
# break
# c,d = yo.pop(0)
# if b>=d:
# flag = 1
# break
# if ha == []:
# flag = 1
# break
#
# ka = ha.pop(0)
#
# ans.append([ka,b])
# ans.append([ka,d])
# yo.append([ka,b])
#
# if flag:
# print(-1)
# else:
# print(len(ans))
# for a,b in ans:
# print(a,b)
def mergeIntervals(arr):
# Sorting based on the increasing order
# of the start intervals
arr.sort(key = lambda x: x[0])
# array to hold the merged intervals
m = []
s = -10000
max = -100000
for i in range(len(arr)):
a = arr[i]
if a[0] > max:
if i != 0:
m.append([s,max])
max = a[1]
s = a[0]
else:
if a[1] >= max:
max = a[1]
#'max' value gives the last point of
# that particular interval
# 's' gives the starting point of that interval
# 'm' array contains the list of all merged intervals
if max != -100000 and [s, max] not in m:
m.append([s, max])
return m
# n = int(input())
# dp = defaultdict(bool)
# n,m,k = map(int,input().split())
# l = []
#
# for i in range(n):
# la = list(map(int,input().split()))
# l.append(la)
# dp = defaultdict(int)
#
# for i in range(n):
# for j in range(m):
# for cnt in range(m//2-1,-1,-1):
# for rem in range(k):
# dp[(i,cnt+1,rem)] = max(dp[(i,cnt+1,rem)],dp[(i,cnt,rem)])
# dp[(i,cnt+1,(rem+l[i][j])%k)] = max(dp[(i,cnt+1,(rem+l[i][j])%k)],dp[(i,cnt,rem)]+l[i][j])
#
#
# print(dp[(n-1,m//2,0)])
#
n,m = map(int,input().split())
l = []
for i in range(n):
la = list(input())
l.append(la)
q = int(input())
qu = []
for i in range(q):
a,b = map(str,input().split())
qu.append([a,int(b)])
ans = []
row = defaultdict(list)
col = defaultdict(list)
for i in range(n):
for j in range(m):
if l[i][j] == '#':
row[i].append(j)
col[j].append(i)
for i in range(n):
for j in range(m):
if l[i][j] != '.' and l[i][j]!='#':
x,y = i,j
flag = 0
for a,b in qu:
z1 = bisect_right(row[x],y)
z2 = bisect_right(col[y],x)
if a == 'N':
x-=b
if x>=n or y>=m or x<0 or y<0:
flag = 1
break
if l[x][y] == '#':
flag = 1
break
if col[y] == []:
continue
if z2 == len(col[y]):
if x<=col[y][z2-1]:
flag = 1
break
elif col[y][z2-1]<x<col[y][z2]:
continue
else:
flag = 1
break
if a == 'S':
x+=b
if x>=n or y>=m or x<0 or y<0:
flag = 1
break
if l[x][y] == '#':
flag = 1
break
if col[y] == []:
continue
if z2 == len(col[y]):
if x<=col[y][z2-1]:
flag = 1
break
elif col[y][z2-1]<x<col[y][z2]:
continue
else:
flag = 1
break
if a == 'E':
y+=b
# print(z1,y,row[x])
if x>=n or y>=m or x<0 or y<0:
flag = 1
break
if l[x][y] == '#':
flag = 1
if row[x] == []:
continue
if z1 == len(row[x]):
if y<=row[x][z1-1]:
flag = 1
break
elif row[x][z1-1]<y<row[x][z1]:
continue
else:
flag = 1
break
if a == 'W':
y-=b
if x>=n or y>=m or x<0 or y<0:
flag = 1
break
if l[x][y] == '#':
flag = 1
if row[x] == []:
continue
if z1 == len(row[x]):
if y<=row[x][z1-1]:
flag = 1
break
elif row[x][z1-1]<y<row[x][z1]:
continue
else:
flag = 1
break
if flag == 0:
ans.append(l[i][j])
print(''.join(ans))
```
No
| 98,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our brave travelers reached an island where pirates had buried treasure. However as the ship was about to moor, the captain found out that some rat ate a piece of the treasure map.
The treasure map can be represented as a rectangle n × m in size. Each cell stands for an islands' square (the square's side length equals to a mile). Some cells stand for the sea and they are impenetrable. All other cells are penetrable (i.e. available) and some of them contain local sights. For example, the large tree on the hills or the cave in the rocks.
Besides, the map also has a set of k instructions. Each instruction is in the following form:
"Walk n miles in the y direction"
The possible directions are: north, south, east, and west. If you follow these instructions carefully (you should fulfill all of them, one by one) then you should reach exactly the place where treasures are buried.
