Repeated string Hackerrank

Repeated String solution in c.

Lilah has a string ,s , of lowercase English letters that she repeated infinitely many times.

Given an integer,n , find and print the number of letter a's in the first  n letters of Lilah's infinite string.

For example, if the string s='abcac'  and n=10 , the substring we consider is abcacabcac, the first 10 characters of her infinite string. There are 4  occurrences of a in the substring.

Function Description

Complete the repeatedString function in the editor below. It should return an integer representing the number of occurrences of a in the prefix of length  n in the infinitely repeating string.

repeatedString has the following parameter(s):

• s: a string to repeat
• n: the number of characters to consider

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 .

Output Format

Print a single integer denoting the number of letter a's in the first  letters of the infinite string created by repeating  infinitely many times.

Sample Input 0

aba
10

Sample Output 0

7

Explanation 0

The first n=10 letters of the infinite string are abaabaabaa. Because there are 7 a's, we print 7 on a new line.

Sample Input 1

a
1000000000000

Sample Output 1

1000000000000

Explanation 1

Because all of the first n=100000000000 letters of the infinite string are a, we print 100000000000 on a new line.

Solutions is below:

#include <assert.h>

#include <limits.h>

#include <math.h>

#include <stdbool.h>

#include <stddef.h>

#include <stdint.h>

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

int main( void ) {

    char s[1000];

    long long n;

    

    scanf("%s %lld", s, &n);

    

    int m = strlen(s);

    

    long long count = 0;

    for( int i = 0; i < m; i++ ) {

        if( s[i] == 'a' ) count++;

    }

    

  count= count*(n/m);

       

    for( int i = 0; i < n % m; i++ ) {

        if( s[i] == 'a' ) count++;

    }

    

    printf("%lld\n", count);

    

    return 0;

}