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int64 97
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int64 194
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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wrote down two strings s of length n and t of length m consisting of small English letters 'a' and 'b'. What is more, he knows that string t has a form "abab...", namely there are letters 'a' on odd positions and letters 'b' on even positions.
Suddenly in the morning, Vasya found that somebody spoiled his string. Some letters of the string s were replaced by character '?'.
Let's call a sequence of positions i, i + 1, ..., i + m - 1 as occurrence of string t in s, if 1 โค i โค n - m + 1 and t1 = si, t2 = si + 1, ..., tm = si + m - 1.
The boy defines the beauty of the string s as maximum number of disjoint occurrences of string t in s. Vasya can replace some letters '?' with 'a' or 'b' (letters on different positions can be replaced with different letter). Vasya wants to make some replacements in such a way that beauty of string s is maximum possible. From all such options, he wants to choose one with the minimum number of replacements. Find the number of replacements he should make.
Input
The first line contains a single integer n (1 โค n โค 105) โ the length of s.
The second line contains the string s of length n. It contains small English letters 'a', 'b' and characters '?' only.
The third line contains a single integer m (1 โค m โค 105) โ the length of t. The string t contains letters 'a' on odd positions and 'b' on even positions.
Output
Print the only integer โ the minimum number of replacements Vasya has to perform to make the beauty of string s the maximum possible.
Examples
Input
5
bb?a?
1
Output
2
Input
9
ab??ab???
3
Output
2
Note
In the first sample string t has a form 'a'. The only optimal option is to replace all characters '?' by 'a'.
In the second sample using two replacements we can make string equal to "aba?aba??". It is impossible to get more than two occurrences.
Submitted Solution:
```
n = int(input())
s = list(input())
t = int(input())
re = s.copy()
kras = 'ab'
kras *= t
kras = kras[:t]
col = s.count('?')
c = 0
v = 0
q = 0
h = 0
for i in range(2 ** col):
shable = bin(i)[2:]
priv = '0' * (col - len(shable))
priv += shable
shable = priv
for el in range(n):
if s[el] == '?':
if shable[q] == '1':
if s[el - 1] == 'b':
s[el] = 'a'
else:
s[el] = 'b'
q += 1
new = ''.join(s).count(kras)
if new > c:
v = shable.count('1')
c = new
s = re.copy()
q = 0
h = 0
print(v)
```
|
instruction
| 0
| 108,368
| 0
| 216,736
|
No
|
output
| 1
| 108,368
| 0
| 216,737
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,395
| 0
| 216,790
|
Tags: implementation
Correct Solution:
```
class CodeforcesTask999BSolution:
def __init__(self):
self.result = ''
self.n = 0
self.code = ''
def read_input(self):
self.n = int(input())
self.code = input()
def process_task(self):
for x in range(1, self.n + 1):
if not self.n % x:
self.code = self.code[:x][::-1] + self.code[x:]
self.result = self.code
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask999BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
```
|
output
| 1
| 108,395
| 0
| 216,791
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,396
| 0
| 216,792
|
Tags: implementation
Correct Solution:
```
a=int(input())
b=input()
f=[]
for i in range(1,a+1):
if a%i==0:
f.append(i)
f.sort()
def aa(a,b):
g=''
for i in range(a,len(b)):
g=g+b[i]
e=[]
for i in range(a):
e.append(b[i])
e.reverse()
t=''
for i in e:
t=t+i
t=t+g
return(t)
for i in f:
b=aa(i,b)
print(b)
```
|
output
| 1
| 108,396
| 0
| 216,793
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,397
| 0
| 216,794
|
Tags: implementation
Correct Solution:
```
n = int(input())
str1 = str(input())
l2 = []
for i in str1:
l2.append(i)
l1 = []
# x = n/1.0
for j in range(1,n):
if n/j == int(n/j):
l1.append(j)
l1.append(int(n))
for i in l1:
l3 = l2[0:i]
l3.reverse()
l2[0:i] = l3[0:i]
strp = ""
for i in l2:
strp+=i
print(strp)
```
|
output
| 1
| 108,397
| 0
| 216,795
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,398
| 0
| 216,796
|
Tags: implementation
Correct Solution:
```
def rev( l , d ) :
ll = l[ d : ]
ll1 = l[ : d]
ll = ll1[::-1] + ll
