Write an algorithm which computes the number of trailing zeros in n factorial.
Example
11! = 39916800, so the out should be 2
Challenge
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一开始没想明白，后来随手列了一些相乘尾数能等于0的数，2*5，4*5，5*6，5*8。发现都是与5相关。所以其实就是求5的数量
class Solution {
public:/** @param n: A long integer* @return: An integer, denote the number of trailing zeros in n!*/long long trailingZeros(long long n) {// write your code here, try to do it without arithmetic operators.long long sum=0;while(n){sum+=n/5;n=n/5;}return sum;}
};