Lintcode 2: 尾部的零

设计一个算法，计算出n阶乘中尾部零的个数

解析：仔细分析这个题，n！=1*2*,,,,,*n要产生0，就要有5，但是坑就在有25,125,,,,,这种多个5组成，考虑这个就ok啦。燃鹅人老了，忘了把计数器定义成long long类型。

c++版本：

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 count = 0;while (n>0){        n=n/5;count =count+ n;}return count;}};

Python版：

class Solution:    """    @param: n: An integer    @return: An integer, denote the number of trailing zeros in n!    """    def trailingZeros(self, n):        # write your code here, try to do it without arithmetic operators.        count=0        while n>0:            n=n/5            count+=n        return count