return ll
n = int( input() )
s = str( input() )
di = []
for i in range( 1 , n + 1 ) :
if n % i == 0 :
di.append( i )
l = list( s )
for i in range( len( di ) ) :
#print( ''.join( l ) )
l = rev( l , di[ i ] )
#print(''.join( l ) , di[ i ] )
print( ''.join(l) )
```
|
output
| 1
| 108,398
| 0
| 216,797
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,399
| 0
| 216,798
|
Tags: implementation
Correct Solution:
```
IL = lambda: list(map(int, input().split()))
n = int(input())
s = list(input())
for i in range(1, n+1):
if n%i:
continue
s[:i] = s[i-1::-1]
print(''.join(s))
```
|
output
| 1
| 108,399
| 0
| 216,799
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,400
| 0
| 216,800
|
Tags: implementation
Correct Solution:
```
n = int(input())
u = [n]
for i in range(n // 2, 1, -1):
if n % i == 0:
u.append(i)
s = list(input())
u.reverse()
for i in u:
d = s[:i]
d.reverse()
s = d + s[i:]
for i in s:
print(i, end = '')
```
|
output
| 1
| 108,400
| 0
| 216,801
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,401
| 0
| 216,802
|
Tags: implementation
Correct Solution:
```
n = int(input())
k = list(input())
l = [i for i in range(2,(n**1//2)+1) if n%i==0 ]
for i in l:
k[:i] = k[i-1::-1]
print(''.join(k[::-1]))
```
|
output
| 1
| 108,401
| 0
| 216,803
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
|
instruction
| 0
| 108,402
| 0
| 216,804
|
Tags: implementation
Correct Solution:
```
n=int(input())
x=input()
for i in range(2,n+1):
# print(n,i)
if n%i==0:
s=x[i-1::-1]
s+=x[i:]
x=s
# print(s)
print(x)
```
|
output
| 1
| 108,402
| 0
| 216,805
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
n = int(input())
s = list(input())
for i in range(1, n+1):
if n % i == 0:
s[0:i] = s[i-1::-1]
print(''.join(s))
```
|
instruction
| 0
| 108,403
| 0
| 216,806
|
Yes
|
output
| 1
| 108,403
| 0
| 216,807
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
n =int(input())
s = input()
def findAllDivisors(k):
lst = []
for i in range(1,k+1):
if k%i == 0:
lst.append(i)
return lst
lst = findAllDivisors(n)
for i in range(0, len(lst)):
newStr = s[0:lst[i]]
newStr = newStr[::-1]
newStr1 = s[lst[i]:len(s)]
s = newStr + newStr1
print(s)
```
|
instruction
| 0
| 108,404
| 0
| 216,808
|
Yes
|
output
| 1
| 108,404
| 0
| 216,809
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
n=int(input())
a=input()
for i in range(2,n):
if n%i==0:
#print(i,a)
#print(a[])
a=a[i-1::-1]+a[i:]
#print(i,a)
a=a[::-1]
print(a)
```
|
instruction
| 0
| 108,405
| 0
| 216,810
|
Yes
|
output
| 1
| 108,405
| 0
| 216,811
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
# http://codeforces.com/contest/999/problem/B
def read_nums():
return [int(x) for x in input().split()]
def main():
n, = read_nums()
s = input()
for i in range(1, n + 1):
if n % i == 0:
one = s[:i][::-1]
two = s[i:]
s = one + two
print(s)
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 108,406
| 0
| 216,812
|
Yes
|
output
| 1
| 108,406
| 0
| 216,813
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
def reverse(y):
p=""
for k in range(len(y)-1,-1,-1):
p=p+y[k]
return p
n=int(input())
s=input()
L=[]
for i in range(1,n+1):
if n%i==0:
L.append(i)
for j in range(len(L)):
a=s[:L[j]]
s=reverse(a)+s[L[j]:]
print(s)
```
|
instruction
| 0
| 108,407
| 0
| 216,814
|
No
|
output
| 1
| 108,407
| 0
| 216,815
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
n = int(input())
string = list(input())
divisors = []
for i in range(1, n+1):
if n % i == 0:
divisors.append(i-1)
for divisor in divisors:
print(divisor)
rev = string[:divisor+1]
rev = rev[::-1]
string = rev + string[divisor+1:]
# print(string)
print("".join(string))
```
|
instruction
| 0
| 108,408
| 0
| 216,816
|
No
|
output
| 1
| 108,408
| 0
| 216,817
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
a = int(input())
s = list(input())
b = []
for i in range(a, 1, -1):
if a % i == 0:
b.append(i)
for i in b:
s[0:i] = reversed(s[0:i])
s = "".join(s)
print(s)
```
|
instruction
| 0
| 108,409
| 0
| 216,818
|
No
|
output
| 1
| 108,409
| 0
| 216,819
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n can be encrypted by the following algorithm:
* iterate over all divisors of n in decreasing order (i.e. from n to 1),
* for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d).