Unfortunately the captain doesn't know the place where to start fulfilling the instructions — as that very piece of the map was lost. But the captain very well remembers that the place contained some local sight. Besides, the captain knows that the whole way goes through the island's penetrable squares.
The captain wants to know which sights are worth checking. He asks you to help him with that.
Input
The first line contains two integers n and m (3 ≤ n, m ≤ 1000).
Then follow n lines containing m integers each — the island map's description. "#" stands for the sea. It is guaranteed that all cells along the rectangle's perimeter are the sea. "." stands for a penetrable square without any sights and the sights are marked with uppercase Latin letters from "A" to "Z". Not all alphabet letters can be used. However, it is guaranteed that at least one of them is present on the map. All local sights are marked by different letters.
The next line contains number k (1 ≤ k ≤ 105), after which k lines follow. Each line describes an instruction. Each instruction possesses the form "dir len", where dir stands for the direction and len stands for the length of the way to walk. dir can take values "N", "S", "W" and "E" for North, South, West and East correspondingly. At that, north is to the top, South is to the bottom, west is to the left and east is to the right. len is an integer from 1 to 1000.
Output
Print all local sights that satisfy to the instructions as a string without any separators in the alphabetical order. If no sight fits, print "no solution" without the quotes.
Examples
Input
6 10
##########
#K#..#####
#.#..##.##
#..L.#...#
###D###A.#
##########
4
N 2
S 1
E 1
W 2
Output
AD
Input
3 4
####
#.A#
####
2
W 1
N 2
Output
no solution
Submitted Solution:
```
def main():
n, m = map(int, input().split())
island = []; features = []
for _ in range(n):
s = input()
for i in range(m):
if s[i] not in "#.":
features.append((i, _))
island.append(s)
k = int(input())
#maxes = [0, 0, 0, 0]
vector = [0, 0]
for __ in range(k):
d, l = input().split()
l = int(l)
if d == "N":
#maxes[0] = max(maxes[0], l)
vector[1] -= l
elif d == "S":
#maxes[1] = max(maxes[1], l)
vector[1] += l
elif d == "E":
#maxes[2] = max(maxes[2], l)
vector[0] += l
else:
#maxes[3] = max(maxes[3], l)
vector[0] -= l
good = []
for feature in features:
endpoint = (feature[0] + vector[0], feature[1] + vector[1])
if endpoint[0] < 1 or endpoint[0] > m - 2 or endpoint[1] < 1 or endpoint[1] > n - 2 or island[endpoint[1]][endpoint[0]] == "#":
continue
good.append(feature)
if good == []:
print("no solution")
else:
print("".join(sorted([island[po[1]][po[0]] for po in good])))
main()
```
No
| 98,624 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
import math
n = int(input())
def lcm(a, b):
while b > 0:
a, b = b, a % b
return a
def pfactors(n):
for i in range(2, math.floor(math.sqrt(n + 1) + 0.5)):
p = 0
while n % i == 0:
n //= i
p += 1
if p > 0:
yield (i, p)
if n > 1:
yield (n, 1)
def factors(n, p=None, i=0, r=[], b=1):
#print("factors", n, p, i, r, b)
if p == None:
p = list(pfactors(n))
if i < len(p):
for x in range(0, p[i][1] + 1):
factors(n, p, i + 1, r, b)
b *= p[i][0]
else:
r.append(b)
if i == 0:
return r
def factors2(n):
large_divisors = []
for i in range(1, int(math.sqrt(n) + 1)):
if n % i == 0:
yield i
if i*i != n:
large_divisors.append(n // i)
for divisor in reversed(large_divisors):
yield divisor
def f(k):
#if k == n:
# return 1
#p = lcm(n, k) # optimize?
