message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Submitted Solution:
```
a , b =(map(int, input().split()))
print(max(a, b))
``` | instruction | 0 | 38,736 | 0 | 77,472 |
No | output | 1 | 38,736 | 0 | 77,473 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given the string S, which consists of lowercase letters and numbers. Follow the steps below to compress the length of string S.
1. Change the order of the characters in the character string to any order.
Example: "0ig3he12fz99"-> "efghiz012399"
2. Perform the following operations any number of times.
* Select a contiguous substring of "abcdefghijklmnopqrstuvwxyz" in the string and replace it with (first character)'-' (last character).
Example: "efghiz012399"-> "e-iz012399"
* Select a string with a tolerance of 1 (a continuous substring of "0123456789") in the string and replace it with (first digit)'-' (last digit).
Example: "e-iz012399"-> "e-iz0-399"
Find the minimum length of the string obtained by compressing the string S.
Constraints
* 1 ≤ | S | ≤ 100
* String S contains only lowercase letters and numbers
Input
The string S is given on one line.
Output
Output the minimum value of the length of the character string obtained by compressing the character string S on one line. If it cannot be compressed, output the length of the original string S.
Examples
Input
0ig3he12fz99
Output
9
Input
1122334455
Output
6 | instruction | 0 | 39,559 | 0 | 79,118 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from collections import defaultdict
S = input()
S = sorted(S)
S = list(map(ord, S))
ans = 0
while len(S) > 0:
mn = S[0]
S.remove(mn)
nxt = mn + 1
succ = 1
while True:
if nxt in S:
S.remove(nxt)
succ += 1
nxt += 1
else:
break
ans += min(3, succ)
print(ans)
``` | output | 1 | 39,559 | 0 | 79,119 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given the string S, which consists of lowercase letters and numbers. Follow the steps below to compress the length of string S.
1. Change the order of the characters in the character string to any order.
Example: "0ig3he12fz99"-> "efghiz012399"
2. Perform the following operations any number of times.
* Select a contiguous substring of "abcdefghijklmnopqrstuvwxyz" in the string and replace it with (first character)'-' (last character).
Example: "efghiz012399"-> "e-iz012399"
* Select a string with a tolerance of 1 (a continuous substring of "0123456789") in the string and replace it with (first digit)'-' (last digit).
Example: "e-iz012399"-> "e-iz0-399"
Find the minimum length of the string obtained by compressing the string S.
Constraints
* 1 ≤ | S | ≤ 100
* String S contains only lowercase letters and numbers
Input
The string S is given on one line.
Output
Output the minimum value of the length of the character string obtained by compressing the character string S on one line. If it cannot be compressed, output the length of the original string S.
Examples
Input
0ig3he12fz99
Output
9
Input
1122334455
Output
6 | instruction | 0 | 39,560 | 0 | 79,120 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
#C
s = list(input())
abc = list("-abcdefghijklmnopqrstuvwxyz-")
num = list("-0123456789-")
abc_cnt = [0]*28
num_cnt = [0]*12
for i in range(len(s)):
for j in range(1,27):
if s[i] == abc[j]:
abc_cnt[j] += 1
for i in range(len(s)):
for j in range(1,11):
if s[i] == num[j]:
num_cnt[j] += 1
ans = 0
abc_sum = sum(abc_cnt)
num_sum = sum(num_cnt)
while abc_sum > 0:
start = 100
current = 1
while current < 27:
if abc_cnt[current-1] == 0 and abc_cnt[current] > 0:
start = min(start,current)
if abc_cnt[current+1] == 0 and abc_cnt[current] > 0:
ans += min(3, current - start + 1)
abc_cnt[current] -= 1
abc_sum -= 1
break
if abc_cnt[current] > 0:
abc_cnt[current] -= 1
abc_sum -= 1
current += 1
while num_sum > 0:
start = 100
current = 1
while current < 11:
if num_cnt[current-1] == 0 and num_cnt[current] > 0:
start = min(start,current)
if num_cnt[current+1] == 0 and num_cnt[current] > 0:
ans += min(3, current - start + 1)
num_cnt[current] -= 1
num_sum -= 1
break
if num_cnt[current] > 0:
num_cnt[current] -= 1
num_sum -= 1
current += 1
print(ans)
``` | output | 1 | 39,560 | 0 | 79,121 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given the string S, which consists of lowercase letters and numbers. Follow the steps below to compress the length of string S.
1. Change the order of the characters in the character string to any order.
Example: "0ig3he12fz99"-> "efghiz012399"
2. Perform the following operations any number of times.
* Select a contiguous substring of "abcdefghijklmnopqrstuvwxyz" in the string and replace it with (first character)'-' (last character).
Example: "efghiz012399"-> "e-iz012399"
* Select a string with a tolerance of 1 (a continuous substring of "0123456789") in the string and replace it with (first digit)'-' (last digit).
Example: "e-iz012399"-> "e-iz0-399"
Find the minimum length of the string obtained by compressing the string S.
Constraints
* 1 ≤ | S | ≤ 100
* String S contains only lowercase letters and numbers
Input
The string S is given on one line.
Output
Output the minimum value of the length of the character string obtained by compressing the character string S on one line. If it cannot be compressed, output the length of the original string S.
Examples
Input
0ig3he12fz99
Output
9
Input
1122334455
Output
6 | instruction | 0 | 39,561 | 0 | 79,122 |
"Correct Solution:
```
# AOJ 1564: String Compression
# Python3 2018.7.13 bal4u
f = [0]*128
ch0, ch9n, cha, chzn = ord('0'), ord('9')+1, ord('a'), ord('z')+1
def act(start, end, n):
while True:
for i in range(start, end):
if f[i] == 0: continue
f[i] -= 1; S.append(i)
n -= 1
if n <= 0: return
S = list(input())
w = len(S)
for s in S: f[ord(s)] += 1
n1 = sum(f[ch0:ch9n]); n2 = w-n1
S = []
if n1 > 0: act(ch0, ch9n, n1)
if n2 > 0: act(cha, chzn, n2)
ans, i = w, 0
while i < w:
j = i+1
while j < w and S[j] == S[j-1]+1: j += 1
if j-i > 3: ans -= j-i-3
i = j
print(ans)
``` | output | 1 | 39,561 | 0 | 79,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Given the string S, which consists of lowercase letters and numbers. Follow the steps below to compress the length of string S.
1. Change the order of the characters in the character string to any order.
Example: "0ig3he12fz99"-> "efghiz012399"
2. Perform the following operations any number of times.
* Select a contiguous substring of "abcdefghijklmnopqrstuvwxyz" in the string and replace it with (first character)'-' (last character).
Example: "efghiz012399"-> "e-iz012399"
* Select a string with a tolerance of 1 (a continuous substring of "0123456789") in the string and replace it with (first digit)'-' (last digit).
Example: "e-iz012399"-> "e-iz0-399"
Find the minimum length of the string obtained by compressing the string S.
Constraints
* 1 ≤ | S | ≤ 100
* String S contains only lowercase letters and numbers
Input
The string S is given on one line.
Output
Output the minimum value of the length of the character string obtained by compressing the character string S on one line. If it cannot be compressed, output the length of the original string S.
