# 172.Factorial Trailing Zeroes

**172.Factorial Trailing Zeroes**

难度:Easy

> 给定一个整数 n，返回 n! 结果尾数中零的数量。

```
示例 1:

输入: 3
输出: 0
解释: 3! = 6, 尾数中没有零。
示例 2:

输入: 5
输出: 1
解释: 5! = 120, 尾数中有 1 个零.
说明: 你算法的时间复杂度应为 O(log n) 。
```

方法：主要是找出5的个数，代码如下：

```
class Solution {
public:
    int trailingZeroes(int n) {
        int t=n/5;
        int total=t;
        while(t>4)
        {
            t/=5;
            total+=t;
        }

        return total;
    }
};
```


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