p = k
x = n // p
#print(k, p, x)
return (2 + (x - 1) * k) * x // 2
#print(f(1), f(2), f(3), f(6))
F = set()
#print(*factors2(340510170))
for x in factors2(n):
if n % x == 0:
F.add(f(x))
print(*sorted(F))
```
| 98,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
class Solution:
def getList(self):
return list(map(int,input().split()))
def solve(self):
n = int(input())
funs = []
i = 1
while i*i <= n:
if n%i == 0:
funs.append(n*(i-1)//2+i)
if i*i<n:
funs.append(n*(n//i-1)//2+n//i)
i+=1
funs.sort()
print(*funs)
x = Solution()
x.solve()
```
| 98,626 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
n=int(input())
b=[]
c=[]
for i in range(1,int(n**(0.5))+1):
if n%i==0:
b.append(i)
b.append(n//i)
for i in range(len(b)):
x=n//b[i]
c.append(x+1 +(x*(x+1)//2)*b[i] -(n+1))
q=list(set(c))
q.sort()
print(*q)
```
| 98,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
from math import sqrt
def apsum(a, r, n):
return n*a + r*n*(n-1)//2
n = int(input())
factors = []
answers = set()
answers.add(1)
for i in range(1, int(sqrt(n))+2):
if n % i == 0:
x, y = i, n//i
answers.add(apsum(1, x, y))
answers.add(apsum(1, y, x))
print(" ".join(str(k) for k in sorted(list(answers))))
```
| 98,628 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
import math
n = int(input())
res = []
for i in range(1, int(math.sqrt(n))+1):
if n%i==0:
sm = (2+n-i)*(n//i)//2
res.append(sm)
if n//i!=i:
sm = (2+n-n//i)*i//2
res.append(sm)
print(*sorted(res))
```
| 98,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
import math
x=int(input())
xr=math.ceil(math.sqrt(x))
LIST=[]
for i in range(1,xr+1):
if x%i==0:
LIST.append(i)
LIST.append(x//i)
LIST=set(LIST)
ANS=[]
for l in LIST:
ANS.append((x-x//l)*l//2+l)
ANS.sort()
for a in ANS:
print(a,end=" ")
```
| 98,630 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
def prime(n):
l=[]
for i in range(1,int(n**(0.5))+1):
if n%i==0:
l.append(i)
l.append(n//i)
return l
n=int(input())
ans=[]
t=prime(n)
for i in t:
elm=n//i
ans.append(((n+2-i)*elm)//2)
ans=list(set(ans))
ans.sort()
print (*ans)
```
| 98,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Tags: math, number theory
Correct Solution:
```
from math import sqrt
n = int(input())
def c(x):
global n
cc = n//x
#print('calc:', n, x, cc + x*(cc-1)*(cc)//2)
return cc + x*(cc-1)*(cc)//2
ans = set()
for i in range(1, int(sqrt(n))+1):
if n%i==0:
ans.add(c(i))
ans.add(c(n//i))
ans = list(ans)
ans.sort()
print(' '.join(map(str, ans)))
```
| 98,632 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
import math
n = int(input())
st = set([])
for i in range(1, int(math.sqrt(n)) + 2):
if not (n % i):
k = i
st.add((k * (n // k - 1) * (n // k)) // 2 + n // k)
k = n // i
st.add((k * (n // k - 1) * (n // k)) // 2 + n // k)
print(*sorted(list(st)))
```
Yes
| 98,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Spyder Editor
This is a temporary script file.
"""
n=int(input())
a=[]
i=1
while i*i<=n:
if n%i==0:
a.append(i)
a.append(n//i)
i+=1
a.sort(reverse=True)
temp=2*n+(n*n)
minus=n
l=[]
l.append(0)
for i in a:
ans=(temp//i-minus)//2
if l[-1]!=ans:
l.append(ans)
for i in l:
if i>0:
print(i,end=' ')
# print(i,end=' ')
```
Yes
| 98,634 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
l = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
a = n//i
b = i
if a!=b :
s = (2+i*(a - 1))*a // 2
l.append(s)
b = (2 + a * (b-1)) * (b) // 2
print(b, end =' ')
for i in reversed(l):
print(i,end = ' ')
```
Yes
| 98,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
import math
n=int(input())
l1=[]
l2=[]
for i in range(1,math.floor(math.sqrt(n))+1):
if n%i==0:
l=1+((n//i)-1)*i
p1=((n//i)*(1+l))//2
l1.append(p1)
l=1+((i-1)*(n//i))
p=int((i/2)*(1+l))
if p1!=p:
l2.append(p)
for j in range(len(l2)):
print(l2[j],end=" ")
for j in range(len(l1)):
print(l1[len(l1)-j-1],end=" ")
```
Yes
| 98,636 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
import sys
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
n = int(input())
fac = factors(n)
sol = []
for f in fac:
if f == 1:
if n%2 == 0:
sq = (n+1)*(n//2)
sol.append(sq)
else:
sq = ((n+1)*(n//2)) + (n+1)//2
sol.append(sq)
elif f == n:
sol.append(1)
else:
end = n-f+1
if end%2 == 0:
sq = (end+1)*(end//2)
sol.append(sq)
else:
sq = ((end+1)*(end//2)) + (end+1)//2
sol.append(sq)
print(' '.join([str(i) for i in sorted(sol)]))
```
No
| 98,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
def gbs(k,n):
c = n
while True:
if c%k == 0 and c%n == 0:
break
c += 1
return (c//k)
cnt = [0]*(n+1)
for k in range(1,n//2+2):
s = 0
c = 1
num = gbs(k,n)
if cnt[num] == 0:
for i in range(1,num):
if (c+k)%n == 0:
s += (c+k)%n+n
else:
s += (c+k)%n
c += k
s += 1
cnt[num] = 1
print(s,end=" ")
print(1,end= " ")
```
No
| 98,638 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
a = set([])
i = 1
while i*i <= n:
if(n%i == 0):
x = i
y = n//i
# 1+x+1+2x+1+....+(y-1)x+1
# y + (y)(y-1)(x)/2
a.add(y+((y)*(y-1)*(x)/2))
x,y = y,x
a.add(y+((y)*(y-1)*(x)/2))
i += 1
a = sorted(list(a))
for i in a:
print(int(i),end=' ')
print()
```
No
| 98,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.
