Trailing Zeros

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

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Example

11! = 39916800, so the out should be 2

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一开始没想明白，后来随手列了一些相乘尾数能等于0的数，2*5，4*5，5*6，5*8。发现都是与5相关。所以其实就是求5的数量

class Solution {
public:
    /*
     * @param n: A long integer
     * @return: An integer, denote the number of trailing zeros in n!
     */
    long long trailingZeros(long long n) {
        // write your code here, try to do it without arithmetic operators.
      long long sum=0;
      while(n){
          sum+=n/5;
          n=n/5;
      }
      return sum;
    }
};