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: Runtime Error)

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

public class Hello
{
    public static void Main()
    {
        var n = int.Parse(Console.ReadLine().Trim());
        var p = GeneratePrime(n/2);
        var a = new int[p.Count()];
        for (int i = 2; i <= n; i++) getA(a, p, i);
        var ans = getAns(a);
        Console.WriteLine(ans);
    }
    static int getAns(int[] a)
    {
        var k = new int[] { 2, 4, 14, 24, 74 };
        var d = new Dictionary<int, int>();
        foreach (var x in k) d[x] = 0;
        foreach (var x in a)
            foreach (var y in k)
                if (x >= y) d[y]++;
        var a1 = d[74];
        var a2 = d[24] * (d[2] - 1);
        var a3 = d[14] * (d[4] - 1);
        var a4 = d[4] * (d[4] - 1) / 2 * (d[2] - 2);
        return a1 + a2 + a3 + a4;
    }
    static void getA(int[] a, List<int> p, int t)
    {
        var tp = t;
        var pc = p.Count();
        for (int i = 0; i < pc; i++)
        {
            if (p[i] > t) break;
            while (true)
                if (tp % p[i] == 0) { tp /= p[i]; a[i]++; }
                else break;
        }
    }
    public static List<int> GeneratePrime(int m)
    {
        var a = new List<int>();
        int p;
        var sqrtMax = Math.Sqrt(m);
        var s = Enumerable.Range(2, m - 1).ToList();
        do
        {
            p = s.First();
            a.Add(p);
            s.RemoveAll(n => n % p == 0);
        }
        while (p < sqrtMax);
        a.AddRange(s);
        return a;
    }
}