The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.
For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1].
Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.
Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.
Input
The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.
Output
Suppose the set of all fun values is f_1, f_2, ..., f_m.
Output a single line containing m space separated integers f_1 through f_m in increasing order.
Examples
Input
6
Output
1 5 9 21
Input
16
Output
1 10 28 64 136
Note
In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21.
<image>
In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.
Submitted Solution:
```
n = int(input())
a = set([])
i = 1
while i*i <= n:
if(n%i == 0):
x = i
y = n//i
# 1+x+1+2x+1+....+(y-1)x+1
# y + (y)(y-1)(x)/2
a.add(y+(((y)*(y-1)*(x))/2))
x,y = y,x
a.add(y+(((y)*(y-1)*(x))/2))
i += 1
a = sorted(list(a))
for i in a:
print(int(i),end=' ')
print()
```
No
| 98,640 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
import sys
input=sys.stdin.buffer.readline
from math import*
n,k=map(int,input().split())
arr=list(map(int,input().split()))
odd=0
for i in range(k-1):
if arr[i]%2==1:
odd+=1
even=k-odd-1
#print(even,odd)
if n%2==0:
if arr[-1]%2==0:
print("even")
else:
print("odd")
else:
if odd%2==0:
if arr[-1]%2==0:
print("even")
else:
print("odd")
else:
if arr[-1]%2==0:
print("odd")
else:
print("even")
```
| 98,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
b, k = map(int, input().split())
a = list(map(int, input().split()))
if b % 2 == 0:
if a[k-1] % 2 == 0:
ans = 'even'
else:
ans = 'odd'
else:
sum = 0
for i in range(k):
sum += a[i]
if sum % 2 == 0:
ans = 'even'
else:
ans = 'odd'
print(ans)
```
| 98,642 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
f=lambda:map(int,input().split())
b,k=f()
b=b%10
l=list(f())
l1=[]
if b%2==0:
print('even' if l[-1]%2==0 else 'odd')
else:
for i in l:
if i%2!=0:
l1.append(1)
print('even' if sum(l1) % 2 == 0 else 'odd')
```
| 98,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
b,k=map(int,input().split())
a=list(map(int,input().split()))
if(b%2==0):
if(a[k-1]%2==0):print('even')
else:print('odd')
else:
cnt=0
for i in range(k):
if(a[i]%2==1):cnt+=1
if(cnt%2==0):print('even')
else:print('odd')
```
| 98,644 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
#!/usr/bin/python3
b, k = map(int, input().split())
a = list(map(int, input().split()))
cur = 0
for c in a:
cur = (cur * b + c) % 2
if cur == 0:
print("even")
else:
print("odd")
```
| 98,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
b,k=map(int,input().split())
a=list(map(int,input().split()))
l=a[-1]
if(b%2==0):
if(l%2==0):
print("even")
else:
print("odd")
else:
for i in range(k-1):
l+=a[i]%2
if(l%2==0):
print("even")
else:
print("odd")
```
| 98,646 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
b,k=map(int,input().split())
l=list(map(int,input().split()))
if b%2==0:
if l[-1]%2==0:
print('even')
else:
print('odd')
else:
c=0
for i in l:
if i%2!=0:
c+=1
if c%2!=0:
print('odd')
else:
print('even')
```
| 98,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Tags: math
Correct Solution:
```
def main():
b,k = map(int,input().split())
a = list(map(int,input().split()))
even = 0
odd = 0
for i in a:
if i%2 == 0:
even += 1
else:
odd += 1
if b%2 == 0:
if a[-1]%2 == 0:
print ('even')
else:
print ('odd')
else:
if odd%2 == 0:
print ('even')
else:
print ('odd')
main()
```
| 98,648 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
x,y=map(int,input().split())
l=list(map(int,input().split()))
if x%2==0:
if l[-1]%2==0:
print("even")
else:
print("odd")
else:
c=0
for i in range(0,len(l)):
if l[i]%2!=0:
c+=1
if c%2==0:
print("even")
else:
print("odd")
```
Yes
| 98,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
def f(a,b):
if b%2==0:
return a[-1]%2==0
nodd = sum([x%2>0 for x in a])
return nodd%2==0
b,_ = list(map(int,input().split()))
a = list(map(int,input().split()))
print('even' if f(a,b) else 'odd')
```
Yes
| 98,650 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
b , k = map(int , input().split())
a = [int(a) for a in input().split()]
odd = 0
if b % 2 == 0:
if a[-1] % 2 == 0:
print('even')
else:
print('odd')
else:
for i in range(k):
if a[i] % 2 != 0:
