(***leetcode_Math)Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

结尾0只能由因素5、10、15得到，注意就是25,125中5的个数！

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