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

namespace ABC114D
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = int.Parse(Console.ReadLine());

            List<int> primeNumbers = new List<int>();
            for (int i = 2; i <= N; i++) primeNumbers.Add(i);
            for (int i = 0; i < primeNumbers.Count; i++)
            {
                primeNumbers = primeNumbers.Where(x => x == primeNumbers[i] || x % primeNumbers[i] != 0).ToList();
            }
            int[] countDivisors = new int[primeNumbers.Count];
            
            for (int i = 2; i <= N; i++)
            {
                int divisioned = i;
                //約数をカウントして各素数の個数に加える
                for(int j = 0; j <primeNumbers.Count; j++)
                {
                    if (divisioned % primeNumbers[j] == 0)
                    {
                        countDivisors[j]++;
                        divisioned /= primeNumbers[j];
                        j--;
                    }
                }
            }

            //
            int overFive = countDivisors.Where(x => x >= 4).ToArray().Count();
            int overThree = countDivisors.Where(x => x >= 2).ToArray().Count();

            int ans = overFive * (overFive - 1) / 2 * (overThree - 2);

            int overSeventyFive = countDivisors.Where(x => x >= 74).ToArray().Count();

            int overFifeen = countDivisors.Where(x => x >= 14).ToArray().Count();

            int overTwentyFive = countDivisors.Where(x => x >= 24).ToArray().Count();

            ans += overSeventyFive + overFifeen * (overFive - 1) + overTwentyFive * (overThree - 1);

            Console.WriteLine(ans);
            Console.ReadKey();
        }
    }
}