odd += 1
if odd % 2 == 0:
print('even')
else:
print('odd')
```
Yes
| 98,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
b,k = list(map(int, input().split(" ")))
a = list(map(int, input().split(" ")))
odd_count = 0
if not b%2:
if a[-1]%2:
print("odd")
else:
print("even")
else:
for i in range(k):
if a[i]%2:
odd_count += 1
if odd_count%2:
print("odd")
else:
print("even")
```
Yes
| 98,652 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
b,k=map(int,input().split(' '))
nums=list(map(int,input().split(' ')))
if b%2==0:
ans=sum(nums)%2
else:
ans=nums[-1]%2
if ans==0:
print('even')
else:
print('odd')
```
No
| 98,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
import re
s=input()
d=re.findall(r'\d+',s)
a=int(d[0])
b=int(d[1])
num=0
if a%2==0 and int(d[-1])%2==0:
print("even")
else:
for j in range (2,len(d)):
num=num+(a**(b-j+1))*int(d[j])
if num%2==0:
print("even")
else:
print("odd")
```
No
| 98,654 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
b, k = [int(i) for i in input().split()]
arr = [int(i) for i in input().split()]
odd = 0
for i in arr:
if b & 1 or i & 1: odd += 1
if odd & 1: print('odd')
else: print('even')
```
No
| 98,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
k, n = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
if len(a) != 1:
if a[len(a) - 1] % 2 == 1:
if k % 2 == 1:
print('even')
else:
print('odd')
else:
if k % 2 == 1:
print('odd')
else:
print('even')
else:
if a[len(a) - 1] % 2 == 0:
print('even')
else:
print('odd')
```
No
| 98,656 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
from math import sqrt
def criba(n):
primos = [True] * n #Inicialmente todos los valores hasta n son marcados como primos
primos[0] = primos[1] = False #Desmarco el 0 y el 1
for i in range(2, int(sqrt(n))+1): #Recorrer solo hasta la raiz de n
if primos[i]:
for j in range(i * i, n, i): #Desmarcar todos los multiplos del primo i
primos[j] = False
return [index for index, item in enumerate(primos) if item] #retornar una lista solo con los primos
a,b = map(int,input().split())
resp = 0
primos = criba(b)
for i in primos:
if (i==2 or i%4==1)and i>=a: #Solo incremento el contador cuando cumplen la condicion de exprsarse como suma de dos cuadrados y son >= que el limite inferior
resp+=1
print(resp)
```
No
| 98,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
primes_under_100 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def isprime(n):
if n <= 100:
return n in primes_under_100
if n % 2 == 0 or n % 3 == 0:
return False
for f in range(5, int(n ** .5), 6):
if n % f == 0 or n % (f + 2) == 0:
return False
return True
s=input()
a=[int(i) for i in s.split()]
b=a[1]
a=a[0]
t=0
for i in range(a,b+1):
if (i-1)%4==0:
if isprime(i):
t=t+1
if a<=2<=b:
t=t+1
print(t)
```
No
| 98,658 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
primes_under_100 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def isprime(n):
if n <= 100:
return n in primes_under_100
if n % 2 == 0 or n % 3 == 0:
return False
for f in range(5, int(n ** .5), 6):
if n % f == 0 or n % (f + 2) == 0:
return False
return True
'''
t=0
for i in range(n):
if n%(i+1)==0:
t=t+1
if t==2:
return True
else:
return False
for i in range(5000):
if isprime(i+1):
print(i+1)'''
s=input()
a=[int(i) for i in s.split()]
b=a[1]
a=a[0]
t=0
if a<=2<=b:
t.append(2)
for i in range(a,b+1):
if (i-1)%4==0:
if isprime(i):
t=t+1
print(t)
```
No
| 98,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Examples
Input
3 5
Output
1
Input
6 66
Output
7
Submitted Solution:
```
primes_under_100 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def isprime(n):
if n <= 100:
return n in primes_under_100
if n % 2 == 0 or n % 3 == 0:
return False
for f in range(5, int(n ** .5), 6):
if n % f == 0 or n % (f + 2) == 0:
return False
return True
s=input()
a=[int(i) for i in s.split()]
b=a[1]
a=a[0]
t=0
for i in range(a,b+1):
if (i-1)%4==0:
if isprime(i):
t=t+1
print(t)
```
No
| 98,660 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
import sys
from math import *
from fractions import gcd
readints=lambda:map(int, input().strip('\n').split())
def hasUnique(s,n,k):
freq = {}
for i in range(n-k+1):
x = s[i:i+k]
if x not in freq: freq[x]=0
freq[x]+=1
for k in freq:
if freq[k] == 1:
