lintcode-easy-Trailing Zeros

Write an algorithm which computes the number of trailing zeros in n factorial.

11! = 39916800, so the out should be 2



这道题的思路比较诡异，结尾要有0的话，必须要有质因数2和质因数5，而任何一个数的阶乘，质因数2的个数会比质因数5多很多，所以只需要算出有多少个质因数5就可以了

class Solution {
/*
* param n: As desciption
* return: An integer, denote the number of trailing zeros in n!
*/
public long trailingZeros(long n) {
// write your code here
if(n <= 0)
return 0;

long result = 0;

for(long i = 5; n / i > 0; i *= 5){
result += n / i;
}

return result;
}
};