Examples
Input
0ig3he12fz99
Output
9
Input
1122334455
Output
6
Submitted Solution:
```
s = list(input())
s.sort()
i = 0
while True:
if chr(ord(s[i]) + 1) in s and chr(ord(s[i]) + 2) in s:
s.append("-")
s.remove(chr(ord(s[i]) + 1))
j = 2
while True:
if chr(ord(s[i]) + j + 1) in s:
s.remove(chr(ord(s[i]) + j))
j += 1
else:break
i += 1
if i >= len(s) or s[i] == "-":
break
print(len(s))
``` | instruction | 0 | 39,562 | 0 | 79,124 |
No | output | 1 | 39,562 | 0 | 79,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,848 | 0 | 79,696 |
Tags: constructive algorithms, strings
Correct Solution:
```
t = int(input())
while (t):
s = input()
if len(set(s)) == 1 and s[0] == "a":
print("NO")
else:
for i in range(len(s)):
if (s[len(s) - i - 1] != "a"):
print("YES")
print(s[:i] + "a" + s[i:])
break;
t -= 1
#bbbb
``` | output | 1 | 39,848 | 0 | 79,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,849 | 0 | 79,698 |
Tags: constructive algorithms, strings
Correct Solution:
```
for i in range(int(input())):
a=input()
if('a'+a!=('a'+a)[::-1]):
print('YES')
print('a'+a)
elif a+'a'!=(a+'a')[::-1]:
print('YES')
print(a+'a')
else:
print('NO')
``` | output | 1 | 39,849 | 0 | 79,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,850 | 0 | 79,700 |
Tags: constructive algorithms, strings
Correct Solution:
```
t=int(input())
for z in range(t):
s=input()
if s.count('a')==len(s):
print("NO")
else:
print("YES")
c='a'+s
if c!=c[::-1]:
print(c)
else:
print(s+'a')
``` | output | 1 | 39,850 | 0 | 79,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,851 | 0 | 79,702 |
Tags: constructive algorithms, strings
Correct Solution:
```
for _ in range(int(input())):
s=list(input())
if len(set(s))==1 and s[0]=='a':
print("NO")
continue
print("YES")
n=len(s)
flag=0
for i in range(n//2+1):
if s[i]!='a':
s.insert(n-i,'a')
print("".join(s))
flag=1
break
if flag==0:
s.append('a')
print("".join(s))
``` | output | 1 | 39,851 | 0 | 79,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,852 | 0 | 79,704 |
Tags: constructive algorithms, strings
Correct Solution:
```
import math
import operator
def solve():
#n,m,x=map(int,input().split())
#n=int(input())
#l=list(map(int,input().split()))
s=input()
c=s.count('a')
if(c==len(s)):
print("NO")
else:
print("YES")
t='a'+s
if(t !=t[::-1]):
print(t)
return
t=s+'a'
if(t !=t[::-1]):
print(t)
return
for i in range(len(s)):
s1=s[:i]
s2=s[i+1:]
t=s1+'a'+s2
if(t !=t[::-1]):
print(t)
return
t=int(input())
while(t>0):
t-=1
solve()
``` | output | 1 | 39,852 | 0 | 79,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,853 | 0 | 79,706 |
Tags: constructive algorithms, strings
Correct Solution:
```
for _ in range(int(input())):
s = input()
if (s+'a')!=(s+'a')[::-1]:
print('YES\n',s+'a')
elif ('a'+s )!=('a'+s[::-1]):
print('YES\n','a'+s)
else:
print('NO')
``` | output | 1 | 39,853 | 0 | 79,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,854 | 0 | 79,708 |
Tags: constructive algorithms, strings
Correct Solution:
```
T_ON = 1
DEBUG_ON = 0
MOD = 998244353
def solve():
s = input()
if set(s) == {'a'}:
NO()
return
n = len(s)
i = 0
while i < n and s[i] == 'a':
i += 1
j = n - 1
while j >= 0 and s[j] == 'a':
j -= 1
left = i
right = n - j
YES()
if left >= right:
print('a' + s)
else:
print(s + 'a')
def main():
T = read_int() if T_ON else 1
for i in range(T):
solve()
def debug(*xargs):
if DEBUG_ON:
print(*xargs)
from collections import *
import math
#---------------------------------FAST_IO---------------------------------------
import os
import sys
from io import BytesIO, IOBase
# 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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#----------------------------------IO_WRAP--------------------------------------
def read_int():
return int(input())
def read_ints():
return list(map(int, input().split()))
def print_nums(nums):
print(" ".join(map(str, nums)))
def YES():
print("YES")
def Yes():
print("Yes")
def NO():
print("NO")
def No():
print("No")
def First():
print("First")
def Second():
print("Second")
#----------------------------------FIB--------------------------------------
def fib(n):
"""
the nth fib, start from zero
"""
a, b = 0, 1
for _ in range(n):
a, b = b, a + b
return a
def fib_ns(n):
"""
the first n fibs, start from zero
"""
assert n >= 1
f = [0 for _ in range(n + 1)]
f[0] = 0
f[1] = 1
for i in range(2, n + 1):
f[i] = f[i - 1] + f[i - 2]
return f
def fib_to_n(n):
"""
return fibs <= n, start from zero
n=8
f=[0,1,1,2,3,5,8]
"""
f = []
a, b = 0, 1
while a <= n:
f.append(a)
a, b = b, a + b
return f
#----------------------------------MOD--------------------------------------
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def xgcd(a, b):
"""return (g, x, y) such that a*x + b*y = g = gcd(a, b)"""
x0, x1, y0, y1 = 0, 1, 1, 0
while a != 0:
(q, a), b = divmod(b, a), a
y0, y1 = y1, y0 - q * y1
x0, x1 = x1, x0 - q * x1
return b, x0, y0
def lcm(a, b):
d = gcd(a, b)
return a * b // d
def is_even(x):
return x % 2 == 0
def is_odd(x):
return x % 2 == 1
def modinv(a, m):
"""return x such that (a * x) % m == 1"""
g, x, _ = xgcd(a, m)
if g != 1:
raise Exception('gcd(a, m) != 1')
return x % m
def mod_add(x, y):
x += y
while x >= MOD:
x -= MOD
while x < 0:
x += MOD
return x
def mod_mul(x, y):
return (x * y) % MOD
def mod_pow(x, y):
if y == 0:
return 1
if y % 2:
return mod_mul(x, mod_pow(x, y - 1))
p = mod_pow(x, y // 2)
return mod_mul(p, p)
def mod_inv(y):
return mod_pow(y, MOD - 2)
def mod_div(x, y):
# y^(-1): Fermat little theorem, MOD is a prime
return mod_mul(x, mod_inv(y))
#---------------------------------PRIME---------------------------------------
def is_prime(n):
if n == 1:
return False
for i in range(2, int(n ** 0.5) + 1):
debug(n, i)
if n % i == 0:
return False
return True
def gen_primes(n):
"""
generate primes of [1..n] using sieve's method
"""
P = [True for _ in range(n + 1)]
P[0] = P[1] = False
for i in range(int(n ** 0.5) + 1):
if P[i]:
for j in range(2 * i, n + 1, i):
P[j] = False
return P
#---------------------------------MISC---------------------------------------
def is_lucky(n):
return set(list(str(n))).issubset({'4', '7'})
#---------------------------------MAIN---------------------------------------
main()
``` | output | 1 | 39,854 | 0 | 79,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A palindrome is a string that reads the same backward as forward. For example, the strings "z", "aaa", "aba", and "abccba" are palindromes, but "codeforces" and "ab" are not. You hate palindromes because they give you déjà vu.
There is a string s. You must insert exactly one character 'a' somewhere in s. If it is possible to create a string that is not a palindrome, you should find one example. Otherwise, you should report that it is impossible.
For example, suppose s= "cbabc". By inserting an 'a', you can create "acbabc", "cababc", "cbaabc", "cbabac", or "cbabca". However "cbaabc" is a palindrome, so you must output one of the other options.
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The only line of each test case contains a string s consisting of lowercase English letters.
The total length of all strings does not exceed 3⋅ 10^5.
Output
For each test case, if there is no solution, output "NO".
Otherwise, output "YES" followed by your constructed string of length |s|+1 on the next line. If there are multiple solutions, you may print any.
You can print each letter of "YES" and "NO" in any case (upper or lower).
Example
Input
6
cbabc
ab
zza
ba
a
nutforajaroftuna
Output
YES
cbabac
YES
aab
YES
zaza
YES
baa
NO
YES
nutforajarofatuna
Note
The first test case is described in the statement.
In the second test case, we can make either "aab" or "aba". But "aba" is a palindrome, so "aab" is the only correct answer.
In the third test case, "zaza" and "zzaa" are correct answers, but not "azza".
In the fourth test case, "baa" is the only correct answer.
In the fifth test case, we can only make "aa", which is a palindrome. So the answer is "NO".
In the sixth test case, "anutforajaroftuna" is a palindrome, but inserting 'a' elsewhere is valid. | instruction | 0 | 39,855 | 0 | 79,710 |
Tags: constructive algorithms, strings
Correct Solution:
```
# from sys import stdin , stdout
for i in range(int(input())):
s = input()
# if s[0] != "a":
# print("YES")
# print(s+"a")
# elif s[-1] != "a":
# print("YES")
# print("a"+s)
# else:
# i,j=1,len(s)-2
# while(i<=j)
if s.count("a") == len(s):
print("NO")
else:
s1 = "a"+s
print("YES")
if s1==s1[::-1]:
print(s+"a")
else:
print(s1)
``` | output | 1 | 39,855 | 0 | 79,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,868 | 0 | 79,736 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
# cook your dish here
from sys import stdin
from sys import stdout
a=stdin.readline()
n=len(a)-1
a=list(a)[:n]
dp=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and a[i]==a[j]):
dp[i][j]=1
else:
temp=dp[i+1][j-1]*(a[i]==a[j])
dp[i][j]=temp
dpr=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dpr[i][i]=1
s=a[::-1]
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and s[i]==s[j]):
dpr[i][j]=1
else:
temp=dpr[i+1][j-1]*(s[i]==s[j])
dpr[i][j]=temp
s=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dp[i][j]
s.append(tempy)
r=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dpr[i][j]
r.append(tempy)
# print(s)
# print(r)
# s=s[::-1]
tot=sum(s)
right=[]
for i in range(n):
right.append(tot)
tot=tot-r[i]
s=s[::-1]
# print(s)
# print(right)
ans=0
for i in range(n):
j=i+1
if(j<n):
ans=ans+s[i]*right[j]
stdout.write(str(ans))
``` | output | 1 | 39,868 | 0 | 79,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,869 | 0 | 79,738 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
from sys import stdin
from sys import stdout
a=stdin.readline()
n=len(a)-1
a=list(a)[:n]
dp=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and a[i]==a[j]):
dp[i][j]=1
else:
temp=dp[i+1][j-1]*(a[i]==a[j])
dp[i][j]=temp
dpr=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dpr[i][i]=1
s=a[::-1]
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and s[i]==s[j]):
dpr[i][j]=1
else:
temp=dpr[i+1][j-1]*(s[i]==s[j])
dpr[i][j]=temp
s=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dp[i][j]
s.append(tempy)
r=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dpr[i][j]
r.append(tempy)
# print(s)
# print(r)
# s=s[::-1]
tot=sum(s)
right=[]
for i in range(n):
right.append(tot)
tot=tot-r[i]
s=s[::-1]
# print(s)
# print(right)
ans=0
for i in range(n):
j=i+1
if(j<n):
ans=ans+s[i]*right[j]
print(ans)
``` | output | 1 | 39,869 | 0 | 79,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,870 | 0 | 79,740 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
from collections import defaultdict
S = input()
N = len(S)
# Step 1: Find the positions of a-z
Pos = defaultdict(list)
for i in range(N):
Pos[S[i]].append(i)
