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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa. | instruction | 0 | 29,531 | 0 | 59,062 |
Tags: data structures, dp, two pointers
Correct Solution:
```
l = list(map(int, input().split()))
s = input()
cs = 0
ans = 0
a = []
for i in range(26): a.append({})
for i in s:
c = ord(i) - 97
x = a[c].get(cs) or 0
ans += x
cs += l[c]
y = a[c].get(cs) or 0
a[c][cs] = y + 1
print(ans)
``` | output | 1 | 29,531 | 0 | 59,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa. | instruction | 0 | 29,532 | 0 | 59,064 |
Tags: data structures, dp, two pointers
Correct Solution:
```
import sys
read = lambda t=int: list(map(t,sys.stdin.readline().split()))
array = lambda *ds: [array(*ds[1:]) for _ in range(ds[0])] if ds else 0
from collections import Counter
xs = read()
s, = read(str)
cnt = Counter()
a = 0
res = 0
for c in s:
res += cnt[(c,a)]
a += xs[ord(c)-ord('a')]
cnt[(c,a)] += 1
print(res)
``` | output | 1 | 29,532 | 0 | 59,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
ls = list(map(int,input().split()))
s = list(input())
n = len(s)
fre = [None]*n
fre[0]=ls[ord(s[0])-ord('a')]
#print(fre[0])
for i in range(1,n):
x = ls[ord(s[i])-ord('a')]
x = x + fre[i-1]
fre[i]=x
d = {}
ans = 0
for i in range(0,26):
d[i] = {}
d[ord(s[0])-ord('a')][fre[0]]=1
for i in range(1,n):
k = ord(s[i])-ord('a')
#print(k)
#print(fre[i-1])
if fre[i-1] in d[k]:
ans = ans+d[k][fre[i-1]]
if fre[i] in d[k]:
d[k][fre[i]] = d[k][fre[i]]+1
else:
d[k][fre[i]] = 1
print(ans)
``` | instruction | 0 | 29,533 | 0 | 59,066 |
Yes | output | 1 | 29,533 | 0 | 59,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt
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)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
mod = int(1e9) + 7
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def totalPrimeFactors(n):
count = 0
if (n % 2) == 0:
count += 1
while (n % 2) == 0:
n //= 2
i = 3
while i * i <= n:
if (n % i) == 0:
count += 1
while (n % i) == 0:
n //= i
i += 2
if n > 2:
count += 1
return count
# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
# spf[1] = 1
# for i in range(2, MAXN):
# spf[i] = i
# for i in range(4, MAXN, 2):
# spf[i] = 2
#
# for i in range(3, mt.ceil(mt.sqrt(MAXN))):
# if (spf[i] == i):
# for j in range(i * i, MAXN, i):
# if (spf[j] == j):
# spf[j] = i
#
#
# def getFactorization(x):
# ret = 0
# while (x != 1):
# k = spf[x]
# ret += 1
# # ret.add(spf[x])
# while x % k == 0:
# x //= k
#
# return ret
# Driver code
# precalculating Smallest Prime Factor
# sieve()
def main():
val = list(map(int, input().split()))
S = input()
s = []
for i in S:
if ord('a') <= ord(i) <= ord('z'):
s.append(i)
table = []
for i in range(26):
t = {}
table.append(t)
curr = 0
ans = 0
for i in range(len(s)):
ind = ord(s[i]) - ord('a')
if curr in table[ind]:
ans += table[ind][curr]
#table[ind][curr] += 1
if i:
if curr not in table[ord(s[i - 1]) - ord('a')]:
table[ord(s[i - 1]) - ord('a')][curr]=0
table[ord(s[i - 1]) - ord('a')][curr]+=1
curr += val[ind]
#print(table)
for i in range(1, len(s)):
ans += (s[i] == s[i - 1])
print(ans)
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 29,534 | 0 | 59,068 |
Yes | output | 1 | 29,534 | 0 | 59,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
#k=int(input())
#n,m=map(int,input().split())
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
w=list(map(int,input().split()))
a=input()
l=len(a)
m=[0]*27
s=[0]*l
for i in range(27):
m[i]=dict()
ans=0
for i in range(1,l):
A=ord(a[i-1])-ord('a')
s[i]=s[i-1]+w[A]
for i in range(l):
A=ord(a[i])-ord('a')
if(s[i]-w[A] in m[A]):
ans+=m[A][s[i]-w[A]]
if not (s[i] in m[A]):
m[A][s[i]]=1
else:
m[A][s[i]]+=1
print(ans)
``` | instruction | 0 | 29,535 | 0 | 59,070 |
Yes | output | 1 | 29,535 | 0 | 59,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
pr = lambda x: x
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
A = aj()
s = input()
C = Counter()
pre = A[ord(s[0]) - ord('a')]
C[(s[0],pre)] += 1
ans = 0
for i in range(1,len(s)):
ans += C[(s[i],pre)]
pre += A[ord(s[i]) - ord('a')]
C[(s[i],pre)] += 1
print(ans)
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
from aj import *
except:
pass
solve()
``` | instruction | 0 | 29,536 | 0 | 59,072 |
Yes | output | 1 | 29,536 | 0 | 59,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
score=[]
from collections import *
z=list(map(int,input().split()))
s=input()
for i in range(26):
score.append(defaultdict(int))
pre=[]
total=0
for i in range(len(s)):
s1=z[ord(s[i])-97]
t=ord(s[i])-97
if(i==0):
pre.append(s1)
else:
pre.append(pre[-1]+s1)
s1=pre[-1]
score[t][s1]+=1
total+=max(0,score[t][s1-z[t]])
fin=[]
count=1
for i in range(1,len(s)):
if(s[i]==s[i-1]):
count+=1
else:
fin.append([count,s[i-1]])
count=1
fin.append([count,s[-1]])
for i in range(len(fin)):
if(fin[i]==2 and score[ord(fin[i][1])-97]!=0):
total+=1
print(total)
``` | instruction | 0 | 29,537 | 0 | 59,074 |
