*Leetcode-Factorial Trailing Zeroes

想了很久才想清楚到底是有多少个5

最开始以为n/5就是0的个数，没有意识到25之中有两个5，所以应该是n一直除以5 直到为0，把得数加起来。

那个0的特殊条件不用考虑 是之前用log的时候残存的。。。。。

Multiple of 5 provides one 5, multiple of 25 provides two 5 and so on.

Note the duplication: multiple of 25 is also multiple of 5, so multiple of 25 only provides one extra 5.

Here is the basic solution:

return n/5 + n/25 + n/125 + n/625 + n/3125+...;

You can easily rewrite it to a loop.