# Step 2: Find Ends here dp
# Minor Change: Add a bool dict to catch the palindrome
Range = defaultdict(bool)
dpE = [1] * N
for i in range(N):
Range[(i,i)] = True
for pos in Pos[S[i]]:
if(pos>=i):
break
else:
if(i - pos == 1):
if(S[i] == S[pos]):
Range[(pos,i)] = True
dpE[i] += 1
elif(Range[(pos+1,i-1)]):
Range[(pos,i)] = True
dpE[i] += 1
# Step 3: Find Till Here dp
dpT = [0] * N
dpT[-1] = 1
for i in range(N-2,-1,-1):
dpT[i] += dpT[i+1] + 1
for j in range(N-1,i,-1):
if(Range[(i,j)]):
dpT[i] += 1
# Step 4: It's Time
Ans = 0
for i in range(N-1):
Ans += dpE[i] * dpT[i+1]
print(Ans)
``` | output | 1 | 39,870 | 0 | 79,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,871 | 0 | 79,742 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
t = input()
n = len(t)
a, b = [1] * n, [1] * n
for x, y in [(i - 1, i + 1) for i in range(n // 2)]:
while x >= 0 and t[x] == t[y]:
a[y] += 1
b[x] += 1
x -= 1
y += 1
for x, y in [(i, i + 1) for i in range(n // 2)]:
while x >= 0 and t[x] == t[y]:
a[y] += 1
b[x] += 1
x -= 1
y += 1
for x, y in [(i - 1, i + 1) for i in range(n // 2, n)]:
while y < n and t[x] == t[y]:
a[y] += 1
b[x] += 1
x -= 1
y += 1
for x, y in [(i, i + 1) for i in range(n // 2, n)]:
while y < n and t[x] == t[y]:
a[y] += 1
b[x] += 1
x -= 1
y += 1
for i in range(1, n):
a[i] += a[i - 1]
print(sum(a[i - 1] * b[i] for i in range(1, n)))
``` | output | 1 | 39,871 | 0 | 79,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,872 | 0 | 79,744 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
import sys
input=sys.stdin.readline
s=list(input().rstrip())
n=len(s)
l=[0]*n # [0,i]
r=[0]*n # [i,n-1]
for i in range(n):
j=0
while 0<=i-j and i+j<n and s[i-j]==s[i+j]:
l[i+j]+=1
r[i-j]+=1
j+=1
j=1
while 0<=i-j+1 and i+j<n and s[i-j+1]==s[i+j]:
l[i+j]+=1
r[i-j+1]+=1
j+=1
ans=0
for i in range(n):
for j in range(i+1,n):
ans+=l[i]*r[j]
print(ans)
``` | output | 1 | 39,872 | 0 | 79,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,873 | 0 | 79,746 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
t = input()
n = len(t)
if n == 2000 and len(set(t)) == 1: print(667333166500)
else:
a, b = [0] + [1] * n, [1] * n
for i in range(n // 2):
x, y = i - 1, i + 1
while x >= 0 and t[x] == t[y]:
a[y + 1] += 1
b[x] += 1
x -= 1
y += 1
x, y = i, i + 1
while x >= 0 and t[x] == t[y]:
a[y + 1] += 1
b[x] += 1
x -= 1
y += 1
for i in range(n // 2, n):
x, y = i - 1, i + 1
while y < n and t[x] == t[y]:
a[y + 1] += 1
b[x] += 1
x -= 1
y += 1
x, y = i, i + 1
while y < n and t[x] == t[y]:
a[y + 1] += 1
b[x] += 1
x -= 1
y += 1
for i in range(1, n):
a[i + 1] += a[i]
print(sum(a[i] * b[i] for i in range(n)))
``` | output | 1 | 39,873 | 0 | 79,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,874 | 0 | 79,748 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
from sys import stdin
from sys import stdout
a=input()
n=len(a)
a=list(a)[:n]
dp=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and a[i]==a[j]):
dp[i][j]=1
else:
temp=dp[i+1][j-1]*(a[i]==a[j])
dp[i][j]=temp
dpr=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dpr[i][i]=1
s=a[::-1]
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and s[i]==s[j]):
dpr[i][j]=1
else:
temp=dpr[i+1][j-1]*(s[i]==s[j])
dpr[i][j]=temp
s=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dp[i][j]
s.append(tempy)
r=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dpr[i][j]
r.append(tempy)
# print(s)
# print(r)
# s=s[::-1]
tot=sum(s)
right=[]
for i in range(n):
right.append(tot)
tot=tot-r[i]
s=s[::-1]
# print(s)
# print(right)
ans=0
for i in range(n):
j=i+1
if(j<n):
ans=ans+s[i]*right[j]
stdout.write(str(ans))
``` | output | 1 | 39,874 | 0 | 79,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36 | instruction | 0 | 39,875 | 0 | 79,750 |
Tags: *special, brute force, dp, strings
Correct Solution:
```
vv=lambda s:ord(s)-ord('a')+1
zz=input()
s=list(map(vv,zz))
prim=31
mod=10**9 +7
ln=len(s)
prefix=[0]*(ln)
pwrs=[0]*(ln+1)
pwrs[0]=1
for n in range(1,ln+1):pwrs[n]=(pwrs[n-1]*prim)%mod
prefix[0]=s[0]
for n in range(1,ln):prefix[n]=(prefix[n-1] + s[n]* pwrs[n])%mod
def bnp(bs,pw):
res=1
while pw:
if pw&1:
res*=bs
res%=mod
bs*=bs
bs%=mod
pw//=2
return res
suffix=[0]*ln
suffix[-1]=s[-1]
bnps=[bnp(n,mod-2) for n in pwrs]
for n in range(ln-2,-1,-1):suffix[n]=(suffix[n+1]+s[n]*pwrs[ln-1-n])%mod
prefix=[0]+prefix+[0]
suffix=[0]+suffix+[0]
BIT=[0]*(10**4)
lnt=len(BIT)
def up(id,val):
while id<lnt:
BIT[id]+=val
id+=id&(-id)
def qr(id):
res=0
while id:
res+=BIT[id]
id-=id&(-id)
return res
res=0
for a in range(1,ln+1):
for b in range(a,ln+1):
p=((prefix[b]-prefix[a-1])*bnps[a-1]%mod)
q=((suffix[a]-suffix[b+1])*bnps[ln-b] %mod)
if p==q:
res+=qr(a-1)
up(b,1)
print(res)
``` | output | 1 | 39,875 | 0 | 79,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
def ispalindrome(s):
if s==s[::-1]:
return True
return False
s = input()
n = len(s)
dp = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
dp[i][i] = 1
for i in range(1,n):
for j in range(i-1,-1,-1):
if s[i]==s[j] and (j+1>i-1 or dp[j+1][i-1]):
dp[j][i] = 1
pref = [0]*n
suff = [0]*n
pref[0] = suff[-1] = 1
for i in range(1,n):
# pref[i] = pref[i-1]
for j in range(i,-1,-1):
if dp[j][i]:
pref[i] += 1
for i in range(n-2,-1,-1):
suff[i] = suff[i+1]
for j in range(i,n):
if dp[i][j]:
suff[i] += 1
# for i in dp:
# print (*i)
# print (pref)
# print (suff)
ans = 0
curr = 0
for i in range(n-1):
ans += suff[i+1]*pref[i]
print (ans)
``` | instruction | 0 | 39,876 | 0 | 79,752 |
Yes | output | 1 | 39,876 | 0 | 79,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
"""
Author - Satwik Tiwari .
4th Oct , 2020 - Sunday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil
from copy import deepcopy
from collections import deque
# from collections import Counter as counter # Counter(list) return a dict with {key: count}
# from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
# from itertools import permutations as permutate
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
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)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
mod = 10**9+7
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def solve(case):
s = list(inp())
n = len(s)
dp = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
dp[i][i] = True
for i in range(n-1,-1,-1):
dp[i][i] = True
for j in range(i+1,n):
if(s[i] == s[j] and (i+1>j-1 or dp[i+1][j-1])):
dp[i][j] = True
pre = [0]*(n+1)
suff =[0]*(n+1)
for i in range(n-1,-1,-1):
suff[i] = suff[i+1]
for j in range(i,n):
if(dp[i][j]):
suff[i]+=1
for i in range(n):
for j in range(i,-1,-1):
if(dp[j][i]):
pre[i]+=1
ans = 0
for i in range(n):
ans+=pre[i]*suff[i+1]
print(ans)
testcase(1)
# testcase(int(inp()))
``` | instruction | 0 | 39,877 | 0 | 79,754 |
Yes | output | 1 | 39,877 | 0 | 79,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
def main():
s = tuple(input())
n = len(s)
a, b = [1] * n, [1] * n
for i in range(n):
for x, y in zip(range(i, -1, -1), range(i + 2, n)):
if s[x] != s[y]:
break
a[y] += 1
b[x] += 1
for x, y in zip(range(i, -1, -1), range(i + 1, n)):
if s[x] != s[y]:
break
a[y] += 1
b[x] += 1
x = 0
for i, y in enumerate(a):
x += y
a[i] = x
print(sum(x * y for x, y in zip(a, b[1:])))
if __name__ == '__main__':
main()
``` | instruction | 0 | 39,878 | 0 | 79,756 |
Yes | output | 1 | 39,878 | 0 | 79,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
s = input()
# s = "aiajadabaa"
n = len(s)
L = [0] * n # L[i] = số lượng xâu đối xứng trong đoạn s[0..i]
R = [0] * n # R[i] = số lượng xâu đối xứng trong đoạn s[i+1..n-1]
for i in range(n):
x = y = i
while x >= 0 and y < n and s[x] == s[y]:
R[x] += 1
L[y] += 1
x -= 1
y += 1
for i in range(1, n):
if s[i - 1] == s[i]:
x, y = i-1, i
while x >= 0 and y < n and s[x] == s[y]:
R[x] += 1
L[y] += 1
x -= 1
y += 1
for i in range(n - 1):
L[i + 1] += L[i]
p = 0
for i in range(1, n):
p += L[i-1] * R[i]
print(p)
``` | instruction | 0 | 39,879 | 0 | 79,758 |
Yes | output | 1 | 39,879 | 0 | 79,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
# cook your dish here
from sys import stdin
from sys import stdout
a=stdin.readline()
n=len(a)
dp=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and a[i]==a[j]):
dp[i][j]=1
else:
temp=dp[i+1][j-1]*(a[i]==a[j])
dp[i][j]=temp
dpr=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dpr[i][i]=1
s=a[::-1]
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and s[i]==s[j]):
dpr[i][j]=1
else:
temp=dpr[i+1][j-1]*(s[i]==s[j])
dpr[i][j]=temp
s=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dp[i][j]
s.append(tempy)
r=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dpr[i][j]
r.append(tempy)
# print(s)
# print(r)
# s=s[::-1]
tot=sum(s)
right=[]
for i in range(n):
right.append(tot)
tot=tot-r[i]
s=s[::-1]
# print(s)
# print(right)
ans=0
for i in range(n):
j=i+1
if(j<n):
ans=ans+s[i]*right[j]
stdout.write(str(a))
``` | instruction | 0 | 39,880 | 0 | 79,760 |
No | output | 1 | 39,880 | 0 | 79,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
# cook your dish here
from sys import stdin
from sys import stdout
a=list(stdin.readline())
n=len(a)
dp=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and a[i]==a[j]):
dp[i][j]=1
else:
temp=dp[i+1][j-1]*(a[i]==a[j])
dp[i][j]=temp
dpr=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dpr[i][i]=1
s=a[::-1]
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and s[i]==s[j]):
dpr[i][j]=1
else:
temp=dpr[i+1][j-1]*(s[i]==s[j])
dpr[i][j]=temp
s=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dp[i][j]
s.append(tempy)
r=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dpr[i][j]
r.append(tempy)
# print(s)
# print(r)
# s=s[::-1]
tot=sum(s)
right=[]
for i in range(n):
right.append(tot)
tot=tot-r[i]
s=s[::-1]
# print(s)
# print(right)
ans=0
for i in range(n):
j=i+1
if(j<n):
ans=ans+s[i]*right[j]
stdout.write(str(ans))
``` | instruction | 0 | 39,881 | 0 | 79,762 |
No | output | 1 | 39,881 | 0 | 79,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
from sys import stdin
from sys import stdout
a=list(stdin.readline())
n=len(a)
dp=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and a[i]==a[j]):
dp[i][j]=1
else:
temp=dp[i+1][j-1]*(a[i]==a[j])
dp[i][j]=temp
dpr=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dpr[i][i]=1
s=a[::-1]
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and s[i]==s[j]):
dpr[i][j]=1
else:
temp=dpr[i+1][j-1]*(s[i]==s[j])
dpr[i][j]=temp
s=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dp[i][j]
s.append(tempy)
r=[]
for i in range(n):
tempy=0
for j in range(i,n):
tempy=tempy+dpr[i][j]
r.append(tempy)
# print(s)
# print(r)
# s=s[::-1]
tot=sum(s)
right=[]
for i in range(n):
right.append(tot)
tot=tot-r[i]
s=s[::-1]
# print(s)
# print(right)
ans=0
for i in range(n):
j=i+1
if(j<n):
ans=ans+s[i]*right[j]
stdout.write(str(ans))
``` | instruction | 0 | 39,882 | 0 | 79,764 |
No | output | 1 | 39,882 | 0 | 79,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".
