[LeetCode] Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

解题思路：只有因子2和因子5相乘会产生10，同时因为因子2的数量大于因子5的数量，所以只需看序列中因子5的个数。它可以通过n/5得到，同时序列中还含有25,125，……这样的因子，其数量可以通过继续除5来得到。

实现代码：

/******************************************************************************  @COPYRIGHT NOTICE*  @Copyright (c) 2015, 楚兴*  @All rights reserved*  @Version  : 1.0*  @Author   : 楚兴*  @Date     : 2015/2/6 16:21*  @Status   : Accepted*  @Runtime  : 7 ms
*****************************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:int trailingZeroes(int n) {int count = 0;while (n){count += n / 5;n /= 5;}return count;}
};int main()
{Solution s;for (int i = 1; i < 100; i++){cout<<s.trailingZeroes(i)<<endl;}system("pause");
}

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