Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

类似于cc：http://www.cnblogs.com/wuchanming/p/4158567.html

5!, 包含1*5, 1个5
10!, 包含1*5,2*5, 2个5
15!, 包含1*5,2*5,3*5, 3个5
20!, 包含1*5,2*5,3*5,4*5, 4个5
25!, 包含1*5,2*5,3*5,4*5,5*5, 6个5

C++实现代码：

#include<iostream>
using namespace std;

class Solution {
public:
    int trailingZeroes(int n) {
        if(n==0)
            return 0;
        int count=0;
        while((n=n/5)>0)
        {
            count+=n;
        }
        return count;
    }
};

int main()
{
    Solution s;
    cout<<s.trailingZeroes(25)<<endl;
}