Input
The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.
Output
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Examples
Input
aa
Output
1
Input
aaa
Output
5
Input
abacaba
Output
36
Submitted Solution:
```
from sys import stdin
from sys import stdout
a=stdin.readline()
n=len(a)
dp=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dp[i][i]=1
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and a[i]==a[j]):
dp[i][j]=1
else:
temp=dp[i+1][j-1]*(a[i]==a[j])
dp[i][j]=temp
dpr=[[0 for j in range(n)] for i in range(n)]
for i in range(n):
dpr[i][i]=1
s=a[::-1]
for l in range(2,n+1):
for i in range(n-l+1):
j=i+l-1
# print(i,j)
if(j-i==1 and s[i]==s[j]):
dpr[i][j]=1
else:
temp=dpr[i+1][j-1]*(s[i]==s[j])
dpr[i][j]=temp
s=[]
for i in range(n):
s.append(sum(dp[:][i]))
r=[]
for i in range(n):
r.append(sum(dpr[:][i]))
# print(s)
# print(r)
# s=s[::-1]
tot=sum(s)
right=[]
for i in range(n):
right.append(tot)
tot=tot-r[i]
s=s[::-1]
ans=0
for i in range(n):
j=i+1
if(j<n):
ans=ans+s[i]*right[j]
stdout.write(str(ans))
``` | instruction | 0 | 39,883 | 0 | 79,766 |
No | output | 1 | 39,883 | 0 | 79,767 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,263 | 0 | 80,526 |
"Correct Solution:
```
s = list(input())
n = len(s)
a = [0] * n
MOD = 998244353
for i in range(n):
a[i] = ord(s[i]) - ord('a')
if max(a) == min(a):
print(1)
exit()
elif n == 3:
def solver(m):
x = m // 100
y = ( m % 100 ) // 10
z = m % 10
if x != y:
c = (3 - x - y) % 3
temp = c * 110 + z
if temp not in ans:
ans.add(temp)
solver(temp)
if y != z:
c = (3 - y - z) % 3
temp = x * 100 + c * 11
if temp not in ans:
ans.add(temp)
solver(temp)
t = a[0] * 100 + a[1] * 10 + a[2]
ans = set()
ans.add(t)
solver(t)
print(len(ans))
exit()
elif n == 2:
print(2)
exit()
dp = [[0,0,0] for _ in range(3)]
dp[0][0] = 1
dp[1][1] = 1
dp[2][2] = 1
for i in range(n-1):
T = [[0,0,0] for _ in range(3)]
T[0][0] = (dp[1][0] + dp[2][0]) % MOD
T[0][1] = (dp[1][1] + dp[2][1]) % MOD
T[0][2] = (dp[1][2] + dp[2][2]) % MOD
T[1][0] = (dp[0][2] + dp[2][2]) % MOD
T[1][1] = (dp[0][0] + dp[2][0]) % MOD
T[1][2] = (dp[0][1] + dp[2][1]) % MOD
T[2][0] = (dp[0][1] + dp[1][1]) % MOD
T[2][1] = (dp[0][2] + dp[1][2]) % MOD
T[2][2] = (dp[0][0] + dp[1][0]) % MOD
dp, T = T, dp
m = sum(a) % 3
ans = 3 ** (n-1) - (dp[0][m] + dp[1][m] + dp[2][m])
check = 1
for i in range(n-1):
if a[i] == a[i+1]:
check = 0
break
ans += check
#print(dp, 2 ** (n-1))
print(ans % MOD)
``` | output | 1 | 40,263 | 0 | 80,527 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,264 | 0 | 80,528 |
"Correct Solution:
```
import itertools
S = input()
N = len(S)
if all(S[0] == c for c in S):
print(1)
exit()
if N == 2:
print(1 if S[0]==S[1] else 2)
exit()
if N == 3:
def another(a,b):
s = set('abc')
s -= set([a,b])
return list(s)[0]
ptn = set()
stack = [S]
while stack:
a = stack.pop()
ptn.add(a)
if a[0] != a[1]:
b = another(a[0],a[1])*2 + a[2]
if not b in ptn:
stack.append(b)
if a[1] != a[2]:
c = a[0] + another(a[1],a[2])*2
if not c in ptn:
stack.append(c)
print(len(ptn))
exit()
#N >= 4
MOD = 998244353
dp = [[[0 for u in range(2)] for l in range(3)] for s in range(3)]
#dp[sum%3][last][exist'xx'?]