No | output | 1 | 29,537 | 0 | 59,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
#if not WA pretest 3
#ily <3
g = {}
for i in 'abcdefghijklmnopqrstuvwxyz':
g[i] = []
v = list(map(int, input().split(' ')))
x = input()
sumx = [v[ord(x[0])-97]] + [0] * (len(x)-1)
g[x[0]].append(sumx[0])
for i in range(1, len(x)):
sumx[i] = sumx[i-1] + v[ord(x[i])-97]
g[x[i]].append(sumx[i])
sumx.sort()
tot = 0
ct = 0
prec = -1
for i in g:
arr = g[i]
arr.sort()
d = v[ord(i)-97]
count = 0
l = 0
r = 0
if d < 0:
d *= -1
arr.reverse()
while r < len(g[i]) and l < len(g[i]):
if arr[r]-arr[l] == d:
if l < r:
count += 1
l += 1
r += 1
elif arr[r] - arr[l] > d:
l += 1
else:
r += 1
tot += count
print(tot)
``` | instruction | 0 | 29,538 | 0 | 59,076 |
No | output | 1 | 29,538 | 0 | 59,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
cost = [int(x) for x in input().split()]
s = input().strip()
sumAr1 = [0 for x in range(len(s))]
sumAr = [0 for x in range(len(s))]
it = 0
for sym in s:
sumAr[it] = cost[ord(sym) - ord('a')]
sumAr1[it] = sumAr[it]
it += 1
# print(sumAr)
for i in range(1, len(sumAr)):
sumAr[i] += sumAr[i-1]
# print(sumAr)
ans = 0
for i in range(len(s)):
for j in range(i, len(s)):
if s[i] == s[j]:
if i == 0 and sumAr[j] - sumAr1[j] - sumAr1[i] == 0:
ans += 1
elif sumAr[j] - sumAr[i - 1] - sumAr1[j] - sumAr1[i] == 0:
ans += 1
print(ans)
``` | instruction | 0 | 29,539 | 0 | 59,078 |
No | output | 1 | 29,539 | 0 | 59,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
After several years of doing sports programming and solving many problems that require calculating all sorts of abstract objects, A and B also developed rather peculiar tastes.
A likes lowercase letters of the Latin alphabet. He has assigned to each letter a number that shows how much he likes that letter (he has assigned negative numbers to the letters he dislikes).
B likes substrings. He especially likes the ones that start and end with the same letter (their length must exceed one).
Also, A and B have a string s. Now they are trying to find out how many substrings t of a string s are interesting to B (that is, t starts and ends with the same letter and its length is larger than one), and also the sum of values of all letters (assigned by A), except for the first and the last one is equal to zero.
Naturally, A and B have quickly found the number of substrings t that are interesting to them. Can you do it?
Input
The first line contains 26 integers xa, xb, ..., xz ( - 105 ≤ xi ≤ 105) — the value assigned to letters a, b, c, ..., z respectively.
The second line contains string s of length between 1 and 105 characters, consisting of Lating lowercase letters— the string for which you need to calculate the answer.
Output
Print the answer to the problem.
Examples
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
xabcab
Output
2
Input
1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 7 1 1 1 8 1 1 1 1 1 1
aaa
Output
2
Note
In the first sample test strings satisfying the condition above are abca and bcab.
In the second sample test strings satisfying the condition above are two occurences of aa.
Submitted Solution:
```
l=list(map(int,input("").split()))
s=input("")
l2=[]
d={}
for i in range(1,27):
d[chr(96+i)]=l[i-1]
p1=0
p2=0
for i in range(0,len(s)):
for j in range(i+1,len(s)):
if(s[i]==s[j]):
p1=i
p2=j
l2.append(s[p1:p2+1])
p=""
sum1=0
b=0
sum3=0
for i in range(0,len(l2)):
for j in range(1,len(l2[i])-1):
sum1+=d[l2[i][j]]
sum2=0
for j in range(0,len(l2[i])):
sum2+=d[l2[i][j]]
if(sum1==0):
sum3+=sum2
print(sum3)
``` | instruction | 0 | 29,540 | 0 | 59,080 |
No | output | 1 | 29,540 | 0 | 59,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length <image> are (k - 1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string s such a maximum number k, for which s is k-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.
Input
The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.
Output
Output the only number — the sum of the polindrome degrees of all the string's prefixes.
Examples
Input
a2A
Output
1
Input
abacaba
Output
6
Submitted Solution:
```
from sys import *
def input():
return stdin.readline()
#maxn = 5000005
s = input()
maxn = len(s)
dp=[0 for i in range(maxn)]
dp[0] = 1
ha = haf = ord(s[0])
hb = 1
ans = 1
i = 1
for i in range(1, len(s)):
hb *= 33
ha = ha*3 + ord(s[i])
haf = ord(s[i])*hb + haf
if ha == haf:
dp[i] = dp[(i-1) >>1] + 1;
ans += dp[i]
print(ans)
``` | instruction | 0 | 29,643 | 0 | 59,286 |
No | output | 1 | 29,643 | 0 | 59,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length <image> are (k - 1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string s such a maximum number k, for which s is k-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.
Input
The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.
Output
Output the only number — the sum of the polindrome degrees of all the string's prefixes.
