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 main () {
  long N;
  cin >> N;
  vector<long> data(101);
  long a;
  for(int i = 1; i <= N; i++){
    a = i;
    for(int j = 2; j <= 100; j++){
      while(a % j == 0){
        a /= j;
        data.at(j)++;
      }
    }
  }
  vector<long> ans(5);
  for(int i = 0; i <= N; i++){
    if(data.at(i) >= 2)ans.at(0)++;
    if(data.at(i) >= 4)ans.at(1)++;
    if(data.at(i) >= 14)ans.at(2)++;
    if(data.at(i) >= 24)ans.at(3)++;
    if(data.at(i) >= 74)ans.at(4)++;
  }
  cout << ans.at(4) + ans.at(3) * (ans.at(0) - 1) + ans.at(2) * (ans.at(1) - 1) + ans.at(1) * (ans.at(1) - 1) * (ans.at(0) - 2) / 2<< endl;
  return 0;
}