LeetCode之172 Factorial Trailing Zeroes

就是找一个数的阶乘后面有多少个零。

逻辑很有意思

原题：

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

Note: Your solution should be in logarithmic time complexity.

AC后的代码：

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