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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

int main() {
    int n;cin>>n;
    int a[100];rep(i,100)a[i]=0;
    
    for(int i=2;i<=n;++i){
        int t=i;
        int j=2;
        while(t>1){
            while(t%j==0){
                ++a[j];
                t/=j;
            }
            ++j;
        }
    }
    int ans=0;
    int c75=0;
    int c25=0;
    int c5=0;
    int c3=0;
    
    rep(i,n){
        if(a[i]>=74)++c75;
        if(a[i]>=24)++c25;
        if(a[i]>= 4)++c5;
        if(a[i]>= 2)++c3;
    }
    ans+=c75;
    ans+=max(c25*(c3-1),0);
    ans+=max(c5*(c5-1)/2*(c3-2),0);

    cout<<ans<<endl;
	return 0;
}