(***leetcode_Math)Factorial Trailing Zeroes
来源：互联网 发布：resset数据库 编辑：程序博客网 时间：2024/06/03 09:42
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
结尾0只能由因素5、10、15得到，注意就是25,125中5的个数！
class Solution {public: int trailingZeroes(int n) { int ret = 0; while(n){ ret += n/5; n /=5; } return ret; }}; 0 0
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