Problem:
Score : 400 points

Problem Statement

You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?

Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.

Note

When a positive integer A divides a positive integer B, A is said to a divisor of B.
For example, 6 has four divisors: 1, 2, 3 and 6.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go numbers that are divisors of N!.

Sample Input 1

9

Sample Output 1

0

There are no Shichi-Go numbers among the divisors of 9! = 1 \times 2 \times ... \times 9 = 362880.

Sample Input 2

10

Sample Output 2

1

There is one Shichi-Go number among the divisors of 10! = 3628800: 32400.

Sample Input 3

100

Sample Output 3

543

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

long int fac(int n){
	if(n>0){
		return n * fac(n-1);}
	else{
		return 1;
	}
}

int main(void)
{
	unsigned long int n;
	scanf("%lud", &n);
	n = fac(n);
	
	int count[2] = {0};

	for(int i=1; i<=n; i++){
		for(int j=1; j<=i; j++){
			if(n % j == 0){
				count[0]++;
			}
		}
		if(count[0] == 75){
			count[1]++;
		}
		count[0]=0;
	}

	printf("%d\n", count[1]);
	return 0;
}