For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" โ "secrofedoc" โ "orcesfedoc" โ "rocesfedoc" โ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).
You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.
Input
The first line of input consists of a single integer n (1 โค n โค 100) โ the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.
Output
Print a string s such that the above algorithm results in t.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
1
z
Output
z
Note
The first example is described in the problem statement.
Submitted Solution:
```
q=lambda:map(int,input().split())
qi=lambda:int(input())
qs=lambda:input().split()
def lol(n):
for i in range(1,n+1):
if n%i==0:
yield i
n=qi()
a=input()
for i in lol(n):
a=''.join(reversed(a[0:i]))+a[i:10]
print(a)
```
|
instruction
| 0
| 108,410
| 0
| 216,820
|
No
|
output
| 1
| 108,410
| 0
| 216,821
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,411
| 0
| 216,822
|
"Correct Solution:
```
#-*- coding:utf-8 -*-
t = input()
print(t.replace("?","D"))
```
|
output
| 1
| 108,411
| 0
| 216,823
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,412
| 0
| 216,824
|
"Correct Solution:
```
T = input()
y = T.replace('?', 'D')
print(y)
```
|
output
| 1
| 108,412
| 0
| 216,825
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,413
| 0
| 216,826
|
"Correct Solution:
```
T = input().replace('?', 'D')
print(T)
```
|
output
| 1
| 108,413
| 0
| 216,827
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,414
| 0
| 216,828
|
"Correct Solution:
```
T = input()
_T = T.replace('?','D')
print(_T)
```
|
output
| 1
| 108,414
| 0
| 216,829
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,415
| 0
| 216,830
|
"Correct Solution:
```
t=input()
for i in t:
print('P' if i=='P' else 'D',end='')
```
|
output
| 1
| 108,415
| 0
| 216,831
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,416
| 0
| 216,832
|
"Correct Solution:
```
t = str(input())
ans = t.replace('?', 'D')
print(ans)
```
|
output
| 1
| 108,416
| 0
| 216,833
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,417
| 0
| 216,834
|
"Correct Solution:
```
T = input()
D = T.replace("?","D")
print(D)
```
|
output
| 1
| 108,417
| 0
| 216,835
|
Provide a correct Python 3 solution for this coding contest problem.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
|
instruction
| 0
| 108,418
| 0
| 216,836
|
"Correct Solution:
```
s = input()
while "?" in s:
s = s.replace("?","D")
print(s)
```
|
output
| 1
| 108,418
| 0
| 216,837
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
Submitted Solution:
```
t=str(input())
t=t.replace('?','D')
print(t)
```
|
instruction
| 0
| 108,420
| 0
| 216,840
|
Yes
|
output
| 1
| 108,420
| 0
| 216,841
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
Submitted Solution:
```
T = input()
s = T.replace("?","D")
print(s)
```
|
instruction
| 0
| 108,421
| 0
| 216,842
|
Yes
|
output
| 1
| 108,421
| 0
| 216,843
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.