for ptn in itertools.product(range(3),repeat=4):
seq = (ptn[0]==ptn[1] or ptn[1]==ptn[2] or ptn[2]==ptn[3])
dp[sum(ptn)%3][ptn[3]][seq] += 1
for n in range(4,N):
dp2 = [[[0 for u in range(2)] for l in range(3)] for s in range(3)]
for s in range(3):
for l in range(3):
for u in range(2):
for l2 in range(3):
s2 = (s+l2)%3
u2 = u or l==l2
dp2[s2][l2][u2] += dp[s][l][u]
dp2[s2][l2][u2] %= MOD
dp = dp2
sm = 0
for c in S:
sm += ord(c) - ord('a')
seq = False
for c1,c2 in zip(S,S[1:]):
if c1 == c2:
seq = True
break
ans = sum([dp[sm%3][i][1] for i in range(3)])
if not seq: ans += 1
print(ans % MOD)
``` | output | 1 | 40,264 | 0 | 80,529 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,265 | 0 | 80,530 |
"Correct Solution:
```
def solve(s):
if all(a == b for a, b in zip(s, s[1:])):
return 1
if len(s) == 2:
return 2
elif len(s) == 3:
if s[0] == s[1] or s[1] == s[2]:
return 6
elif s[0] == s[2]:
return 7
else:
return 3
# dp[has succession=0][mod 3][last char], dp[has succession=1][mod 3]
dp = [[[0] * 3 for _ in range(3)], [0] * 3]
dp[0][0][0] = 1
dp[0][1][1] = 1
dp[0][2][2] = 1
MOD = 998244353
for _ in range(len(s) - 1):
ndp = [[[0] * 3 for _ in range(3)], [0] * 3]
dp0, dp1 = dp
ndp0, ndp1 = ndp
sdp1 = sum(dp1)
ndp0[0][0] = (dp0[0][1] + dp0[0][2]) % MOD
ndp0[1][0] = (dp0[1][1] + dp0[1][2]) % MOD
ndp0[2][0] = (dp0[2][1] + dp0[2][2]) % MOD
ndp0[0][1] = (dp0[2][0] + dp0[2][2]) % MOD
ndp0[1][1] = (dp0[0][0] + dp0[0][2]) % MOD
ndp0[2][1] = (dp0[1][0] + dp0[1][2]) % MOD
ndp0[0][2] = (dp0[1][0] + dp0[1][1]) % MOD
ndp0[1][2] = (dp0[2][0] + dp0[2][1]) % MOD
ndp0[2][2] = (dp0[0][0] + dp0[0][1]) % MOD
ndp1[0] = (dp0[0][0] + dp0[1][2] + dp0[2][1] + sdp1) % MOD
ndp1[1] = (dp0[1][0] + dp0[2][2] + dp0[0][1] + sdp1) % MOD
ndp1[2] = (dp0[2][0] + dp0[0][2] + dp0[1][1] + sdp1) % MOD
dp = ndp
return (dp[1][sum(map(ord, s)) % 3] + all(a != b for a, b in zip(s, s[1:]))) % MOD
s = input()
print(solve(s))
``` | output | 1 | 40,265 | 0 | 80,531 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,266 | 0 | 80,532 |
"Correct Solution:
```
#!/usr/bin/env python3
M = 998244353
def powmod(a, x, m = M):
y = 1
while 0 < x:
if x % 2 == 1:
y *= a
y %= m
x //= 2
a = a ** 2
a %= m
return y
def solve(s):
n = len(s)
nb = nc = 0
ch = s[0]
if ch == 'b':
nb += 1
elif ch == 'c':
nc += 1
sf = True
tf = True
left = ch
for ch in s[1:]:
if ch == 'b':
nb += 1
elif ch == 'c':
nc += 1
if ch == left:
sf = False
else:
tf = False
left = ch
if tf:
return 1
if n == 3:
if (nb + nc * 2) % 3:
return 7 if sf else 6
else:
return 3
if n % 3:
return (powmod(3, n - 1) + M - powmod(2, n - 1) + (1 if sf else 0)) % M
else:
if (nb + nc * 2) % 3:
return (powmod(3, n - 1) + M - (powmod(2, n - 1) - powmod(2, n // 3 - 1)) + (1 if sf else 0)) % M
else:
return (powmod(3, n - 1) + M - (powmod(2, n // 3) + 4 * powmod(8, n // 3 - 1)) + (1 if sf else 0)) % M
def main():
s = input()
print(solve(s))
if __name__ == '__main__':
main()
``` | output | 1 | 40,266 | 0 | 80,533 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,267 | 0 | 80,534 |
"Correct Solution:
```
def solve(s):
if all(a == b for a, b in zip(s, s[1:])):
return 1
if len(s) == 2:
return 2
elif len(s) == 3:
if s[0] == s[1] or s[1] == s[2]:
return 6
elif s[0] == s[2]:
return 7
else:
return 3
# dp[has succession][mod 3][last char]
dp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
dp[0][0][0] = 1
dp[0][1][1] = 1
dp[0][2][2] = 1
MOD = 998244353
for _ in range(len(s) - 1):
ndp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
dp0, dp1 = dp
ndp0, ndp1 = ndp
sdp10, sdp11, sdp12 = sum(dp1[0]), sum(dp1[1]), sum(dp1[2])
ndp0[0][0] = (dp0[0][1] + dp0[0][2]) % MOD
ndp0[1][0] = (dp0[1][1] + dp0[1][2]) % MOD
ndp0[2][0] = (dp0[2][1] + dp0[2][2]) % MOD
ndp0[0][1] = (dp0[2][0] + dp0[2][2]) % MOD
ndp0[1][1] = (dp0[0][0] + dp0[0][2]) % MOD
ndp0[2][1] = (dp0[1][0] + dp0[1][2]) % MOD
ndp0[0][2] = (dp0[1][0] + dp0[1][1]) % MOD
ndp0[1][2] = (dp0[2][0] + dp0[2][1]) % MOD
ndp0[2][2] = (dp0[0][0] + dp0[0][1]) % MOD
ndp1[0][0] = (dp0[0][0] + sdp10) % MOD
ndp1[1][0] = (dp0[1][0] + sdp11) % MOD
ndp1[2][0] = (dp0[2][0] + sdp12) % MOD
ndp1[0][1] = (dp0[2][1] + sdp12) % MOD
ndp1[1][1] = (dp0[0][1] + sdp10) % MOD
ndp1[2][1] = (dp0[1][1] + sdp11) % MOD
ndp1[0][2] = (dp0[1][2] + sdp11) % MOD
ndp1[1][2] = (dp0[2][2] + sdp12) % MOD
ndp1[2][2] = (dp0[0][2] + sdp10) % MOD
dp = ndp
return (sum(dp[1][sum(map(ord, s)) % 3]) + all(a != b for a, b in zip(s, s[1:]))) % MOD
s = input()
print(solve(s))
``` | output | 1 | 40,267 | 0 | 80,535 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,268 | 0 | 80,536 |
"Correct Solution:
```
def main():
md = 998244353
s = input()
n = len(s)
al = True # allすべて同じ文字
an = 1 # any連続が存在すると0
for c0, c1 in zip(s, s[1:]):
if c0 == c1:
an = 0
else:
al = False
if an == 0 and not al: break
if al:
print(1)
exit()
if n == 2:
print(2)
exit()
if n == 3:
if s[0] == s[-1]:
print(7)
elif s[0] == s[1] or s[1] == s[2]:
print(6)
else:
print(3)
exit()
#print(n)
#print(an)
ord0=ord("a")
r = sum(ord(c) - ord0 for c in s) % 3
if n%3==0:
d=pow(2,n//3-1,md)
if r==0:
an-=d*2
else:
an+=d
print((pow(3, n - 1, md) - pow(2, n - 1, md) + an) % md)
main()
``` | output | 1 | 40,268 | 0 | 80,537 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,269 | 0 | 80,538 |
"Correct Solution:
```
def solve(s):
if len(s) == 2:
return 1 if s[0] == s[1] else 2
elif len(s) == 3:
if s[0] == s[1] == s[2]:
return 1
elif s[0] == s[1] or s[1] == s[2]:
return 6
elif s[0] == s[2]:
return 7
else:
return 3
if all(a == b for a, b in zip(s, s[1:])):
return 1
dp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
dp[0][0][0] = 1
dp[0][1][1] = 1
dp[0][2][2] = 1
MOD = 998244353
for _ in range(len(s) - 1):
ndp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
ndp[0][0][0] = (dp[0][1][0] + dp[0][2][0]) % MOD
ndp[0][0][1] = (dp[0][1][1] + dp[0][2][1]) % MOD
ndp[0][0][2] = (dp[0][1][2] + dp[0][2][2]) % MOD
ndp[0][1][0] = (dp[0][0][2] + dp[0][2][2]) % MOD
ndp[0][1][1] = (dp[0][0][0] + dp[0][2][0]) % MOD
ndp[0][1][2] = (dp[0][0][1] + dp[0][2][1]) % MOD
ndp[0][2][0] = (dp[0][0][1] + dp[0][1][1]) % MOD
ndp[0][2][1] = (dp[0][0][2] + dp[0][1][2]) % MOD
ndp[0][2][2] = (dp[0][0][0] + dp[0][1][0]) % MOD
ndp[1][0][0] = (dp[0][0][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
ndp[1][0][1] = (dp[0][0][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][0][2] = (dp[0][0][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][1][0] = (dp[0][1][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][1][1] = (dp[0][1][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
ndp[1][1][2] = (dp[0][1][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][2][0] = (dp[0][2][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][2][1] = (dp[0][2][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][2][2] = (dp[0][2][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
dp = ndp
return (sum(dp[1][x][sum(map(ord, s)) % 3] for x in range(3)) + all(a != b for a, b in zip(s, s[1:]))) % MOD
s = input()
print(solve(s))
``` | output | 1 | 40,269 | 0 | 80,539 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497 | instruction | 0 | 40,270 | 0 | 80,540 |
"Correct Solution:
```
from collections import defaultdict
import sys
S=input()
N=len(S)
if N==3:
if S[0]==S[1]:
if S[1]==S[2]:
print(1)
else:
print(6)
else:
if S[0]==S[2]:
print(7)
elif S[1]==S[2]:
print(6)
else:
print(3)
sys.exit()
a=S[0]
t=1
for i in range(N):
if S[i]!=a:
t=0
break
if t==1:
print(1)
sys.exit()
mod=998244353
M=0
for s in S:
if s=='b':
M+=1
elif s=='c':
M+=2
M%=3
dp=[]
for i in range(2):
for k in range(3):
for l in range(3):
dp.append((i,k,l))
dd=[[[0]*3 for i in range(3)] for i in range(2)]
for i in range(3):
dd[0][i][i]=1
for k in range(2,N+1):
nd=[[[0]*3 for i in range(3)] for i in range(2)]
for p in dp:
con,moji,m=p
n=dd[con][moji][m]
#print(con,moji,m)
if con==0:
for k in range(3):
if moji==k:
nd[1][k][ (m + k) % 3] += n
nd[1] [k ][(m + k) % 3] %=mod
else:
nd[0][k][(m+k)%3]+=n
nd[0][ k] [(m+k) % 3] %=mod
else:
for k in range(3):
nd[1][k][(m + k) % 3] += n
nd[1][k][(m + k) % 3] %=mod
#print(nd)
dd=nd
flag=1
for i in range(N-1):
if S[i]==S[i+1]:
flag=0
ans=flag
for k in range(3):
ans+=dd[1][k][M]
ans%=mod
print(ans)
``` | output | 1 | 40,270 | 0 | 80,541 |
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 `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497
Submitted Solution:
```
s = input()
n = len(s)
flg = 0
for i in range(1,n):
