Factorial Trailing Zeroes -- leetcode

Factorial Trailing Zeroes -- leetcode

 

Given an integer n, return the number of trailing zeroes in n !.

Note: Your solution shocould be in logarithmic time complexity.

 

Basic Ideas:

Count n! The number of tails 0.

The ending number 0 is obtained by the product of 2, 5.

To count the number of 0, you need to count the number of 2, 5 factors. Because the number of 2 is more than 5, you only need to count the number of 5 factors.

For example, 5, 10, 15, 15, 20 n * 5, contains a 5.

25, 50, 75, n * 25, containing 2 5.

N * 125 contains 3 5.

...

Because the latter has an intersection with the former, the former has been counted. Therefore, you only need to add the latter one more time.

 

return n/5 + n/25 + n/125 + n/625 + n/3125+...;

 

Or write:

N/5 + (n/5)/5 + (n/5/5)/5 + (n/5/5/5)/5

 

 

class Solution {public:    int trailingZeroes(int n) {        int ans = 0;        int count = 0;        while (n) {            n /= 5;            ans += n;        }        return ans;    }};

 

return n/5 + n/25 + n/125 + n/625 + n/3125+...;