Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
（1）
class Solution { public: int trailingZeroes(int n) { int ans = 0; for(long long i = 5; n / i; i *= 5) { ans += n / i; } return ans; } };
（2）
class Solution { public: int trailingZeroes(int n) { int ans = 0; while(n) { int t = n / 5; ans += t; n = t; } return ans; } };
172. Factorial Trailing Zeroes -- 求n的阶乘末尾有几个0
原文：http://www.cnblogs.com/argenbarbie/p/5800403.html