if s[i-1] != s[i]:
flg += 1
if flg == 0:
print(1)
exit()
par = (s.count("a")+s.count("b")*2+2)%3
dp = [[0 for i in range(9)] for j in range(n)]
mod = 998244353
dp[0][0] = 1
dp[0][4] = 1
dp[0][8] = 1
for i in range(1,n):
dp[i][0] = dp[i-1][5]+dp[i-1][8]
dp[i][1] = dp[i-1][3]+dp[i-1][6]
dp[i][2] = dp[i-1][4]+dp[i-1][7]
dp[i][3] = dp[i-1][1]+dp[i-1][7]
dp[i][4] = dp[i-1][2]+dp[i-1][8]
dp[i][5] = dp[i-1][0]+dp[i-1][6]
dp[i][6] = dp[i-1][0]+dp[i-1][3]
dp[i][7] = dp[i-1][1]+dp[i-1][4]
dp[i][8] = dp[i-1][2]+dp[i-1][5]
for j in range(9):
dp[i][j] %= mod
ans = pow(3,n-1,mod)-dp[-1][par]-dp[-1][par+3]-dp[-1][par+6]
if flg == n-1:
ans += 1
if n == 3 and flg == n-1 and par == 2:
ans -= 1
print(ans%mod)
``` | instruction | 0 | 40,271 | 0 | 80,542 |
Yes | output | 1 | 40,271 | 0 | 80,543 |
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 `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497
Submitted Solution:
```
def main():
md = 998244353
s = input()
n = len(s)
if s.count(s[0]) == n:
print(1)
exit()
if n == 2:
print(2)
exit()
if n == 3:
if s[0] == s[-1]:
print(7)
elif s[0] == s[1] or s[1] == s[2]:
print(6)
else:
print(3)
exit()
ord0 = ord("a")
a = [ord(c) - ord0 for c in s]
r = sum(a) % 3
dp = [[[[0] * 3 for _ in range(2)] for _ in range(3)] for _ in range(n)]
for m in range(3):
dp[0][m][0][m] = 1
# print(a)
# print(dp)
for i in range(n - 1):
for j in range(3):
for k in range(2):
for m in range(3):
pre = dp[i][j][k][m] # iけた j余り k連続有無 m前の位
for nm in range(3):
nj = (j + nm) % 3
if nm == m:
dp[i + 1][nj][1][nm] = (dp[i + 1][nj][1][nm] + pre) % md
else:
dp[i + 1][nj][k][nm] = (dp[i + 1][nj][k][nm] + pre) % md
d = 1
for c0, c1 in zip(s, s[1:]):
if c0 == c1:
d = 0
break
print((sum(dp[n - 1][r][1]) + d) % md)
main()
``` | instruction | 0 | 40,272 | 0 | 80,544 |
Yes | output | 1 | 40,272 | 0 | 80,545 |
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 `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497
Submitted Solution:
```
from collections import Counter
import sys
sys.setrecursionlimit(10**6)
MOD = 998244353
ABC = "abc".index
*S, = map(ABC, input())
N = len(S)
def bruteforce(S):
used = set()
def dfs(s):
key = tuple(s)
if key in used:
return
used.add(key)
for i in range(len(s)-1):
if s[i] != s[i+1]:
a = s[i]; b = s[i+1]
s[i] = s[i+1] = 3 - a - b
dfs(s)
s[i] = a; s[i+1] = b
dfs(S)
return len(used)
if len(S) <= 3:
print(bruteforce(S))
else:
c = Counter(S)
if c[0] == N or c[1] == N or c[2] == N:
print(1)
else:
P = [[[0,0] for i in range(3)] for j in range(3)]
Q = [[[0,0] for i in range(3)] for j in range(3)]
P[0][0][0] = P[1][1][0] = P[2][2][0] = 1
for i in range(N-1):
for p in range(3):
for q in range(3):
for r in range(2):
Q[p][q][r] = 0
for p in range(3):
for q in range(3):
for r in range(2):
for k in range(3):
Q[k][(q+k)%3][r | (p == k)] += P[p][q][r] % MOD
P, Q = Q, P
ans = 0
r = sum(S) % 3
for i in range(3):
ans += P[i][r][1]
if all(S[i] != S[i+1] for i in range(N-1)):
ans += 1
print(ans % MOD)
``` | instruction | 0 | 40,273 | 0 | 80,546 |
Yes | output | 1 | 40,273 | 0 | 80,547 |
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 `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497
Submitted Solution:
```
def solve(s):
if len(s) == 2:
return 1 if s[0] == s[1] else 2
elif len(s) == 3:
if s[0] == s[1] == s[2]:
return 1
elif s[0] == s[1] or s[1] == s[2]:
return 6
elif s[0] == s[2]:
return 7
else:
return 3
dp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
dp[0][0][0] = 1
dp[0][1][1] = 1
dp[0][2][2] = 1
MOD = 998244353
for _ in range(len(s) - 1):
ndp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
ndp[0][0][0] = (dp[0][1][0] + dp[0][2][0]) % MOD
ndp[0][0][1] = (dp[0][1][1] + dp[0][2][1]) % MOD
ndp[0][0][2] = (dp[0][1][2] + dp[0][2][2]) % MOD
ndp[0][1][0] = (dp[0][0][2] + dp[0][2][2]) % MOD
ndp[0][1][1] = (dp[0][0][0] + dp[0][2][0]) % MOD
ndp[0][1][2] = (dp[0][0][1] + dp[0][2][1]) % MOD
ndp[0][2][0] = (dp[0][0][1] + dp[0][1][1]) % MOD
ndp[0][2][1] = (dp[0][0][2] + dp[0][1][2]) % MOD
ndp[0][2][2] = (dp[0][0][0] + dp[0][1][0]) % MOD
ndp[1][0][0] = (dp[0][0][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
ndp[1][0][1] = (dp[0][0][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][0][2] = (dp[0][0][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][1][0] = (dp[0][1][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][1][1] = (dp[0][1][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
ndp[1][1][2] = (dp[0][1][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][2][0] = (dp[0][2][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][2][1] = (dp[0][2][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][2][2] = (dp[0][2][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
dp = ndp
return (sum(dp[1][x][sum(map(ord, s)) % 3] for x in range(3)) + all(a != b for a, b in zip(s, s[1:]))) % MOD
s = input()
print(solve(s))
``` | instruction | 0 | 40,274 | 0 | 80,548 |
No | output | 1 | 40,274 | 0 | 80,549 |
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 `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497
Submitted Solution:
```
def solve(s):
if len(s) == 2:
return 1 if s[0] == s[1] else 2
elif len(s) == 3:
if s[0] == s[1] == s[2]:
return 1
elif s[0] == s[1] or s[1] == s[2]:
return 6
elif s[0] == s[2]:
return 7
else:
return 3
dp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
dp[0][0][0] = 1
dp[0][1][1] = 1
dp[0][2][2] = 1
MOD = 998244353
for _ in range(len(s) - 1):
ndp = [[[0] * 3 for _ in range(3)] for _ in range(2)]
ndp[0][0][0] = (dp[0][1][0] + dp[0][2][0]) % MOD
ndp[0][0][1] = (dp[0][1][1] + dp[0][2][1]) % MOD
ndp[0][0][2] = (dp[0][1][2] + dp[0][2][2]) % MOD
ndp[0][1][0] = (dp[0][0][2] + dp[0][2][2]) % MOD
ndp[0][1][1] = (dp[0][0][0] + dp[0][2][0]) % MOD
ndp[0][1][2] = (dp[0][0][1] + dp[0][2][1]) % MOD
ndp[0][2][0] = (dp[0][0][1] + dp[0][1][1]) % MOD
ndp[0][2][1] = (dp[0][0][2] + dp[0][1][2]) % MOD
ndp[0][2][2] = (dp[0][0][0] + dp[0][1][0]) % MOD
ndp[1][0][0] = (dp[0][0][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
ndp[1][0][1] = (dp[0][0][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][0][2] = (dp[0][0][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][1][0] = (dp[0][1][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][1][1] = (dp[0][1][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
ndp[1][1][2] = (dp[0][1][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][2][0] = (dp[0][2][1] + dp[1][0][1] + dp[1][1][1] + dp[1][2][1]) % MOD
ndp[1][2][1] = (dp[0][2][2] + dp[1][0][2] + dp[1][1][2] + dp[1][2][2]) % MOD
ndp[1][2][2] = (dp[0][2][0] + dp[1][0][0] + dp[1][1][0] + dp[1][2][0]) % MOD
dp = ndp
return sum(dp[1][x][sum(map(ord, s)) % 3] for x in range(3)) + all(a != b for a, b in zip(s, s[1:]))
s = input()
print(solve(s))
``` | instruction | 0 | 40,275 | 0 | 80,550 |
No | output | 1 | 40,275 | 0 | 80,551 |
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 `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497
Submitted Solution:
```
# coding: utf-8
# Your code here!
MOD = 998244353
import sys
S=input()
if len(S)==2:
if S[0]==S[1]:
print(1)
else:
print(2)
sys.exit()
flag1=0
flag2=0
for i in range(len(S)):
if S[i] != S[0]:
flag1 = 1
break
for i in range(len(S)-1):
if S[i+1] == S[i]:
flag2 = 1
break
c=0
for i in range(len(S)):
if S[i]=='a':
c+=1
elif S[i]=='b':
c+=2
if flag1==0:
print(1)
sys.exit()
L = len(S)
dp = [[[0 for i in range(3)] for j in range(3)] for k in range(L+1)]
for a in range(3):
dp[0][a][a]=1
for i in range(1,L):
for p in range(3):
for a in range(3):
# print(i,p,a)
dp[i][(p+a+1)%3][(a+1)%3] += dp[i-1][p][a]
dp[i][(p+a+1)%3][(a+1)%3] %= MOD
dp[i][(p+a+2)%3][(a+2)%3] += dp[i-1][p][a]
dp[i][(p+a+2)%3][(a+2)%3] %= MOD
#print(dp)
print((pow(3,(len(S)-1),MOD) - dp[L-1][c%3][0] - dp[L-1][c%3][1] - dp[L-1][c%3][2] + (1-flag2))%MOD)
``` | instruction | 0 | 40,276 | 0 | 80,552 |
No | output | 1 | 40,276 | 0 | 80,553 |
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 `a`,`b` and `c`. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo 998244353:
* Choose an integer i such that 1\leq i\leq |S|-1 and the i-th and (i+1)-th characters in S are different. Replace each of the i-th and (i+1)-th characters in S with the character that differs from both of them (among `a`, `b` and `c`).
Constraints
* 2 \leq |S| \leq 2 × 10^5
* S consists of `a`, `b` and `c`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo 998244353.
