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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;

public class Remainder
{
    static int[] PrimeFactors;

    static public void Main()
    {
        var N = int.Parse(Console.ReadLine());
        PrimeFactors = new int[N+1];
        var L = new List<long>();
        //それぞれの数について素因数を辞書に登録する
        for (int i = 2; i <= N; i++)
        {
            PublicFactorization(i);
        }
        
      
        var count3 = PrimeFactors.Count(n => n >= 2);
        var count5 = PrimeFactors.Count(n => n >= 4);
        var count15 = PrimeFactors.Count(n => n >= 14);
        var count25 = PrimeFactors.Count(n => n >= 24);
        var count75 = PrimeFactors.Count(n => n >= 74);
      
        var sum = 0L;
      
        //75個の約数がある場合をカウント
        sum += count75;
      
        //5と15のペアをカウント
        if(count5 > 0){
          sum += (count5-1)*count15;
        }
      
        //3と25のペアをカウント
        if(count3 > 0){
          sum += (count3-1)*count25;
        }
      
        //3,5,5のペアをカウント
        if(count5 >= 2){
          sum += (count5*(count5-1)/2)*(count3-2);
        }
        

        Console.WriteLine(sum);

    }

    static void PublicFactorization(int n)
    {
        int i = 2;
        int tmp = n;

        while(i*i <= n)
        {
            if(tmp % i == 0)
            {
                tmp /= i;
                PrimeFactors[i]++;
            }
            else
            {
                i++;
            }
        }

        if(tmp != 1)
        {
            PrimeFactors[tmp]++;
        }
    }

}