return True
return False
def good(s,n,k):
for i in range(1,k):
if hasUnique(s,n,i):
return False
freq={}
for i in range(n-k+1):
x = s[i:i+k]
if x not in freq: freq[x]=0
freq[x]+=1
st = set()
for x in freq:
if freq[x]==1:
if len(st)>0: return False
st.add(x)
return len(st) == 1
def gen(i,s,n,k):
if i==n:
if good(s,n,k):
print(s)
else:
gen(i+1,s+'0',n,k)
gen(i+1,s+'1',n,k)
#gen(0,'',5,3)
# for n in range(3,9):
# for k in range(1,n+1):
# if (n%2 != k%2):
# print(n,k,'skip')
# print()
# continue
# print(n,k)
# gen(0,'',n,k)
# print()
n,k = readints()
if n == k:
print('0' * n)
sys.exit(0)
s = ''
while len(s) < n:
r = int((n-k)//2)
s = s + (('0'*r) + '1')
s = s[:n]
print(s)
```
| 98,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
n, k = map(int, input().split())
if k == 1:
print("1", end = "")
for i in range(1, n):
print("0", end = "")
print()
exit()
a = (n + k - 2) // 2
ans = ""
if (a - k + 2) % 2:
for i in range (a - k + 2):
ans += chr(i % 2 + 48)
else:
ans += chr(48)
for i in range(1, a - k + 2):
ans += chr(49)
for i in range(a - k + 2, n):
ans += ans[i - a + k - 2]
print(ans)
```
| 98,662 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
N, K = map(int, input().split())
if K == 0 or N == K:
print("1" * N)
elif K == 1:
print("0" + "1" * (N - 1))
elif K <= 2//N:
if K & 1:
print("10" * (K//2+1) + "1" * (N-2-K//2*2))
else:
print("0" + "10" * (K//2) + "1" * (N-1-K//2*2))
else:
print((("0"+"1"*((N-K)//2))*(N//(((N-K)//2))+1))[:N])
```
| 98,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
n, k = list(map(int,input().split()))
chuj_twojej_starej = (n - k) // 2 + 1
i = 1
while True:
if i % chuj_twojej_starej == 0:
print(0, end = "")
else:
print(1, end = "")
if i == n:
break
i += 1
```
| 98,664 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
n,k=map(int,input().split())
d=(n-k)//2
s=0
while s!=n:
if (s+1)%(d+1)==0:
print("1",end="")
else :
print("0",end="")
s+=1
```
| 98,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
N, K = map(int, input().split())
if N == K:
print("0"*N)
elif K == 1:
print("0"*(N-1) + "1")
elif K == 3:
print("1" + "0"*(N-4) + "101")
else:
res = ["0"]*N
for i in range(0, N, N//2-K//2+1):
res[i] = "1"
print(''.join(res))
```
| 98,666 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
n,k=map(int,input().split())
x=(n-(k-1)+1)//2
STR="0"*(x-1)+"1"
ANS=STR*(n//x+1)
print(ANS[:n])
```
| 98,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Tags: constructive algorithms, math, strings
Correct Solution:
```
#------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
import math
n,k=map(int,input().split())
a=(n-k)//2
s=""
while(len(s)<n):
s=s+"0"*(a)+"1"
print(s[:n])
```
| 98,668 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n,k=[int(x) for x in input().split()]
a=(n-k)//2
tot=''
for i in range(n):
if (i+1)%(a+1)==0:
tot+='1'
else:
tot+='0'
print(tot)
```
Yes
| 98,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
from sys import stdin
n,m=map(int,stdin.readline().strip().split())
x=n-m
x=x//2
s=""
y=0
if m==1:
print("0"+(n-1)*"1")
exit(0)
while len(s)<n:
if y==1:
s+="1"*x
else:
s+="0"
y+=1
y=y%2
print(s[0:n])
```
Yes
| 98,670 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n, k = map(int, input().split())
# while making the string, k = (n - 2 * length + 2)
length = (n - k + 2) // 2 # length of a cycle
string = "0" * (length - 1) + "1" # make the cycle
answer = string * (n // length + 1) # make the string with length >= n
print(answer[ : n])
```
Yes
| 98,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n, k = map(int, input().split())
s, v = (n - k) // 2 * '0' + '1', ''
while len(v) < n:
v += s
print(v[:n])
```
Yes
| 98,672 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
n,k=map(int,input().strip().split())
d=n-k//2+1
x=['1' if (i+1)%d==0 else '0' for i in range(n)]
print(''.join(x))
```
No
| 98,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
def solve(n, k):
if n // 3 >= k:
string1 = ["0"] + ["1"] * (k-2)
string2 = ["1"] * (n + 2 - 2*k)
answer = (string2 + string1 + string1)
return ("".join(answer))
else:
key = (n - k) // 2
a, c = key + 1, key - 1
b = n // a
string1 = ["1"] * (a-1) + ["0"]
string2 = ["1"] * (n - b*a)
answer = (string1 * b + string2)
return ("".join(answer))
