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