We have a string T consisting of `P`, `D`, and `?`.
Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
Constraints
* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.
Input
Input is given from Standard Input in the following format:
T
Output
Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.
Examples
Input
PD?D??P
Output
PDPDPDP
Input
P?P?
Output
PDPD
Submitted Solution:
```
T = input()
ans = T[-1]
if T[-1] == '?':
ans = 'D'
for i in range(1, len(T)):
s, t = T[-1-i], T[-i]
if s == '?':
if t == 'D' :
ans += 'P'
elif t == 'P':
ans += 'D'
else:
if ans[i-1] == 'D':
ans += 'P'
else:
ans += 'D'
else:
ans += s
print(ans[::-1])
```
|
instruction
| 0
| 108,426
| 0
| 216,852
|
No
|
output
| 1
| 108,426
| 0
| 216,853
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
For a non-empty string S, we will define f(S) as the shortest even string that can be obtained by appending one or more characters to the end of S. For example, f(`abaaba`)=`abaababaab`. It can be shown that f(S) is uniquely determined for a non-empty string S.
You are given an even string S consisting of lowercase English letters. For each letter in the lowercase English alphabet, find the number of its occurrences from the l-th character through the r-th character of f^{10^{100}} (S).
Here, f^{10^{100}} (S) is the string f(f(f( ... f(S) ... ))) obtained by applying f to S 10^{100} times.
Constraints
* 2 \leq |S| \leq 2\times 10^5
* 1 \leq l \leq r \leq 10^{18}
* S is an even string consisting of lowercase English letters.
* l and r are integers.
Input
Input is given from Standard Input in the following format:
S
l r
Output
Print 26 integers in a line with spaces in between. The i-th integer should be the number of the occurrences of the i-th letter in the lowercase English alphabet from the l-th character through the r-th character of f^{10^{100}} (S).
Examples
Input
abaaba
6 10
Output
3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Input
xx
1 1000000000000000000
Output
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000000000000000 0 0
Input
vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000
Output
87167725689669676 0 0 0 0 0 282080685775825810 0 0 0 87167725689669676 0 87167725689669676 0 0 87167725689669676 0 0 0 0 87167725689669676 141040342887912905 0 141040342887912905 0 0
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from collections import Counter
"""
ใปๅฎ้จใใใจใไฝๅใใใใจFibonacci๏ผๆช่จผๆ๏ผ
"""
S = readline().rstrip()
l,r = map(int,read().split())
def double_naive(S):
L = len(S)
for i in range(1,L+1):
T = S[:i] * 2
if len(T) > L and T[:L] == S:
return T
def Z_algorithm(S):
# ๅ
ฑ้ๆฅ้ ญ่พใฎ้ทใใ่ฟใ
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
def F(S):
N = len(S)
opt_len = N
Z = Z_algorithm(S)
for x in range(N//2+1,N):
z = Z[x]
if x + z >= N:
opt_len = x
break
return S[:opt_len] * 2
S = F(S)
T = F(S)
# S,T,TS,TST,...
L = [len(S), len(T)]
for _ in range(100):
L.append(L[-1] + L[-2])
ctr = [Counter(S), Counter(T)]
for _ in range(100):
ctr.append(ctr[-1] + ctr[-2])
def get_dist(N,R):
# S_NใซใใใR็ช็ฎไปฅไธ
if N == 0:
return Counter(S[:R])
if N == 1:
return Counter(T[:R])
if L[N-1] >= R:
return get_dist(N-1,R)
return ctr[N-1] + get_dist(N-2,R-L[N-1])
ctr = get_dist(99,r) - get_dist(99,l-1)
answer = [ctr[x + ord('a')] for x in range(26)]
print(*answer)
```
|
instruction
| 0
| 108,512
| 0
| 217,024
|
No
|
output
| 1
| 108,512
| 0
| 217,025
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
For a non-empty string S, we will define f(S) as the shortest even string that can be obtained by appending one or more characters to the end of S. For example, f(`abaaba`)=`abaababaab`. It can be shown that f(S) is uniquely determined for a non-empty string S.
