calculate factor count of k in n!

Define a function Z(n, k), Z(n, k) is the factor count of k exist in n!.
For example, Z(4, 2) = 3, as there are three factors of 2 in 4!.

In a naive calculation, the time complexity to calculate Z(n, k) is O(nlogn).

int Z(int n, int k) {
  int result = 0;
  for (int i = 2; i <= n; ++i) {
    int temp = i;
    while (temp % k == 0) {
       temp /= k;
       ++result;
    }
  }
  return result;
}

But there is a simple way to generate the result. in 1...n, there are n/k numbers which are dividable by k, there are n/(k^2) numbers which are dividable by k^2, and so on. So we can generate Z(n, k) like below, in O(log(n)) time.

int Z(int n, int k) {
  int result = 0;
  int temp = k;
  while (n / temp >= k) { // Take care temp shouldn't become negative or overflow.
     result += n / temp;
     temp *= k;
  }
  result += n / temp;
  return result;
}