题目描述：
Given an integer n, return the number of trailing zeroes in n!.
Example 1:
Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.
Example 2:
Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.
Note: Your solution should be in logarithmic time complexity.
想要尾数为0，必须要有2、5或者10相乘才行，而5的个数肯定小于2，所以只需要统计1~n的范围内，尾数为5的个数和10的个数。
class Solution {
public:
int trailingZeroes(int n) {
int result=0;
while(n>0)
{
result+=n/5;
n/=5;
}
return result;
}
};
