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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int main(){
	int n,i,yak[101]={0},j;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		int icp=i;
		for(j=1;j<=icp;j++){
			if(icp%j==0&&j!=1){
				yak[j]++;
				icp=icp/j;
				j=1;
			}
		}
	}
	//for(i=1;i<=100;i++)printf("%d,",yak[i]);
	int thr=0,fv=0,thfv=0,tf=0,ft=0,sf=0;
	for(i=0;i<=100;i++){
		//if(yak[i]>=2)thr++;
		if(yak[i]<=23&&yak[i]>=4)fv++;
		if(yak[i]>=2&&yak[i]<=3)thfv++;
		if(yak[i]>=24)tf++;
		if(yak[i]>=14)ft++;
		if(yak[i]>=74)sf++;
		
	}
	int ans=0;
	if(thfv>0)ans+=thfv*(tf+fv-1)*(tf+fv)/2;
	if(tf+fv>=3)ans+=(tf+fv)*(tf+fv-1)*(tf+fv-2)/2;
	if(tf>0)ans+=tf*(fv+thfv+tf-1);
	if(ft>0)ans+=ft*(tf+fv-1);
	ans+=sf;
	printf("%d\n",ans>0?ans:0);
	return 0;
}