Examples
Input
abc
Output
3
Input
abbac
Output
65
Input
babacabac
Output
6310
Input
ababacbcacbacacbcbbcbbacbaccacbacbacba
Output
148010497
Submitted Solution:
```
from collections import defaultdict
import sys
S=input()
N=len(S)
if N==3:
if S[0]==S[1]:
if S[1]==S[2]:
print(1)
else:
print(6)
else:
if S[0]==S[2]:
print(7)
elif S[1]==S[2]:
print(6)
else:
print(3)
sys.exit()
mod=998244353
M=0
for s in S:
if s=='b':
M+=1
elif s=='c':
M+=2
M%=3
dd=defaultdict(int)
for i in range(3):
dd[(0,i,i)]=1
for k in range(2,N+1):
nd=defaultdict(int)
for p,n in dd.items():
con,moji,m=p
#print(con,moji,m)
if con==0:
for k in range(3):
if moji==k:
nd[(1, k, (m + k) % 3)] += n
nd[(1, k, (m + k) % 3)] %=mod
else:
nd[(0,k,(m+k)%3)]+=n
nd[(0, k, (m+k) % 3)] %=mod
else:
for k in range(3):
nd[(1, k, (m + k) % 3)] += n
nd[(1, k, (m + k) % 3)] %=mod
#print(nd)
dd=nd
flag=1
for i in range(N-1):
if S[i]==S[i+1]:
flag=0
ans=flag
for p, n in dd.items():
con, moji, m = p
if m==M and con==1:
ans+=n
ans%=mod
print(ans)
``` | instruction | 0 | 40,277 | 0 | 80,554 |
No | output | 1 | 40,277 | 0 | 80,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two strings s and t, consisting only of letters a and b. You can make the following operation several times: choose a prefix of s, a prefix of t and swap them. Prefixes can be empty, also a prefix can coincide with a whole string.
Your task is to find a sequence of operations after which one of the strings consists only of a letters and the other consists only of b letters. The number of operations should be minimized.
Input
The first line contains a string s (1 ≤ |s| ≤ 2·105).
The second line contains a string t (1 ≤ |t| ≤ 2·105).
Here |s| and |t| denote the lengths of s and t, respectively. It is guaranteed that at least one of the strings contains at least one a letter and at least one of the strings contains at least one b letter.
Output
The first line should contain a single integer n (0 ≤ n ≤ 5·105) — the number of operations.
Each of the next n lines should contain two space-separated integers ai, bi — the lengths of prefixes of s and t to swap, respectively.
If there are multiple possible solutions, you can print any of them. It's guaranteed that a solution with given constraints exists.
Examples
Input
bab
bb
Output
2
1 0
1 3
Input
bbbb
aaa
Output
0
Note
In the first example, you can solve the problem in two operations:
1. Swap the prefix of the first string with length 1 and the prefix of the second string with length 0. After this swap, you'll have strings ab and bbb.
2. Swap the prefix of the first string with length 1 and the prefix of the second string with length 3. After this swap, you'll have strings bbbb and a.
In the second example, the strings are already appropriate, so no operations are needed. | instruction | 0 | 40,424 | 0 | 80,848 |
Tags: constructive algorithms, strings
Correct Solution:
```
from sys import exit
def blokovi(x):
ret = [0]
for i in range(len(x) - 1):
if x[i] != x[i + 1]:
ret.append(i + 1)
return ret + [len(x)]
s = input()
t = input()
ss = blokovi(s)
tt = blokovi(t)
if s[-1] == 'a':
s += 'b'
else:
s += 'a'
if t[-1] == 'a':
t += 'b'
else:
t += 'a'
def greedy(x, y, rev=False):
i, j = len(x) - 1, len(y) - 1
swaps = []
while True:
while i >= 0 and x[i] == 'a':
i -= 1
while j >= 0 and y[j] == 'b':
j -= 1
if i < 0 and j < 0:
break
x, y = y, x
if rev:
swaps.append((j + 1, i + 1))
else:
swaps.append((i + 1, j + 1))
i, j = j, i
return swaps
def solve(x, y):
p = greedy(x, y)
q = greedy(y, x, True)
if len(p) < len(q):
return p
return q
probao = set()
total = len(ss) + len(tt)
sol = solve(s[:-1], t[:-1])
for b, i in enumerate(ss):
for c in range((2 * b + len(tt) - len(ss)) // 2 - 2, (2 * b + len(tt) - len(ss) + 1) // 2 + 3):
if 0 <= c < len(tt):
j = tt[c]
bs = b + len(tt) - c - 1
bt = c + len(ss) - b - 1
if abs(bs - bt) > 2:
continue
proba = (bs, bt, s[i], t[j])
if proba in probao:
continue
probao.add(proba)
s2 = t[:j] + s[i:-1]
t2 = s[:i] + t[j:-1]
if i + j > 0:
if i + j == len(s) + len(t) - 2:
cand = solve(t2, s2)
else:
cand = [(i, j)] + solve(s2, t2)
else:
cand = solve(s2, t2)
if len(cand) < len(sol):
sol = cand
print(len(sol))
for i, j in sol:
print(i, j)
``` | output | 1 | 40,424 | 0 | 80,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two strings s and t, consisting only of letters a and b. You can make the following operation several times: choose a prefix of s, a prefix of t and swap them. Prefixes can be empty, also a prefix can coincide with a whole string.
Your task is to find a sequence of operations after which one of the strings consists only of a letters and the other consists only of b letters. The number of operations should be minimized.
Input
The first line contains a string s (1 ≤ |s| ≤ 2·105).
The second line contains a string t (1 ≤ |t| ≤ 2·105).
Here |s| and |t| denote the lengths of s and t, respectively. It is guaranteed that at least one of the strings contains at least one a letter and at least one of the strings contains at least one b letter.
Output
The first line should contain a single integer n (0 ≤ n ≤ 5·105) — the number of operations.
Each of the next n lines should contain two space-separated integers ai, bi — the lengths of prefixes of s and t to swap, respectively.
If there are multiple possible solutions, you can print any of them. It's guaranteed that a solution with given constraints exists.
Examples
Input
bab
bb
Output
2
1 0
1 3
Input
bbbb
aaa
Output
0
Note
In the first example, you can solve the problem in two operations:
1. Swap the prefix of the first string with length 1 and the prefix of the second string with length 0. After this swap, you'll have strings ab and bbb.
2. Swap the prefix of the first string with length 1 and the prefix of the second string with length 3. After this swap, you'll have strings bbbb and a.
In the second example, the strings are already appropriate, so no operations are needed.
Submitted Solution:
```
from sys import exit
def blokovi(x):
ret = [0]
for i in range(len(x) - 1):
if x[i] != x[i + 1]:
ret.append(i + 1)
return ret + [len(x)]
s = input()
t = input()
ss = blokovi(s)
tt = blokovi(t)
if s[-1] == 'a':
s += 'b'
else:
s += 'a'
if t[-1] == 'a':
t += 'b'
else:
t += 'a'
def greedy(x, y, rev=False):
i, j = len(x) - 1, len(y) - 1
swaps = []
while True:
while i >= 0 and x[i] == 'a':
i -= 1
while j >= 0 and y[j] == 'b':
j -= 1
if i < 0 and j < 0:
break
x, y = y, x
if rev:
swaps.append((j + 1, i + 1))
else:
swaps.append((i + 1, j + 1))
i, j = j, i
return swaps
def solve(x, y):
p = greedy(x, y)
q = greedy(y, x, True)
if len(p) < len(q):
return p
return q
probao = set()
total = len(ss) + len(tt)
sol = solve(s[:-1], t[:-1])
for b, i in enumerate(ss):
for c in range((len(tt) - len(ss)) // 2 - 2, (len(tt) - len(ss) + 1) // 2 + 3):
if 0 <= c < len(tt):
j = tt[c]
if s[i] == t[j]:
continue
bs = b + len(tt) - c - 1
bt = c + len(ss) - b - 1
if abs(bs - bt) > 2:
continue
proba = (bs, bt, s[i])
if proba in probao:
continue
probao.add(proba)
s2 = t[:j] + s[i:-1]
t2 = s[:i] + t[j:-1]
if i + j > 0 and i + j < len(s) + len(t) - 2:
cand = [(i, j)] + solve(s2, t2)
else:
cand = solve(s2, t2)
if len(cand) < len(sol):
sol = cand
print(len(sol))
for i, j in sol:
print(i, j)
``` | instruction | 0 | 40,425 | 0 | 80,850 |
No | output | 1 | 40,425 | 0 | 80,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two strings s and t, consisting only of letters a and b. You can make the following operation several times: choose a prefix of s, a prefix of t and swap them. Prefixes can be empty, also a prefix can coincide with a whole string.
Your task is to find a sequence of operations after which one of the strings consists only of a letters and the other consists only of b letters. The number of operations should be minimized.
Input
The first line contains a string s (1 ≤ |s| ≤ 2·105).
The second line contains a string t (1 ≤ |t| ≤ 2·105).
Here |s| and |t| denote the lengths of s and t, respectively. It is guaranteed that at least one of the strings contains at least one a letter and at least one of the strings contains at least one b letter.
Output
The first line should contain a single integer n (0 ≤ n ≤ 5·105) — the number of operations.
Each of the next n lines should contain two space-separated integers ai, bi — the lengths of prefixes of s and t to swap, respectively.
If there are multiple possible solutions, you can print any of them. It's guaranteed that a solution with given constraints exists.
Examples
Input
bab
bb
Output
2
1 0
1 3
Input
bbbb
aaa
Output
0
Note
In the first example, you can solve the problem in two operations:
1. Swap the prefix of the first string with length 1 and the prefix of the second string with length 0. After this swap, you'll have strings ab and bbb.
2. Swap the prefix of the first string with length 1 and the prefix of the second string with length 3. After this swap, you'll have strings bbbb and a.
In the second example, the strings are already appropriate, so no operations are needed.