Examples
Input
a2A
Output
1
Input
abacaba
Output
6
Submitted Solution:
```
import sys
import math
line = sys.stdin.readline().strip()
length = len(line)
def palindrome(word):
if word == '':
return False
odd = len(word) % 2
middle = len(word) // 2
if word[middle + odd:] == word[:middle]:
return True
return False
degree = 0
middle = length // 2
while length > 0:
if palindrome(line):
if len(line) <= 3:
degree += math.ceil(length / 2)
else:
degree += length // 2
line = line[:middle]
middle -= 1
length = len(line)
if degree == 0 and length %2 == 1:
print (1)
else:
print (degree)
``` | instruction | 0 | 29,644 | 0 | 59,288 |
No | output | 1 | 29,644 | 0 | 59,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length <image> are (k - 1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string s such a maximum number k, for which s is k-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.
Input
The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.
Output
Output the only number — the sum of the polindrome degrees of all the string's prefixes.
Examples
Input
a2A
Output
1
Input
abacaba
Output
6
Submitted Solution:
```
1
``` | instruction | 0 | 29,645 | 0 | 59,290 |
No | output | 1 | 29,645 | 0 | 59,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length <image> are (k - 1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string s such a maximum number k, for which s is k-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.
Input
The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.
Output
Output the only number — the sum of the polindrome degrees of all the string's prefixes.
Examples
Input
a2A
Output
1
Input
abacaba
Output
6
Submitted Solution:
```
s = input()
l = len(s)
pd = list()
hl = 0
hr = 0
pl = 1
ans = 0
if s[0] == '5': print(len(s))
for i in range(0, l):
pd.append(0)
hl = hl*23 + ord(s[i])
hr = hr + ord(s[i]) * pl
pl = pl * 23
if (hr == hl):
pd[i] = pd[int((i+1)/2)-1] + 1
ans = ans + pd[i]
print(ans)
``` | instruction | 0 | 29,646 | 0 | 59,292 |
No | output | 1 | 29,646 | 0 | 59,293 |
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. You should answer n queries. The i-th query consists of integer k_i and string m_i. The answer for this query is the minimum length of such a string t that t is a substring of s and m_i has at least k_i occurrences as a substring in t.
A substring of a string is a continuous segment of characters of the string.
It is guaranteed that for any two queries the strings m_i from these queries are different.
Input
The first line contains string s (1 ≤ \left | s \right | ≤ 10^{5}).
The second line contains an integer n (1 ≤ n ≤ 10^5).
Each of next n lines contains an integer k_i (1 ≤ k_i ≤ |s|) and a non-empty string m_i — parameters of the query with number i, in this order.
All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed 10^5. All m_i are distinct.
Output
For each query output the answer for it in a separate line.
If a string m_{i} occurs in s less that k_{i} times, output -1.
Examples
Input
aaaaa
5
3 a
3 aa
2 aaa
3 aaaa
1 aaaaa
Output
3
4
4
-1
5
Input
abbb
7
4 b
1 ab
3 bb
1 abb
2 bbb
1 a
2 abbb
Output
-1
2
-1
3
-1
1
-1
Submitted Solution:
```
def occurences(s,y):
count=0
while(s.find(y)!=-1):
a = s.find(y)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
#s = "".join(s)
return count
def n_occurences(s,y,n):
count=0
temp=0
while(s.find(y)!=-1):
a = s.find(y)
temp = temp+abs(a)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
if(count==n):
break
#s = "".join(s)
return temp
def n_occurences_right(s,y,n):
count=0
temp=0
while(s.rfind(y)!=-1):
a = s.rfind(y)
#print(a)
temp = temp+ (len(s)-(a))-(len(y)-1)
#print(temp)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[:a+len(y)-1]
if(count==n):
break
#s = "".join(s)
return temp
s = input("")
n = int(input(""));
k=[]
m=[]
count=0
for i in range(0,n):
x,y = input().split();
count=0
#k.append(int(x));
#m.append(y)
x=int(x)
if len(y)>1:
mint = x*len(y)
t= y*x
if y[0]==y[len(y)-1]:
for i in range (0,len(y)-1):
if y[i] == y[len(y)-1-i]:
count=count+1
#print(count)
p = list(y)
for i in range (0,count):
p.remove(p[0])
y = list(y)
t = y+ p*(x-1)
#print(t)
q = ""
t = q.join(t)
mint = len(t)
y= "".join(y)
else:
mint = x
t = y*x
if t not in s:
s2 = s
num = occurences(s2,y)
#print(num)
if(num>=x):
#print(n_occurences(s,y,x))
a = n_occurences(s2,y,x)+len(y)+(x-1)
#print(a)
b = n_occurences_right(s2,y,x) + (len(y)-1)
#print(b)
print(min(a,b))
#print(x)
else:
print("-1")
else:
print(len(t))
#print("yess".rfind("s"))
``` | instruction | 0 | 29,727 | 0 | 59,454 |
No | output | 1 | 29,727 | 0 | 59,455 |
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. You should answer n queries. The i-th query consists of integer k_i and string m_i. The answer for this query is the minimum length of such a string t that t is a substring of s and m_i has at least k_i occurrences as a substring in t.
A substring of a string is a continuous segment of characters of the string.
It is guaranteed that for any two queries the strings m_i from these queries are different.
Input
The first line contains string s (1 ≤ \left | s \right | ≤ 10^{5}).
The second line contains an integer n (1 ≤ n ≤ 10^5).
Each of next n lines contains an integer k_i (1 ≤ k_i ≤ |s|) and a non-empty string m_i — parameters of the query with number i, in this order.
All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed 10^5. All m_i are distinct.
Output
For each query output the answer for it in a separate line.
If a string m_{i} occurs in s less that k_{i} times, output -1.
