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>
#include <map>
using namespace std;

int count(int n, map<int, int> mp)
{
  int ret = 0;
  for (auto m : mp)
  {
    if (n <= m.second)
      ret++;
  }
  return ret;
}

int main()
{
  int N;
  cin >> N;
  map<int, int> prime;
  for (int i = 2; i <= N; i++)
  {
    int value = i;
    for (int j = 2; j * j <= value; j++)
    {
      while (value % j == 0)
      {
        if (prime.find(j) == prime.end())
          prime[j] = 1;
        else
          prime[j] = prime[j] + 1;
        value /= j;
      }
    }
    if (value != 1)
    {
      if (prime.find(value) == prime.end())
        prime[value] = 1;
      else
        prime[value] = prime[value] + 1;
    }
  }
  int ans = 0;
  // p^74
  ans += count(74, prime);
  // p^24 * q^2
  ans += count(24, prime) * (count(2, prime) - 1);
  // p^14 * p^4
  ans += count(14, prime) * (count(4, prime) - 1);
  // p^4 * q^4 * r^2
  ans += count(4, prime) * (count(4, prime) - 1) * (count(2, prime) - 2) / 2;
  cout << ans << endl;
  return 0;
}