设计一个算法，计算出n阶乘中尾部零的个数

样例 1:
输入: 11
输出: 2

样例解释: 
11! = 39916800, 结尾的0有2个。

样例 2:
输入: 5
输出: 1

样例解释: 
5! = 120， 结尾的0有1个。

public class Solution {
    /*
     * @param n: An integer
     * @return: An integer, denote the number of trailing zeros in n!
     */
    public long trailingZeros(long n) {
        // write your code here, try to do it without arithmetic operators.
     long count=0;
	long temp=n/5;
	while (temp!=0) 
	{
		count+=temp;
		temp/=5;
	}
	return count;
    }
}