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 <bits/stdc++.h>
using namespace std;
int N, ans = 0;
vector<int> p_vec, pf_vec;

bool is_primitive(int n) {
    for (int j = 2; j * j <= n; j++) {
        if(n % j == 0) return false;
    }
    return true;
}

int main(void) {
    cin >> N;

    for (int i = 2; i <= N; i++) {
        if(is_primitive(i)){
            p_vec.push_back(i);
        }
    }
  
    for (int i = 0; i < p_vec.size(); i++) {
        int n = N;
        pf_vec.push_back(0);
        while(n > 1){
            n /= p_vec[i];
            pf_vec[i] += n;
        }
    }

	// p^2 * q^4 * r^4
    int p = 0, q = 0;
    for (int i = 0; i < pf_vec.size(); i++) {
        if (pf_vec.at(i) >= 2) p++;
        if (pf_vec.at(i) >= 4) q++;
    }
    ans += q * (q - 1) / 2 * (p - 2);
    
    // p^2 * q^24
    p = q = 0;
    for (int i = 0; i < pf_vec.size(); i++) {
        if (pf_vec.at(i) >= 2) p++;
        if (pf_vec.at(i) >= 24) q++;
    }
    ans += (p -1) * q;
    
    // p^4 * q^14
    p = q = 0;
    for (int i = 0; i < pf_vec.size(); i++) {
        if (pf_vec.at(i) >= 4) p++;
        if (pf_vec.at(i) >= 14) q++;
    }
    ans += (p -1) * q;
    
    // p^74
    p = 0;
    for (int i = 0; i < pf_vec.size(); i++) {
        if (pf_vec.at(i) >= 74) p++;
    }
    ans += p;

    cout << ans << endl;
}