# 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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