LeetCode : Factorial Trailing Zeroes
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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.
解释：
对n!做质因数分解n!=2x*3y*5z*…
显然0的个数等于min(x,z)，并且min(x,z)==z
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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