Examples
Input
aaaaa
5
3 a
3 aa
2 aaa
3 aaaa
1 aaaaa
Output
3
4
4
-1
5
Input
abbb
7
4 b
1 ab
3 bb
1 abb
2 bbb
1 a
2 abbb
Output
-1
2
-1
3
-1
1
-1
Submitted Solution:
```
def occurences(s,y):
count=0
while(s.find(y)!=-1):
a = s.find(y)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
#s = "".join(s)
return count
def n_occurences(s,y,n):
count=0
temp=0
while(s.find(y)!=-1):
a = s.find(y)
temp = temp+abs(a)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
if(count==n):
break
#s = "".join(s)
return temp
def n_occurences_right(s,y,n):
count=0
temp=0
while(s.rfind(y)!=-1):
a = s.rfind(y)
#print(a)
temp = temp+ (len(s)-(a))-(len(y)-1)
#print(temp)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[:a+len(y)-1]
if(count==n):
break
#s = "".join(s)
return temp
s = input("Enter the string: ")
n = int(input("Enter the number of queries: "));
k=[]
m=[]
count=0
for i in range(0,n):
x,y = input().split();
count=0
#k.append(int(x));
#m.append(y)
x=int(x)
if len(y)>1:
mint = x*len(y)
t= y*x
if y[0]==y[len(y)-1]:
for i in range (0,len(y)-1):
if y[i] == y[len(y)-1-i]:
count=count+1
#print(count)
p = list(y)
for i in range (0,count):
p.remove(p[0])
y = list(y)
t = y+ p*(x-1)
#print(t)
q = ""
t = q.join(t)
mint = len(t)
y= "".join(y)
else:
mint = x
t = y*x
if t not in s:
s2 = s
num = occurences(s2,y)
#print(num)
if(num>=x):
#print(n_occurences(s,y,x))
a = n_occurences(s2,y,x)+len(y)+(x-1)
#print(a)
b = n_occurences_right(s2,y,x) + (len(y)-1)
#print(b)
print(min(a,b))
#print(x)
else:
print("-1")
else:
print(len(t))
#print("yess".rfind("s"))
``` | instruction | 0 | 29,728 | 0 | 59,456 |
No | output | 1 | 29,728 | 0 | 59,457 |
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. You should answer n queries. The i-th query consists of integer k_i and string m_i. The answer for this query is the minimum length of such a string t that t is a substring of s and m_i has at least k_i occurrences as a substring in t.
A substring of a string is a continuous segment of characters of the string.
It is guaranteed that for any two queries the strings m_i from these queries are different.
Input
The first line contains string s (1 ≤ \left | s \right | ≤ 10^{5}).
The second line contains an integer n (1 ≤ n ≤ 10^5).
Each of next n lines contains an integer k_i (1 ≤ k_i ≤ |s|) and a non-empty string m_i — parameters of the query with number i, in this order.
All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed 10^5. All m_i are distinct.
Output
For each query output the answer for it in a separate line.
If a string m_{i} occurs in s less that k_{i} times, output -1.
Examples
Input
aaaaa
5
3 a
3 aa
2 aaa
3 aaaa
1 aaaaa
Output
3
4
4
-1
5
Input
abbb
7
4 b
1 ab
3 bb
1 abb
2 bbb
1 a
2 abbb
Output
-1
2
-1
3
-1
1
-1
Submitted Solution:
```
def occurences(s,y):
count=0
while(s.find(y)!=-1):
a = s.find(y)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
#s = "".join(s)
return count
def n_occurences(s,y,n):
count=0
temp=0
while(s.find(y)!=-1):
a = s.find(y)
if count!=0:
temp = temp+abs(a)
else:
temp = temp
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
if(count==n):
break
#s = "".join(s)
return temp
def n_occurences_right(s,y,n):
count=0
temp=0
while(s.rfind(y)!=-1):
a = s.rfind(y)
#print(a)
if count==0:
#temp = temp+ (((len(s)-(a))-(len(y)-1)) - ((len(s)-(a+1))+(len(y)-1)))
temp=temp+1
else:
temp = temp+ (len(s)-(a))-(len(y)-1)
#print(temp)
#print(temp)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[:a+len(y)-1]
if(count==n):
break
#s = "".join(s)
return temp
s = input("")
n = int(input(""));
k=[]
m=[]
count=0
for i in range(0,n):
s2 = s
x,y = input().split();
count=0
#k.append(int(x));
#m.append(y)
x=int(x)
if len(y)>1:
mint = x*len(y)
t= y*x
if y[0]==y[len(y)-1]:
for i in range (0,(len(y)-1)):
if y[i] == y[len(y)-1-i]:
if i!=(len(y)-1-i) or (count == i and y[i]==y[i-1]):
count=count+1
#print(y[i])
else:
break
#print(count)
#print(count)
p = list(y)
for i in range (0,count):
p.remove(p[0])
y = list(y)
t = y+ p*(x-1)
#print(t)
q = ""
t = q.join(t)
mint = len(t)
y= "".join(y)
#print(t)
else:
mint = x
t = y*x
if t not in s or occurences(s2,y)<x:
s2 = s
num = occurences(s2,y)
#print(num)
if(num>=x):
#print(n_occurences(s,y,x))
a = n_occurences(s2,y,x)+len(y)+(x-1)
#print(a)
b = n_occurences_right(s2,y,x) + (len(y)-1)
#print(b)
print(min(a,b))
#print(x)
else:
print("-1")
else:
print(len(t))
#print("yess".rfind("s"))
``` | instruction | 0 | 29,729 | 0 | 59,458 |
No | output | 1 | 29,729 | 0 | 59,459 |
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. You should answer n queries. The i-th query consists of integer k_i and string m_i. The answer for this query is the minimum length of such a string t that t is a substring of s and m_i has at least k_i occurrences as a substring in t.
A substring of a string is a continuous segment of characters of the string.
It is guaranteed that for any two queries the strings m_i from these queries are different.
Input
The first line contains string s (1 ≤ \left | s \right | ≤ 10^{5}).
