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

Example 1:

Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:

Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.

Note: Your solution should be in logarithmic time complexity.

方法1：看n里面有多少个5,2的个数比5的个数多（逢二进一>逢五进一）

class Solution {
public:
    int trailingZeroes(int n) {
        int num=0;
        while(n>1){
            num+=n/5;
            n=n/5;
        }
        return num;
    }
};