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>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> tii;
#define rep(i,s,n) for(int i=(int)(s);i<(int)(n);i++)

int p[15] = {2,3,5,7,11,13,17,19,23,27,31,37,41,43,47};
int main(){
  int N; cin >> N;
  map<int,int> mp;
  int n2 = 0,n4 = 0,n14 = 0, n24 = 0, n74 = 0;
  for(int i=0;i<15;i++){
    int w = p[i];
    for(int j=1;j<=N;j++){
      int x = j;
      while(x%w==0){
        x /= w;
        mp[w]++;
      }
    }
    if(mp[w]>=2) n2++;
    if(mp[w]>=4) n4++;
    if(mp[w]>=14) n14++;
    if(mp[w]>=24) n24++;
    if(mp[w]>=74) n74++;
  }
  int ans = n4*(n4-1)*(n2-2)/2+n14*(n4-1)+n24*(n2-1)+n74;
  cout << ans  << endl;
    
  
  
}