172. 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.

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给出n，求n的阶乘的尾部0的个数。转换成统计5因子的个数。

代码：

class Solution {
public:
    int trailingZeroes(int n) {
    	int res = 0;
    	long long m = 5;
    	while(m <= n)
    	{
    		res += (n / m);
    		m *= 5;
		}
		return res;
    }
};