return
n, k = map(int, input().split())
print (solve(n, k))
```
No
| 98,674 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
def solve(n, k):
if n // 2 >= k:
string1 = ["0"] + ["1"] * (k-2)
string2 = ["1"] * (n + 2 - 2*k)
answer = (string2 + string1 + string1)
return ("".join(answer))
else:
key = (n - k) // 2
a, c = key + 1, key - 1
b = n // a
string1 = ["1"] * (a-1) + ["0"]
string2 = ["1"] * (n - b*a)
answer = (string1 * b + string2)
return ("".join(answer))
return
n, k = map(int, input().split())
print (solve(n, k))
```
No
| 98,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l.
For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.
Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.
You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.
Input
The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).
Output
Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.
Examples
Input
4 4
Output
1111
Input
5 3
Output
01010
Input
7 3
Output
1011011
Note
In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.
In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.
In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
Submitted Solution:
```
if __name__ == "__main__":
N, K = [int(x) for x in input().split()]
if K * 2 <= N:
print('0' * K + '1' * (N - K))
elif N == K:
print('0' * N)
elif N % 2 == 0 and K == N // 2 + 2:
print('01' * (N // 2))
elif N % 2 == 1 and K == N // 2 + 1:
print('01' * (N // 2) + '0')
else:
print('orzJumpmelonAKCTS2019')
```
No
| 98,676 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
n = int(input())
if n == 1:
print(1)
else:
print( int((2*n-1)**2 - 4*(n*(n-1)/2) ))
```
| 98,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
# your code goes here
import math
n=int(input())
sum=1
for i in range(2,n+1):
sum+=4*(i-1)
print(sum)
```
| 98,678 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
def ans(n):
if n==1:
return 1
else:
return ((n-1)*4) + ans(n-1)
n = int(input())
print(ans(n))
```
| 98,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
n = int(input())
print(2 * n**2 - 2 * n + 1)
################################################################
```
| 98,680 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
n=int(input())
a=[]
l=1;
n=2*(n**2)-2*n+1
print(n)
```
| 98,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
ar=[0 for i in range(0,101)]
def res():
ar[0]=1
s=1
for i in range(1,101):
ar[i]=ar[i-1]+4*s
s+=1
n=int(input())
res()
print(ar[n-1])
```
| 98,682 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
def solve(n):
if n == 1: return 1
if n == 2: return 5
return solve(n-1) + 4*(n-1)
def main():
n = int(input())
print(solve(n))
main()
```
| 98,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Tags: dp, implementation, math
Correct Solution:
```
n=int(input())
a=2*n*(n-1)+1
print(a)
```
| 98,684 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
n = int(input())
sum = 1
for i in range(1, n):
sum = sum + 4 * i
print(sum)
```
Yes
| 98,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
dp = dict()
dp[1] = 1
def testCase():
# pass
n = int(input())
for i in range(2, n+1):
dp[i] = dp[i-1] + (i-1)*4
print("{}".format(dp[n]))
# main
#test = int(input())
test = 1
while test:
testCase()
test -= 1
```
Yes
| 98,686 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
n = int(input())
if n % 2 == 0:
print(((n-1)**2) + 4*(((n*(n+1))//2) - ((n)*(n+2))//4))
else:
print((n**2) + 4*((((n-2)*(n-1))//2) - (((n-2)//2)*(((n-2)//2)+1))))
```
Yes
| 98,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
n=int(input())
from math import ceil
x=((n*2 - 1)**2)/2
print(ceil(x))
```
Yes
| 98,688 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
n = int(input())
if n == 1:
print(1)
if n == 2:
print(5)
else:
print((2*n-3)**2+4)
```
No
| 98,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
num = int(input())
s = 1
for i in range(1,num):
s+=2**(i+1)
print(s)
```
No
| 98,690 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
k = int(input())
if k == 1:
print(1)
else:
print(4*(2**(k-1)-1)+1)
```
No
| 98,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While playing with geometric figures Alex has accidentally invented a concept of a n-th order rhombus in a cell grid.