The second line contains an integer n (1 ≤ n ≤ 10^5).
Each of next n lines contains an integer k_i (1 ≤ k_i ≤ |s|) and a non-empty string m_i — parameters of the query with number i, in this order.
All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed 10^5. All m_i are distinct.
Output
For each query output the answer for it in a separate line.
If a string m_{i} occurs in s less that k_{i} times, output -1.
Examples
Input
aaaaa
5
3 a
3 aa
2 aaa
3 aaaa
1 aaaaa
Output
3
4
4
-1
5
Input
abbb
7
4 b
1 ab
3 bb
1 abb
2 bbb
1 a
2 abbb
Output
-1
2
-1
3
-1
1
-1
Submitted Solution:
```
def occurences(s,y):
count=0
while(s.find(y)!=-1):
a = s.find(y)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
#s = "".join(s)
return count
def n_occurences(s,y,n):
count=0
temp=0
while(s.find(y)!=-1):
a = s.find(y)
if count!=0:
temp = temp+abs(a)
else:
temp = temp
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[a+1:]
if(count==n):
break
#s = "".join(s)
return temp
def n_occurences_right(s,y,n):
count=0
temp=0
while(s.rfind(y)!=-1):
a = s.rfind(y)
#print(a)
if count==0:
#temp = temp+ (((len(s)-(a))-(len(y)-1)) - ((len(s)-(a+1))+(len(y)-1)))
temp=temp+1
else:
temp = temp+ (len(s)-(a))-(len(y)-1)
#print(temp)
#print(temp)
count=count+1
#s=list(s)
"""for i in range(0,a+1):
s.remove(s[i])"""
s = s[:a+len(y)-1]
if(count==n):
break
#s = "".join(s)
return temp
s = input("Enter the string: ")
n = int(input("Enter the number of queries: "));
k=[]
m=[]
count=0
for i in range(0,n):
s2 = s
x,y = input().split();
count=0
#k.append(int(x));
#m.append(y)
x=int(x)
if len(y)>1:
mint = x*len(y)
t= y*x
if y[0]==y[len(y)-1]:
for i in range (0,(len(y)-1)):
if y[i] == y[len(y)-1-i]:
if i!=(len(y)-1-i) or (count == i and y[i]==y[i-1]):
count=count+1
#print(y[i])
else:
break
#print(count)
#print(count)
p = list(y)
for i in range (0,count):
p.remove(p[0])
y = list(y)
t = y+ p*(x-1)
#print(t)
q = ""
t = q.join(t)
mint = len(t)
y= "".join(y)
#print(t)
else:
mint = x
t = y*x
if t not in s or occurences(s2,y)<x:
s2 = s
num = occurences(s2,y)
#print(num)
if(num>=x):
#print(n_occurences(s,y,x))
a = n_occurences(s2,y,x)+len(y)+(x-1)
#print(a)
b = n_occurences_right(s2,y,x) + (len(y)-1)
#print(b)
print(min(a,b))
#print(x)
else:
print("-1")
else:
print(len(t))
#print("yess".rfind("s"))
``` | instruction | 0 | 29,730 | 0 | 59,460 |
No | output | 1 | 29,730 | 0 | 59,461 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,851 | 0 | 59,702 |
"Correct Solution:
```
from collections import Counter
a = input()
n = len(a)
c = Counter()
ans = 1
c[a[0]] = 1
for i in range(1, n):
ans += i - c[a[i]]
c[a[i]] += 1
print(ans)
``` | output | 1 | 29,851 | 0 | 59,703 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,852 | 0 | 59,704 |
"Correct Solution:
```
import collections
a = list(input())
c = collections.Counter(a)
n = len(a)
ans = n*(n-1)//2 + 1
for v in c.values():
ans -= v*(v-1)//2
print(ans)
``` | output | 1 | 29,852 | 0 | 59,705 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,853 | 0 | 59,706 |
"Correct Solution:
```
from collections import Counter
s = input()
n = len(s)
print(n * ~-n // 2 + 1 - sum(i * ~-i // 2 for i in Counter(s).values()))
``` | output | 1 | 29,853 | 0 | 59,707 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,854 | 0 | 59,708 |
"Correct Solution:
```
import collections
A = input()
n = len(A)
c = collections.Counter(A)
ans = 0
for e in c:
ans += c[e] * (n - c[e])
ans = ans // 2 + 1
print(ans)
``` | output | 1 | 29,854 | 0 | 59,709 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,855 | 0 | 59,710 |
"Correct Solution:
```
a=input()
l=len(a)
b=l*(l-1)//2+1
for i in range(97,123):
c=a.count(chr(i))
d=(c*(c-1))//2
b-=d
print(b)
``` | output | 1 | 29,855 | 0 | 59,711 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,856 | 0 | 59,712 |
"Correct Solution:
```
A=input()
dp=[0]*26
r=0
for i,a in enumerate(A):
c=ord(a)-97
r+=i-dp[c]
dp[c]+=1
print(r+1)
``` | output | 1 | 29,856 | 0 | 59,713 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,857 | 0 | 59,714 |
"Correct Solution:
```
A = input()
_memo = {}
ans = 1
for i,a in enumerate(A):
ans += i - _memo.setdefault(a,0)
_memo[a] += 1
#print(ans)
#print(_memo)
print(ans)
``` | output | 1 | 29,857 | 0 | 59,715 |
Provide a correct Python 3 solution for this coding contest problem.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44 | instruction | 0 | 29,858 | 0 | 59,716 |
"Correct Solution:
```
a = list(input())
l = len(a)
alphcnt = [0] * 26
for i in a:
