Nyoj84 factorial 0

0 time limit of factorial: 3000 MS | memory limit: 65535 kb difficulty: 3

Description

Calculate n! In decimal format.

Input

Enter an integer N in the first line to indicate the number of groups of test data (1 <= n <= 100)
Each group of test data occupies one row, with only one integer m (0 <= m <= 10000000)

Output

Number of the last 0 in the decimal number of the factorial output M
For example, 5! = 120 then the number of the last 0 is 1

Sample Input

63601001024234568735373

Sample output

0142425358612183837

Solution: in prime numbers, a "0" is displayed at the end only when 2 and 5 are multiplied, so as long as M is decomposed into prime factors, and then the number of 2 and 5 is counted, the smaller one is the answer.
Further, after M decomposition, the number of 2 is definitely more than 5,
Then the problem is further simplified, as long as the number of five of all prime factors is counted.

For example:

1 --> multiples of 5 in 100

There are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.

20 5

However, after dividing by 5, there are still 5

5 10 15 20

At this time, there are 4 5

That is to say, 1 --> 100 has a total of 24 5; that is, the number of 0.

The Code uses recursion to calculate the number of 5:

# Include <stdio. h> int COUNT = 0; // count int factorial (INT m) {If (M = 0) return count; else {count + = m/5; return factorial (M/5) ;}} int main () {int m; int N; scanf ("% d", & N); While (n --) {COUNT = 0; scanf ("% d", & M); printf ("% d \ n", factorial (m )); // recursive operation Implementation} return 0 ;}

Nyoj84 factorial 0