You are given an even string S consisting of lowercase English letters. For each letter in the lowercase English alphabet, find the number of its occurrences from the l-th character through the r-th character of f^{10^{100}} (S).
Here, f^{10^{100}} (S) is the string f(f(f( ... f(S) ... ))) obtained by applying f to S 10^{100} times.
Constraints
* 2 \leq |S| \leq 2\times 10^5
* 1 \leq l \leq r \leq 10^{18}
* S is an even string consisting of lowercase English letters.
* l and r are integers.
Input
Input is given from Standard Input in the following format:
S
l r
Output
Print 26 integers in a line with spaces in between. The i-th integer should be the number of the occurrences of the i-th letter in the lowercase English alphabet from the l-th character through the r-th character of f^{10^{100}} (S).
Examples
Input
abaaba
6 10
Output
3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Input
xx
1 1000000000000000000
Output
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000000000000000 0 0
Input
vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000
Output
87167725689669676 0 0 0 0 0 282080685775825810 0 0 0 87167725689669676 0 87167725689669676 0 0 87167725689669676 0 0 0 0 87167725689669676 141040342887912905 0 141040342887912905 0 0
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from collections import Counter
"""
ใปๅฎ้จใใใจใไฝๅใใใใจFibonacci๏ผๆช่จผๆ๏ผ
"""
S = readline().rstrip()
l,r = map(int,read().split())
def double_naive(S):
L = len(S)
for i in range(1,L+1):
T = S[:i] * 2
if len(T) > L and T[:L] == S:
return T
def Z_algorithm(S):
# ๅ
ฑ้ๆฅ้ ญ่พใฎ้ทใใ่ฟใ
N=len(S)
arr = [0]*N
arr[0] = N
i,j = 1,0
while i<N:
while i+j<N and S[j]==S[i+j]:
j += 1
arr[i]=j
if not j:
i += 1
continue
k = 1
while i+k<N and k+arr[k]<j:
arr[i+k] = arr[k]
k += 1
i += k; j -= k
return arr
def F(S):
N = len(S)
opt_len = N
Z = Z_algorithm(S)
for x in range(N//2+1,N):
z = Z[x]
if x + z >= N:
opt_len = x
break
return S[:opt_len] * 2
S = F(S)
S = F(S)
T = F(S)
# S,T,TS,TST,...
L = [len(S), len(T)]
for _ in range(100):
L.append(L[-1] + L[-2])
ctr = [Counter(S), Counter(T)]
for _ in range(100):
ctr.append(ctr[-1] + ctr[-2])
def get_dist(N,R):
# S_NใซใใใR็ช็ฎไปฅไธ
if N == 0:
return Counter(S[:R])
if N == 1:
return Counter(T[:R])
if L[N-1] >= R:
return get_dist(N-1,R)
return ctr[N-1] + get_dist(N-2,R-L[N-1])
ctr = get_dist(99,r) - get_dist(99,l-1)
answer = [ctr[x + ord('a')] for x in range(26)]
print(*answer)
```
|
instruction
| 0
| 108,513
| 0
| 217,026
|
No
|
output
| 1
| 108,513
| 0
| 217,027
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
For a non-empty string S, we will define f(S) as the shortest even string that can be obtained by appending one or more characters to the end of S. For example, f(`abaaba`)=`abaababaab`. It can be shown that f(S) is uniquely determined for a non-empty string S.
You are given an even string S consisting of lowercase English letters. For each letter in the lowercase English alphabet, find the number of its occurrences from the l-th character through the r-th character of f^{10^{100}} (S).
Here, f^{10^{100}} (S) is the string f(f(f( ... f(S) ... ))) obtained by applying f to S 10^{100} times.
Constraints
* 2 \leq |S| \leq 2\times 10^5
* 1 \leq l \leq r \leq 10^{18}
* S is an even string consisting of lowercase English letters.
* l and r are integers.
Input
Input is given from Standard Input in the following format:
S
l r
Output
Print 26 integers in a line with spaces in between. The i-th integer should be the number of the occurrences of the i-th letter in the lowercase English alphabet from the l-th character through the r-th character of f^{10^{100}} (S).