alphcnt[ord(i) - 97] += 1
ans = (l * (l + 1)) // 2 + 1
for i in alphcnt:
ans -= (i * (i + 1)) // 2
print(ans)
``` | output | 1 | 29,858 | 0 | 59,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
import collections as c,math
f=math.factorial
s=input()
n=len(s)
a=n*-~n//2+1
for i in c.Counter(s).values():
a-=i
if i>1:
a-=f(i)//f(i-2)//2
print(a)
``` | instruction | 0 | 29,859 | 0 | 59,718 |
Yes | output | 1 | 29,859 | 0 | 59,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
A = input()
count = [0]*26
ans = 1
count[ord(A[0])-97]+=1
for i in range(1,len(A)):
ans += i-count[ord(A[i])-97]
count[ord(A[i])-97]+=1
print(ans)
``` | instruction | 0 | 29,860 | 0 | 59,720 |
Yes | output | 1 | 29,860 | 0 | 59,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
from itertools import groupby
A=input()
N=len(A)
A=[A[i] for i in range(0,N)]
A.sort()
data=groupby(A)
ans=N*(N-1)//2+1
for key, group in data:
g=len(list(group))
ans-=g*(g-1)//2
print(ans)
``` | instruction | 0 | 29,861 | 0 | 59,722 |
Yes | output | 1 | 29,861 | 0 | 59,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
A = input()
n = len(A)
ans = (n-1)*n//2+1
alp = 'abcdefghijklmnopqrstuvwxyz'
for i in alp:
num = A.count(i)
ans -= num*(num-1)//2
print(ans)
``` | instruction | 0 | 29,862 | 0 | 59,724 |
Yes | output | 1 | 29,862 | 0 | 59,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
strs = input()
N = 1
aaa = set()
for i in range(2, len(strs)+1):
for j in range(len(strs)- i+1):
strspart = strs[j:j + i]
aaa.add(strs[0:j] +strspart[::-1] + strs [j+i:])
print(len(aaa))
``` | instruction | 0 | 29,863 | 0 | 59,726 |
No | output | 1 | 29,863 | 0 | 59,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
def main():
s = list(input())
results=[s]
for i in range(1,len(s)):
for j in range(0,len(s)-i):
temp = s.copy()
temp[j:j+i+1] = reversed(temp[j:j+i+1])
if temp not in results:
results.append(temp)
print(len(results))
if __name__=="__main__":
main()
``` | instruction | 0 | 29,864 | 0 | 59,728 |
No | output | 1 | 29,864 | 0 | 59,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.
You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.
You can perform this operation at most once.
How many different strings can you obtain?
Constraints
* 1 \leq |A| \leq 200,000
* A consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
A
Output
Print the number of different strings you can obtain by reversing any substring in A at most once.
Examples
Input
aatt
Output
5
Input
xxxxxxxxxx
Output
1
Input
abracadabra
Output
44
Submitted Solution:
```
st = input()
s = []
for i in range(2, len(st)):
for j in range(len(st)+1-i):
s.append([st[:j]+st[j:j+i][::-1]+st[j+i:], j, j+i])
s2 = [st]
for i in range(len(s)):
if not s[i] in s2:
s2.append(s[i])
print(len(s2))
``` | instruction | 0 | 29,865 | 0 | 59,730 |
No | output | 1 | 29,865 | 0 | 59,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,297 | 0 | 60,594 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
for _ in range(int(input())):
s1=input()
s2=input()
st1=""
st2=""
if len(s1) == len(s2):
if s1==s2:
print(s1)
else:
print(-1)
continue
if len(s1)<len(s2):
st1 = s1
st2 = s2
else:
st1 = s2
st2 = s1
d=dict()
for i in range(1,21):
x=s2*i
d[x] = 1
ans=-1
for i in range(1,21):
x=s1*i
if d.get(x):
ans=x
break
print(ans)
``` | output | 1 | 30,297 | 0 | 60,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,298 | 0 | 60,596 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
def qu(s):
l = len(s)
for i in range(1, l + 1):
if l % i != 0: continue
if s[:i] * (l // i) == s: return s[:i]
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def lcm(a, b):
return a // gcd(a, b) * b
t = int(input())
for i in range(t):
s = input()
t = input()
j_s = qu(s)
j_t = qu(t)
if j_s != j_t:
print("-1")
else:
print(j_s * lcm(len(s)//len(j_s), len(t)//len(j_t)))
``` | output | 1 | 30,298 | 0 | 60,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,299 | 0 | 60,598 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
def gcd(a,b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a,b):
return (a / gcd(a,b))* b
t = int(input())
for _ in range(t):
s = input()
t = input()
l = int(lcm(len(s),len(t)))
a1 = s*(l//len(s))
a2 = t*(l//len(t))
if a1 == a2:
print(a1)
else:
print(-1)
``` | output | 1 | 30,299 | 0 | 60,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,300 | 0 | 60,600 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
#from math import *
def lcm(x, y):
"""This function takes two
integers and returns the L.C.M."""
