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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static int[] divisor;
	public static int n;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		divisor = new int[n+1];
		for (int i = 2; i <= n; i++) {
			int k = i;
			for (int j = 2; j <= i; j++) {
				while(k%j==0) {
					divisor[j]++;
					k /= j;
				}
			}
		}
		int ans = 0;
		ans += count(74);
		ans += count(24)*(count(2)-1);
		ans += count(14)*(count(4)-1);
		ans += count(4)*(count(4)-1)/2*(count(2)-2);
		System.out.println(ans);
		sc.close();
	}
	public static int count(int x) {
		int res = 0;
		for (int i = 0; i <= n; i++) {
			if(divisor[i]>=x) res++;
		}
		return res;
	}
}