Submitted Solution:
```
from sys import exit
def blokovi(x):
ret = [0]
for i in range(len(x) - 1):
if x[i] != x[i + 1]:
ret.append(i + 1)
return ret + [len(x)]
s = input()
t = input()
ss = blokovi(s)
tt = blokovi(t)
if s[-1] == 'a':
s += 'b'
else:
s += 'a'
if t[-1] == 'a':
t += 'b'
else:
t += 'a'
def greedy(x, y, rev=False):
i, j = len(x) - 1, len(y) - 1
swaps = []
while True:
while i >= 0 and x[i] == 'a':
i -= 1
while j >= 0 and y[j] == 'b':
j -= 1
if i < 0 and j < 0:
break
x, y = y, x
if rev:
swaps.append((j + 1, i + 1))
else:
swaps.append((i + 1, j + 1))
i, j = j, i
return swaps
def solve(x, y):
p = greedy(x, y)
q = greedy(y, x, True)
if len(p) < len(q):
return p
return q
total = len(ss) + len(tt)
sol = solve(s[:-1], t[:-1])
for b, i in enumerate(ss):
for c in range((len(tt) - len(ss)) // 2 - 1, (len(tt) - len(ss) + 1) // 2 + 2):
if 0 <= c < len(tt):
j = tt[c]
if s[i] == t[j]:
continue
bs = b + len(tt) - c - 1
bt = c + len(ss) - b - 1
if bs == bt:
s2 = t[:j] + s[i:-1]
t2 = s[:i] + t[j:-1]
if i + j > 0:
cand = [(i, j)] + solve(s2, t2)
else:
cand = solve(s2, t2)
if len(cand) < len(sol):
sol = cand
elif abs(bs - bt) == 1:
s2 = t[:j] + s[i:]
t2 = s[:i] + t[j:]
if i + j > 0:
cand = [(i, j)] + solve(s2, t2)
else:
cand = solve(s2, t2)
if len(cand) < len(sol):
sol = cand
print(len(sol))
for i, j in sol:
print(i, j)
``` | instruction | 0 | 40,426 | 0 | 80,852 |
No | output | 1 | 40,426 | 0 | 80,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two strings s and t, consisting only of letters a and b. You can make the following operation several times: choose a prefix of s, a prefix of t and swap them. Prefixes can be empty, also a prefix can coincide with a whole string.
Your task is to find a sequence of operations after which one of the strings consists only of a letters and the other consists only of b letters. The number of operations should be minimized.
Input
The first line contains a string s (1 ≤ |s| ≤ 2·105).
The second line contains a string t (1 ≤ |t| ≤ 2·105).
Here |s| and |t| denote the lengths of s and t, respectively. It is guaranteed that at least one of the strings contains at least one a letter and at least one of the strings contains at least one b letter.
Output
The first line should contain a single integer n (0 ≤ n ≤ 5·105) — the number of operations.
Each of the next n lines should contain two space-separated integers ai, bi — the lengths of prefixes of s and t to swap, respectively.
If there are multiple possible solutions, you can print any of them. It's guaranteed that a solution with given constraints exists.
Examples
Input
bab
bb
Output
2
1 0
1 3
Input
bbbb
aaa
Output
0
Note
In the first example, you can solve the problem in two operations:
1. Swap the prefix of the first string with length 1 and the prefix of the second string with length 0. After this swap, you'll have strings ab and bbb.
2. Swap the prefix of the first string with length 1 and the prefix of the second string with length 3. After this swap, you'll have strings bbbb and a.
In the second example, the strings are already appropriate, so no operations are needed.
Submitted Solution:
```
from sys import exit
def blokovi(x):
ret = [0]
for i in range(len(x) - 1):
if x[i] != x[i + 1]:
ret.append(i + 1)
return ret + [len(x)]
s = input()
t = input()
ss = blokovi(s)
tt = blokovi(t)
if s[-1] == 'a':
s += 'b'
else:
s += 'a'
if t[-1] == 'a':
t += 'b'
else:
t += 'a'
def greedy(x, y, rev=False):
i, j = len(x) - 1, len(y) - 1
swaps = []
while True:
while i >= 0 and x[i] == 'a':
i -= 1
while j >= 0 and y[j] == 'b':
j -= 1
if i < 0 and j < 0:
break
x, y = y, x
if rev:
swaps.append((j + 1, i + 1))
else:
swaps.append((i + 1, j + 1))
i, j = j, i
return swaps
def solve(x, y):
p = greedy(x, y)
q = greedy(y, x, True)
if len(p) < len(q):
return p
return q
total = len(ss) + len(tt)
sol = None
for b, i in enumerate(ss):
for c in range((len(tt) - len(ss)) // 2 - 1, (len(tt) - len(ss) + 1) // 2 + 2):
if 0 <= c < len(tt):
j = tt[c]
if s[i] == t[j]:
continue
bs = b + len(tt) - c - 1
bt = c + len(ss) - b - 1
if bs == bt:
s2 = t[:j] + s[i:-1]
t2 = s[:i] + t[j:-1]
if i + j > 0:
sol = [(i, j)] + solve(s2, t2)
else:
sol = solve(s2, t2)
print(len(sol))
for i, j in sol:
print(i, j)
exit(0)
elif abs(bs - bt) == 1 and sol is None:
s2 = t[:j] + s[i:]
t2 = s[:i] + t[j:]
if i + j > 0:
sol = [(i, j)] + solve(s2, t2)
else:
sol = solve(s2, t2)
print(len(sol))
for i, j in sol:
print(i, j)
``` | instruction | 0 | 40,427 | 0 | 80,854 |
No | output | 1 | 40,427 | 0 | 80,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two strings s and t, consisting only of letters a and b. You can make the following operation several times: choose a prefix of s, a prefix of t and swap them. Prefixes can be empty, also a prefix can coincide with a whole string.
Your task is to find a sequence of operations after which one of the strings consists only of a letters and the other consists only of b letters. The number of operations should be minimized.
Input
The first line contains a string s (1 ≤ |s| ≤ 2·105).
The second line contains a string t (1 ≤ |t| ≤ 2·105).
Here |s| and |t| denote the lengths of s and t, respectively. It is guaranteed that at least one of the strings contains at least one a letter and at least one of the strings contains at least one b letter.
Output
The first line should contain a single integer n (0 ≤ n ≤ 5·105) — the number of operations.
Each of the next n lines should contain two space-separated integers ai, bi — the lengths of prefixes of s and t to swap, respectively.
If there are multiple possible solutions, you can print any of them. It's guaranteed that a solution with given constraints exists.
Examples
Input
bab
bb
Output
2
1 0
1 3
Input
bbbb
aaa
Output
0
Note
In the first example, you can solve the problem in two operations:
1. Swap the prefix of the first string with length 1 and the prefix of the second string with length 0. After this swap, you'll have strings ab and bbb.
2. Swap the prefix of the first string with length 1 and the prefix of the second string with length 3. After this swap, you'll have strings bbbb and a.
In the second example, the strings are already appropriate, so no operations are needed.
Submitted Solution:
```
from sys import exit
def blokovi(x):
ret = [0]
for i in range(len(x) - 1):
if x[i] != x[i + 1]:
ret.append(i + 1)
return ret + [len(x)]
s = input()
t = input()
ss = blokovi(s)
tt = blokovi(t)
if s[-1] == 'a':
s += 'b'
else:
s += 'a'
if t[-1] == 'a':
t += 'b'
else:
t += 'a'
def greedy(x, y, rev=False):
i, j = len(x) - 1, len(y) - 1
swaps = []
while True:
while i >= 0 and x[i] == 'a':
i -= 1
while j >= 0 and y[j] == 'b':
j -= 1
if i < 0 and j < 0:
break
x, y = y, x
if rev:
swaps.append((j + 1, i + 1))
else:
swaps.append((i + 1, j + 1))
i, j = j, i
return swaps
def solve(x, y):
p = greedy(x, y)
q = greedy(y, x, True)
if len(p) < len(q):
return p
return q
probao = set()
total = len(ss) + len(tt)
sol = solve(s[:-1], t[:-1])
for b, i in enumerate(ss):
for c in range((len(tt) - len(ss)) // 2 - 2, (len(tt) - len(ss) + 1) // 2 + 3):
if 0 <= c < len(tt):
j = tt[c]
if s[i] == t[j]:
continue
bs = b + len(tt) - c - 1
bt = c + len(ss) - b - 1
if abs(bs - bt) > 2:
continue
proba = (bs, bt, s[i])
if proba in probao:
continue
probao.add(proba)
s2 = t[:j] + s[i:-1]
t2 = s[:i] + t[j:-1]
if i + j > 0:
if i + j == len(s) + len(t) - 2:
cand = solve(t2, s2)
else:
cand = [(i, j)] + solve(s2, t2)
else:
cand = solve(s2, t2)
if len(cand) < len(sol):
sol = cand
print(len(sol))
for i, j in sol:
print(i, j)
``` | instruction | 0 | 40,428 | 0 | 80,856 |
No | output | 1 | 40,428 | 0 | 80,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,527 | 0 | 81,054 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=tf-8
#
"""
"""
from operator import itemgetter
from collections import Counter
from sys import stdin
input = stdin.readline
def solve(s):
alphs = "abcdefghijklmnopqrstuvwxyz"
cumsum_dict = {a:[0] for a in alphs}
for i,ss in enumerate(s):
for a in alphs:
cumsum_dict[a].append(cumsum_dict[a][i])
cumsum_dict[ss][i+1] += 1
q = int(input())
for i in range(q):
l, r = map(int,input().split())
ls = s[l-1]
rs = s[r-1]
al_set = set([])
for a in alphs:
if cumsum_dict[a][r] - cumsum_dict[a][l-1] > 0:
al_set.add(a)
al_num = len(al_set)
if l == r or (ls is not rs) or (al_num >= 3):
print("Yes")
else:
print("No")
def main():
s= input().rstrip()
solve(s)
if __name__ == "__main__":
main()
``` | output | 1 | 40,527 | 0 | 81,055 |
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