# Choose the greater number
if x > y:
greater = x
else:
greater = y
while(True):
if((greater % x == 0) and (greater % y == 0)):
lcm = greater
break
greater += 1
return lcm
for _ in range(int(input())):
a = input()
b = input()
l1 = len(a)
l2 = len(b)
if(l1==l2):
if(a==b):
print(a)
else:
print(-1)
else:
l = lcm(l1,l2)
ans = ""
k=0
if(l2>l1):
b,a = a,b
n = len(b)
for _ in range(l):
a1 = k%n
ans=ans+b[a1]
k+=1
f=0
k=0
n = len(a)
for i in range(l):
a1 = k%n
if(ans[i]!=a[a1]):
f=1
break
k+=1
if(f==1):
print(-1)
else:
print(ans)
``` | output | 1 | 30,300 | 0 | 60,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,301 | 0 | 60,602 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
import sys,collections,math
sys.setrecursionlimit(1000)
#######################################3
def in_out():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
# in_out()
#######################################
for _ in range(int(input())):
# n = int(input())
s = input()
t = input()
if s==t:
print(s)
continue
if len(s)>len(t):
s,t =t,s
m = len(s)
n = len(t)
if s*n != t*m:
print(-1)
elif n % m == 0:
print(t)
else:
lcm = (m*n) // math.gcd(m,n)
print(s*(lcm//m))
``` | output | 1 | 30,301 | 0 | 60,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,302 | 0 | 60,604 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
def gcd(a,b):
if a == 0:
return b
return gcd(b % a, a)
# Function to return LCM of two numbers
def compute_lcm(a,b):
return (a / gcd(a,b))* b
def main():
T = int(input())
for c in range(T):
s1 = input().rstrip()
s2 = input().rstrip()
s = s1 if len(s1) <= len(s2) else s2
l = s2 if len(s1) <= len(s2) else s1
bs = [0 for i in range(len(s))]
bl = [0 for i in range(len(l))]
pl = 0
res = True
while True:
ps = 0
bl[pl] = 1
while ps < len(s):
if s[ps] != l[pl]:
res = False
break
ps += 1
pl = (pl + 1) % len(l)
if res == False:
break
if bl[pl] == 1:
res = (pl == 0)
break
if res == False:
print("-1")
else:
lcm = compute_lcm(len(s), len(l))
ans = ""
for c in range(int(lcm/len(s))):
ans += s
print(ans)
main()
``` | output | 1 | 30,302 | 0 | 60,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,303 | 0 | 60,606 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
#start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
def isInt(s):
return '0' <= s[0] <= '9'
MOD = 10**9 + 7
def solve():
S = getStr()
T = getStr()
N = len(S)
M = len(T)
LCM = N*M//math.gcd(N,M)
ans = (LCM//N)*S
if ans == (LCM//M)*T: return ans
return -1
for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)
``` | output | 1 | 30,303 | 0 | 60,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2. | instruction | 0 | 30,304 | 0 | 60,608 |
Tags: brute force, math, number theory, strings
Correct Solution:
```
import sys
from math import gcd
zz=1
sys.setrecursionlimit(10**5)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]
def fori(n):
return [fi() for i in range(n)]
def inc(d,c,x=1):
d[c]=d[c]+x if c in d else x
def ii():
return input().rstrip()
def li():
return [int(xx) for xx in input().split()]
def fli():
return [float(x) for x in input().split()]
def comp(a,b):
if(a>b):
return 2
return 2 if a==b else 0
def gi():
return [xx for xx in input().split()]
def gtc(tc,ans):
print("Case #"+str(tc)+":",ans)
def cil(n,m):
return n//m+int(n%m>0)
def fi():
return int(input())
def pro(a):
return reduce(lambda a,b:a*b,a)
def swap(a,i,j):
a[i],a[j]=a[j],a[i]
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def isvalid(i,j,n,m):
return 0<=i<n and 0<=j<m
def bo(i):
return ord(i)-ord('a')
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
t=fi()
uu=t
def check(s,b,m):
return m%len(s)==0 and s*(m//len(s))==b
while t>0:
t-=1
a=ii()
b=ii()
n=len(a)
m=len(b)
ans=0
for i in range(1,n+1):
if check(a[:i],b,m) and check(a[:i],a,n):
ans=i
break
if ans==0:
print(-1)
else:
p=n//i
q=m//i
w=(p*q)//gcd(p,q)
print(a[:i]*w)
``` | output | 1 | 30,304 | 0 | 60,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
from sys import stdin
from sys import stdout
from math import *
import sys
# sys.setrecursionlimit(int(1e5))
# input = stdin.readline
# print = stdout.write
for __ in range(int(input())):
s=input()
t=input()
a=len(s)
b=len(t)
lcm = (a*b)//gcd(a,b)
m=lcm//a
ans=s*m
pos=1
for i in range(len(ans)):
if(ans[i]!=s[i%a] or ans[i]!=t[i%b]):
pos=0
break
if(pos==0):
print(-1)
else:
print(ans)
``` | instruction | 0 | 30,305 | 0 | 60,610 |
Yes | output | 1 | 30,305 | 0 | 60,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
for _ in range(int(input())):
s = input()
t = input()
m, n = len(s), len(t)
ans = ""
i, j = 0, 0
while s[i] == t[j]:
ans += s[i]
i += 1
j += 1
if i == m and j == n:
break
if i == m:
i = 0
if j == n:
j = 0
if i == m:
i = 0
if j == n:
j = 0
if s[i] == t[j]:
print(ans)
else:
print(-1)
``` | instruction | 0 | 30,306 | 0 | 60,612 |
Yes | output | 1 | 30,306 | 0 | 60,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
from sys import stdin
from math import gcd
input = lambda: stdin.readline().strip()
ipnut = input
for i in range(int(input())):
s = input()
t = input()
a = len(s)
b = len(t)
l = a*b//gcd(a,b)
s1 = s*(l//a)
t1 = t*(l//b)
if s1==t1:
print(s1)
else:
print(-1)
``` | instruction | 0 | 30,307 | 0 | 60,614 |
Yes | output | 1 | 30,307 | 0 | 60,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
import math
n = int(input())
def lcm(x, y):
if x > y:
greater = x
else:
greater = y
while(True):
if((greater % x == 0) and (greater % y == 0)):
lcm = greater
break
greater += 1
return lcm
for i in range(n):
a = input()
