**Repeated String**

### Problem Statement :

There is a string, s, of lowercase English letters that is repeated infinitely many times. Given an integer, n, find and print the number of letter a's in the first n letters of the infinite string. Example s = 'abcac' n = 10 The substring we consider is abcacabcac, the first 10 characters of the infinite string. There are 4 occurrences of a in the substring. Function Description Complete the repeatedString function in the editor below. repeatedString has the following parameter(s): s: a string to repeat n: the number of characters to consider Returns int: the frequency of a in the substring Input Format The first line contains a single string, s. The second line contains an integer, n. Constraints 1 <= | s | <= 100 1 <= n <= 10^12 For 25% of the test cases, n <= 10^6 .

### Solution :

` ````
Solution in C :
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
char* s = (char *)malloc(512000 * sizeof(char));
scanf("%s",s);
long n,o,p,i;
scanf("%ld",&n);
o=0;
for(i=0;s[i]!='\0';i++)
{
if(s[i]=='a')
o++;
}
p=n%i;
n=n/i;
o=o*n;
n=0;
for(i=0;i<p;i++)
if(s[i]=='a')
n++;
printf("%ld",o+n);
return 0;
}
```

` ````
Solution in C++ :
In C ++ :
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <set>
#include <string>
#include <map>
#include <ctime>
#include <cstdlib>
#include <unordered_set>
using namespace std;
typedef long long ll;
const int MaxN = 100;
char s[MaxN + 1];
int main() {
ll n;
scanf ("%s", s);
int len = (int)strlen(s);
scanf ("%lld", &n);
ll tot = n / len;
ll as = 0;
for (int i = 0; i < len; ++i)
if (s[i] == 'a')
++as;
ll res = tot * as;
for (int i = 0; i < n % len; ++i)
if (s[i] == 'a')
++res;
printf ("%lld", res);
}
```

` ````
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String s = in.next();
long n = in.nextLong();
long num = n/s.length();
long rem = n%s.length();
long ans = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i)=='a') {
ans += num;
if (i < rem)
ans++;
}
}
System.out.println(ans);
}
}
```

` ````
Solution in Python :
In Python3 :
#!/bin/python3
import math
import os
import random
import re
import sys
# Complete the repeatedString function below.
def repeatedString(s, n):
return (s.count('a') * (n//len(s))+s[:n%len(s)].count('a'))
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
s = input()
n = int(input())
result = repeatedString(s, n)
fptr.write(str(result) + '\n')
fptr.close()
```

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