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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
using namespace std;

int B[15], C[]={1, 3, 5, 15, 25, 75};
int se(int i, int key){
    if(i==14){
        if(key<=B[i]) return 1;
        else return 0;
    }
    else{
        int ret=0;
        for(int j=0; j<6; ++j){
            if(B[i]<C[j]) break;
            if(key%C[j]<1) ret += se(i+1, key/C[j]);
        }
        return ret;
    }
}
int main() {
	int N; cin >> N;
	int A[]={2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47};
	for(int i=0; i<15; ++i){
	    int p=N;
	    B[i]=1;
	    while(p>0){
	        B[i] += p/A[i];
	        p /= A[i];
	    }
	}
	cout << se(0, 75) << endl;
	return 0;
}