Examples
Input
abaaba
6 10
Output
3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Input
xx
1 1000000000000000000
Output
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000000000000000 0 0
Input
vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000
Output
87167725689669676 0 0 0 0 0 282080685775825810 0 0 0 87167725689669676 0 87167725689669676 0 0 87167725689669676 0 0 0 0 87167725689669676 141040342887912905 0 141040342887912905 0 0
Submitted Solution:
```
#!/usr/bin/env python3
M = 10 ** 9 + 7
a = ord('a')
def check_cycle(t, p):
len_t = len(t)
if len_t % p != 0:
return False
pre = t[: p]
for i in range(p, len_t, p):
if t[i: i + p] != pre:
return False
return True
def solve_cycle(t, l, r):
len_t = len(t)
ql = (l - 1) // len_t
rl = (l - 1) % len_t
qr = r // len_t
rr = r % len_t
ans = [0] * 26
for ch in t:
ans[ord(ch) - a] += 1
for j in range(26):
ans[j] *= (qr - ql)
for ch in t[: rr + 1]:
ans[ord(ch) - a] += 1
for ch in t[: rl + 1]:
ans[ord(ch) - a] -= 1
return ans
def calc_part(tp, tm, len_p, len_m, dp0, dp1, r, ir):
num_p = 0
num_m = 0
for i in range(ir, -1, -1):
n = dp0[i] * len_p + dp1[i] * len_m
if n <= r:
num_p += dp0[i]
num_m += dp1[i]
r -= n
vec = [0] * 26
tr = tp
if len_p + len_m < r:
num_p += 1
num_m += 1
r -= len_p + len_m
elif len_p < r:
num_p += 1
r -= len_p
tr = tm
for ch in tp:
vec[ord(ch) - a] += num_p
for ch in tm:
vec[ord(ch) - a] += num_m
for ch in tr[: r]:
vec[ord(ch) - a] += 1
return vec
def solve_part(t, len_p, l, r):
len_t = len(t)
len_m = len_t - 2 * len_p
dp0 = [0] * 121
dp1 = [0] * 121
dp0[0] = 2
dp1[0] = 1
for i in range(1, 121):
dp0[i] = dp0[i - 1] + dp1[i - 1]
dp1[i] = dp0[i - 1]
n = dp0[i] * len_p + dp1[i] * len_m
if l - 1 <= n:
il = i - 1
if r <= n:
ir = i - 1
break
vec_l = calc_part(t[: len_p], t[len_p: len_p + len_m], len_p, len_m, dp0, dp1, l - 1, il)
vec_r = calc_part(t[: len_p], t[len_p: len_p + len_m], len_p, len_m, dp0, dp1, r, ir)
ans = [vec_r[j] - vec_l[j] for j in range(26)]
return ans
def solve(s, l, r):
t = s[: len(s) // 2]
len_t = len(t)
if len_t == 1:
return solve_cycle(t, l, r)
len_p = 0
ph = 0
sh = 0
base = 1
for i in range(len_t // 2):
ph *= 26
ph += ord(s[i]) - a
ph %= M
sh += base * (ord(s[-1 - i]) - a)
sh %= M
if ph == sh:
if check_cycle(t, i + 1):
return sovle_cycle(t[: i + 1], l, r)
elif t[: i + 1] == t[-1 - i:]:
len_p = i + 1
base *= 26
if len_p == 0:
return solve_cycle(t, l, r)
return solve_part(t, len_p, l, r)
def main():
s = input()
l, r = input().split()
l = int(l)
r = int(r)
print(' '.join(list(map(str, solve(s, l, r)))))
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 108,514
| 0
| 217,028
|
No
|
output
| 1
| 108,514
| 0
| 217,029
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will call a string that can be obtained by concatenating two equal strings an even string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.
For a non-empty string S, we will define f(S) as the shortest even string that can be obtained by appending one or more characters to the end of S. For example, f(`abaaba`)=`abaababaab`. It can be shown that f(S) is uniquely determined for a non-empty string S.
