Use Python 3:

s, n = input().strip(), int(input().strip())
print(s.count("a") * (n // len(s)) + s[:n % len(s)].count("a"))

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    String s = in.next();
    long n = in.nextLong();
    long count =0;
    for(char c : s.toCharArray())
        if(c == 'a')
        count++;

     long factor = (n/s.length());
     long rem = (n%s.length());
     count =  factor*count  ;
    for(int i=0;i<rem;i++)
        if(s.charAt(i)=='a')
                count++;        
    System.out.println(count);



}

Java Solution

static long count(String s, long n) {
    long numOfS = n/s.length();
    long rest = n % s.length();

    if(!s.contains("a")) return 0;
    return s.length()>n? aCounter(s, rest) 
            : numOfS*aCounter(s, s.length()) + aCounter(s, rest);
}

    private static long aCounter(String s, long end) {
        int a=0;
        for (int i = 0; i < end; i++) {
            if (s.charAt(i) == 'a') a++;
        }
        return a;
}

Efficient JavaScript solution

The algorithm doesn't count the string a second time, both best and worst case is O(n) where n = length of the string

From my repo https://github.com/gabrielgiordan/HackerRank

function repeatedString(s, n) {
  let c = 0,
      ca = 0,
      r = n % s.length

  for (let i = s.length; i-- > 0;) {
    if (s.charAt(i) == 'a') {
      ++c

      if (i < r)
        ++ca
    }
  }

  return ((n - r) / s.length * c) + ca
}

my python3 code easy to understand

def repeatedString(s, n):
    c = s.count('a')
    div=n//len(s)
    if n%len(s)==0:
        c= c*div
    else:
        m = n%len(s)
        c= c*div+s[:m].count('a')
    return c