【Leetcode】 Factorial Trailing Zeroes #172
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Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
求末尾0的个数，即为求各factor中5的个数，因为每有一个5必有一个2与之组成10。其中factor 5^n 应被计算n次，因为其贡献了n个5.
5的倍数有：a = n/5
则25的倍数有：b = a/5
125的倍数有：c=b/5
....
根据此思路，得：
class Solution {public: int trailingZeroes(int n) { int number_of_trailing_zero = 0; int num_of_5 = n/5; while(num_of_5 != 0) { number_of_trailing_zero += num_of_5; num_of_5 /= 5; } return number_of_trailing_zero; }}; 0 0
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