A 1-st order rhombus is just a square 1 × 1 (i.e just a cell).
A n-th order rhombus for all n ≥ 2 one obtains from a n-1-th order rhombus adding all cells which have a common side with it to it (look at the picture to understand it better).
<image>
Alex asks you to compute the number of cells in a n-th order rhombus.
Input
The first and only input line contains integer n (1 ≤ n ≤ 100) — order of a rhombus whose numbers of cells should be computed.
Output
Print exactly one integer — the number of cells in a n-th order rhombus.
Examples
Input
1
Output
1
Input
2
Output
5
Input
3
Output
13
Note
Images of rhombus corresponding to the examples are given in the statement.
Submitted Solution:
```
n = int(input())
print((n+1)**2+4)
```
No
| 98,692 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Tags: implementation
Correct Solution:
```
n,x,y=map(int,input().split())
a=*map(int,input().split()),
for k in range(n):
v=1
for l in range(max(0,k-x),k):v&=a[k]<a[l]
for l in range(k+1,min(k+y+1,n)):v&=a[k]<a[l]
if v:print(k+1);break
```
| 98,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Tags: implementation
Correct Solution:
```
n,x,y = map(int,input().split())
l = list(map(int,input().split()))
for i in range(n):
if l[i] <= min(l[max(0,i-x):i+y+1]):
break
print(i+1)
```
| 98,694 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Tags: implementation
Correct Solution:
```
def solve(days, x, y, n):
window = dict()
win_beg = 0
win_end = 0
res = 0
pre = 0
while pre < n:
while win_beg < pre - x:
del window[days[win_beg]]
win_beg += 1
while win_end < min(pre + y + 1, n):
window[days[win_end]] = True
if days[win_end] < days[res]:
res = win_end
win_end += 1
if pre == res:
return pre + 1
pre = res
return -1
if __name__ == "__main__":
n, x, y = [int(ele) for ele in input().split()]
days = [int(ele) for ele in input().split()]
print(solve(days, x, y, n))
```
| 98,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Tags: implementation
Correct Solution:
```
n,x,y=map(int,input().split())
l=list(map(int,input().split()))
for i in range(n):
j=i-x
k=i+y
if j<0:
j=0
if k>=n:
k=n-1
if all(l[i]<arr for arr in l[j:i]) and all(l[i]<brr for brr in l[i+1:k+1]):
print(i+1)
break
```
| 98,696 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Tags: implementation
Correct Solution:
```
N, BEFORE, AFTER = map(int, input().split())
days = list(map(int, input().split()))
def goodbefore(j):
for diff in range(1, BEFORE+1):
k = j - diff
if k < 0:
continue
if days[k] <= days[j]:
return False
return True
def goodafter(j):
for diff in range(1, AFTER+1):
k = j + diff
if k >= N:
continue
if days[k] <= days[j]:
return False
return True
for i in range(N):
if goodbefore(i) and goodafter(i):
print(i+1)
break
```
| 98,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Tags: implementation
Correct Solution:
```
n, x, y = map(int, input().split())
inf = 10**10
a = [inf]*7 + list(map(int, input().split())) + [inf]*7
for i in range(7, n+7):
if a[i] < (min(a[i-x:i]) if x else inf) and a[i] < (min(a[i+1:i+y+1]) if y else inf):
print(i-6)
exit()
```
| 98,698 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, a_i millimeters of rain will fall. All values a_i are distinct.
The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if a_d is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, a_d < a_j should hold for all d - x ≤ j < d and d < j ≤ d + y. Citizens only watch the weather during summer, so we only consider such j that 1 ≤ j ≤ n.
Help mayor find the earliest not-so-rainy day of summer.
Input
The first line contains three integers n, x and y (1 ≤ n ≤ 100 000, 0 ≤ x, y ≤ 7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i denotes the rain amount on the i-th day.
Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.
Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.
In the second example day 3 is not not-so-rainy, because 3 + y = 6 and a_3 > a_6. Thus, day 8 is the answer. Note that 8 + y = 11, but we don't consider day 11, because it is not summer.
Tags: implementation
Correct Solution:
```
n, x, y = map(int, input().split());
arr = list(map(int, input().split()));
for i in range(n):
if ((not arr[max(i-1, 0):i] or min(arr[i-x:i] + [2e9]) > arr[i]) and (not arr[i+1: min(n, i+y+1)] or min(arr[i+1:i+y+1]+[2e9]) > arr[i])):
print(i+1)
exit()
```
| 98,699 |
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