You are given an even string S consisting of lowercase English letters. For each letter in the lowercase English alphabet, find the number of its occurrences from the l-th character through the r-th character of f^{10^{100}} (S).
Here, f^{10^{100}} (S) is the string f(f(f( ... f(S) ... ))) obtained by applying f to S 10^{100} times.
Constraints
* 2 \leq |S| \leq 2\times 10^5
* 1 \leq l \leq r \leq 10^{18}
* S is an even string consisting of lowercase English letters.
* l and r are integers.
Input
Input is given from Standard Input in the following format:
S
l r
Output
Print 26 integers in a line with spaces in between. The i-th integer should be the number of the occurrences of the i-th letter in the lowercase English alphabet from the l-th character through the r-th character of f^{10^{100}} (S).
Examples
Input
abaaba
6 10
Output
3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Input
xx
1 1000000000000000000
Output
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000000000000000 0 0
Input
vgxgpuamkvgxgvgxgpuamkvgxg
1 1000000000000000000
Output
87167725689669676 0 0 0 0 0 282080685775825810 0 0 0 87167725689669676 0 87167725689669676 0 0 87167725689669676 0 0 0 0 87167725689669676 141040342887912905 0 141040342887912905 0 0
Submitted Solution:
```
def getSS(S):
helfLength = int(len(S) / 2) + 1
while S[:int(len(S[helfLength:]))] != S[helfLength:]:
helfLength += 1
return S[:helfLength] * 2
result = ""
S = input()
SS = S
lr = [int(i) for i in input().split()]
i = 0
while i < 10:
SS = getSS(SS)
i += 1
cList = [chr(i) for i in range(97, 97+26)]
for c in cList:
result += str(SS[lr[0]-1:lr[1]].count(c)) + " "
print(result.rstrip())
```
|
instruction
| 0
| 108,515
| 0
| 217,030
|
No
|
output
| 1
| 108,515
| 0
| 217,031
|
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ M $ type characters. Use them to create a string of length $ N $. How many strings are used that have $ K $ or more? Find too much divided by $ 998244353 $.
Here, the difference between two strings of length $ N $ is defined as follows.
* If the two strings are $ S = S_1S_2 \ ldots S_N $, $ T = T_1T_2 \ ldots T_N $, then $ i $$ (1 \ leq i \ leq N) becomes $ S_i \ neq T_i $ ) $ Exists.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq 10 ^ {18} $
* $ 1 \ leq N \ leq 10 ^ {18} $
* $ 1 \ leq K \ leq 1000 $
* All inputs are integers
Input
The input is given in the following format.
$ M $ $ N $ $ K $
$ M, N, K $ are given on one line, separated by blanks.
Output
Output the remainder of the number of characters used in a string of length $ N $ that is greater than or equal to $ K $ divided by $ 998244353 $.
Examples
Input
2 10 1
Output
1024
Input
1 1 2
Output
0
Input
5 10 3
Output
9755400
Input
7 8 3
Output
5759460
Input
1000000000000000000 1000000000000000000 1000
Output
133611974
|
instruction
| 0
| 108,599
| 0
| 217,198
|
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
res = list(sys.stdin.readline())
if res[-1] == "\n":
return res[:-1]
return res
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 998244353
def solve():
m,n,k = LI()
if n < k:
print(0)
return
if m < k:
print(0)
return
ans = pow(m,n,mod)
p = [pow(i,n,mod) for i in range(k+1)]
c = m
comb = [[0 for i in range(i+1)] for i in range(k+1)]
comb[0][0] = 1
for i in range(k):
for j in range(i+1):
comb[i+1][j] += comb[i][j]
comb[i+1][j+1] += comb[i][j]
for i in range(1,k):
k = 0
for j in range(1,i+1)[::-1]:
if (i+j)&1:
k -= comb[i][j]*p[j]
else:
k += comb[i][j]*p[j]
k *= c
c *= (m-i)
c *= pow(i+1,mod-2,mod)
c %= mod
ans -= k
ans %= mod
print(ans)
return
if __name__ == "__main__":
solve()
```
|
output
| 1
| 108,599
| 0
| 217,199
|
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