How many 0 are there at the end of the factorial of n?

In the joint iflab of the new question and answer on the book to see such a question, the effect is:

Give an integer n, ask N! How many 0 are at the end of (N's factorial)?

Because when n is very big, it is forced to calculate n! is impossible, so must find another way to solve the problem.

First of all, why would there be 0 at the end? Because 2*5 = 10,0 just came. So just ask out this n! How many of the 2 how many 5 times better, because 2 of the number of occurrences is certainly greater than 5, so only ask how many 5 to multiply the good.

Because the factorial of N is asked, and n! = 1*2*3*....*n

So: the number of N number that can be divisible by 5 = N/5

For example, n = 50, divisible by 5 has 5 10 15 20 25 30 35 40 45 50 total 10, i.e. 50/5=10

But don't forget 25 and 50, they can be split into 5*5 and 5*5*2, that is, the number that can be divisible by 5^2 will bring 2 0 (and so on 5 m-time will bring M 0 at the end).

So 50! At the end there is a 10+2=12 "0".

So the additional 5 of the number of M-order, we can from 5 to 5*5 to 5*5*5 ... Slowly looking for the big.

So the specific code is as follows (using the Web, 2L answer)

1 intTotalzero (intN)2 {3     intTotal =0;4      while(N >5)5     {6n = (n-(N)5)) /5;7Total + =N;8     }9     returnTotal ;Ten}

The nth cycle here is equivalent to the number of 5^n divisible by the sum, and the result is added.

The effect of the code is equivalent

First: number that can be divisible by 5 = N/5

Second time: number divisible by 25 = N/5/5

The third time .....

The fourth time ....

Although the number of n is changed after each cycle, the overall effect is equivalent to the number of 5^n divisible by the sum.

How many 0 are there at the end of the factorial of n?