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;

namespace abc114_d
{
	class Program
	{
		static void Main(string[] args)
		{
			int n = int.Parse(Console.ReadLine());
			int[] arr = new int[25];
			int[] prime = new int[] { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 };
			for (int i = 2; i <= n; i++)
			{
				int x = i;
				for (int j = 0; j < 25; j++)
				{
					while (x % prime[j] == 0) { arr[j]++; x /= prime[j]; }
				}
			}
			int ret = 0;
			for (int i = 0; i < 25; i++)
			{
				for (int j = i + 1; j < 25; j++)
				{
					for (int k = 0; k < 25; k++)
					{
						if (i == k || j == k) continue;
						if (arr[i] >= 4 && arr[j] >= 4 && arr[k] >= 2)
							ret++;
					}
				}
			}
			for (int i = 0; i < 25; i++)
			{
				for (int j = 0; j < 25; j++)
				{
					if (i == j) continue;
					if (arr[i] >= 4 && arr[j] >= 14) ret++;
					if (arr[i] >= 2 && arr[j] >= 24) ret++;
				}
			}
			for (int i = 0; i < 25; i++) if (arr[i] >= 74) ret++;
			Console.WriteLine(ret);
		}
	}
}