Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

The idea is:

The ZERO comes from 10.
The 10 comes from 2 x 5
And we need to account for all the products of 5 and 2. likes 4×5 = 20 ...
So, if we take all the numbers with 5 as a factor, we'll have way more than enough even numbers to pair with them to get factors of 10
Example One

How many multiples of 5 are between 1 and 23? There is 5, 10, 15, and 20, for four multiples of 5. Paired with 2's from the even factors, this makes for four factors of 10, so: 23! has 4 zeros.

Example Two

How many multiples of 5 are there in the numbers from 1 to 100?

because 100 ÷ 5 = 20, so, there are twenty multiples of 5 between 1 and 100.

but wait, actually 25 is 5×5, so each multiple of 25 has an extra factor of 5, e.g. 25 × 4 = 100，which introduces extra of zero.

So, we need know how many multiples of 25 are between 1 and 100? Since 100 ÷ 25 = 4, there are four multiples of 25 between 1 and 100.

Finally, we get 20 + 4 = 24 trailing zeroes in 100!

The above example tell us, we need care about 5, 5×5, 5×5×5, 5×5×5×5 ....

Example Three

By given number 4617.

5^1 : 4617 ÷ 5 = 923.4, so we get 923 factors of 5

5^2 : 4617 ÷ 25 = 184.68, so we get 184 additional factors of 5

5^3 : 4617 ÷ 125 = 36.936, so we get 36 additional factors of 5

5^4 : 4617 ÷ 625 = 7.3872, so we get 7 additional factors of 5

5^5 : 4617 ÷ 3125 = 1.47744, so we get 1 more factor of 5

5^6 : 4617 ÷ 15625 = 0.295488, which is less than 1, so stop here.

Then 4617! has 923 + 184 + 36 + 7 + 1 = 1151 trailing zeroes.

class Solution {
public:
    int trailingZeroes(int n) {
        if(n < 0)   return 0;

        int res = 0;
        while(n != 0){
            res += n/5;
            n /= 5;
        }

        return res;
    }
};

求n！中二进制表示中最低位1所在的位置，最后一个1所在位置取决于1~n的数中因子2有多少个。

int rightOne(int num){
  int two = 0;

  while(num >= 2){
    num >>= 1;
    two+num;
  }
  return res;
}