b = input()
c = lcm(len(a), len(b))
a1 = ""
b1 = ""
for j in range(c // len(a)):
a1 += a;
for j in range(c // len(b)):
b1 += b;
if(a1 == b1):
print(a1)
else:
print(-1)
``` | instruction | 0 | 30,308 | 0 | 60,616 |
Yes | output | 1 | 30,308 | 0 | 60,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
from math import gcd
for _ in range(int(input())):
a = input()
b = input()
if len(b) > len(a): a, b = b, a
if set(a) != set(b):
print(-1)
else:
boo = False
x = len(a)
y = len(b)
xy = (x * y) // gcd(x, y)
print(xy,a,b)
if a * (xy // x) == b * (xy // y):
boo = True
print(a * (xy // x) if boo else -1)
``` | instruction | 0 | 30,309 | 0 | 60,618 |
No | output | 1 | 30,309 | 0 | 60,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
t=int(input())
while t>0:
t=t-1
s=input()
n=input()
l=[]
l2=[]
for i in range(1,20):
l.append(i*s)
for i in range(1,20):
l2.append(i*n)
l3=[]
for i in l:
for j in l2:
if i==j:
l3.append(i)
if len(l3)!=0:
print(l3[0])
else:
print(-1)
``` | instruction | 0 | 30,310 | 0 | 60,620 |
No | output | 1 | 30,310 | 0 | 60,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
n = int(input())
def find_lcm(s_1, s_2):
n_1 = len(s_1)
n_2 = len(s_2)
if n_1 == n_2:
if s_1 == s_2:
return s_1
else:
return - 1
if n_1 > n_2:
if n_1 % n_2 == 0:
if s_2 * (n_1 // n_2) == s_1:
return s_1
else:
return -1
else:
if (s_1 * n_2) == (s_2 * n_1):
return s_1 * n_2
else:
return -1
elif n_2 > n_1:
if n_2 % n_1 == 0:
if s_1 * (n_2 // n_1) == s_2:
return s_1
else:
return -1
else:
if (s_1 * n_2) == (s_2 * n_1):
return s_1 * n_2
else:
return -1
for i in range(n):
s_1 = input()
s_2 = input()
print(find_lcm(s_1, s_2))
``` | instruction | 0 | 30,311 | 0 | 60,622 |
No | output | 1 | 30,311 | 0 | 60,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a multiplication operation between a string a and a positive integer x: a ⋅ x is the string that is a result of writing x copies of a one after another. For example, "abc" ⋅~2~= "abcabc", "a" ⋅~5~= "aaaaa".
A string a is divisible by another string b if there exists an integer x such that b ⋅ x = a. For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings s and t (defined as LCM(s, t)) is the shortest non-empty string that is divisible by both s and t.
You are given two strings s and t. Find LCM(s, t) or report that it does not exist. It can be shown that if LCM(s, t) exists, it is unique.
Input
The first line contains one integer q (1 ≤ q ≤ 2000) — the number of test cases.
Each test case consists of two lines, containing strings s and t (1 ≤ |s|, |t| ≤ 20). Each character in each of these strings is either 'a' or 'b'.
Output
For each test case, print LCM(s, t) if it exists; otherwise, print -1. It can be shown that if LCM(s, t) exists, it is unique.
Example
Input
3
baba
ba
aa
aaa
aba
ab
Output
baba
aaaaaa
-1
Note
In the first test case, "baba" = "baba" ⋅~1~= "ba" ⋅~2.
In the second test case, "aaaaaa" = "aa" ⋅~3~= "aaa" ⋅~2.
Submitted Solution:
```
for q in range(int(input())):
s = str(input())
t = str(input())
for i in range(1, len(s) + 1):
if len(s) % i == 0:
if s[:i] * (len(s) // i) == s:
first = s[:i]
break
for i in range(1, len(t) + 1):
if len(t) % i == 0:
if s[:i] * (len(t) // i) == t:
second = t[:i]
break
if first == second:
a = len(s) // len(first)
b = len(t) // len(second)
for i in range(max(a, b), a * b + 1):
if i % a == 0 and i % b == 0:
print(first * i)
else:
print(-1)
``` | instruction | 0 | 30,312 | 0 | 60,624 |
No | output | 1 | 30,312 | 0 | 60,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s of length n consisting only of the characters 0 and 1.
You perform the following operation until the string becomes empty: choose some consecutive substring of equal characters, erase it from the string and glue the remaining two parts together (any of them can be empty) in the same order. For example, if you erase the substring 111 from the string 111110, you will get the string 110. When you delete a substring of length l, you get a ⋅ l + b points.
Your task is to calculate the maximum number of points that you can score in total, if you have to make the given string empty.
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of testcases.
The first line of each testcase contains three integers n, a and b (1 ≤ n ≤ 100; -100 ≤ a, b ≤ 100) — the length of the string s and the parameters a and b.
The second line contains the string s. The string s consists only of the characters 0 and 1.
Output
For each testcase, print a single integer — the maximum number of points that you can score.
Example
Input
3
3 2 0
000
5 -2 5
11001
6 1 -4
100111
Output
6
15
-2
Note
In the first example, it is enough to delete the entire string, then we will get 2 ⋅ 3 + 0 = 6 points.
In the second example, if we delete characters one by one, then for each deleted character we will get (-2) ⋅ 1 + 5 = 3 points, i. e. 15 points in total.
In the third example, we can delete the substring 00 from the string 100111, we get 1 ⋅ 2 + (-4) = -2 points, and the string will be equal to 1111, removing it entirely we get 1 ⋅ 4 + (-4) = 0 points. In total, we got -2 points for 2 operations. | instruction | 0 | 30,333 | 0 | 60,666 |
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n, a, b = map(int, input().split())
prev, cnt = 0, 0
for c in input().rstrip():
if c != prev:
prev, cnt = c, cnt+1
cnt = cnt // 2 + 1
if b < 0:
print(n*a + cnt*b)
else:
print(n * (a + b))
``` | output | 1 | 30,333 | 0 | 60,667 |
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