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.*;
public class Main {
		
	public static void main(String[] args) {		
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		sc.close();
	
		int prime[] = new int[N+1];
		Arrays.fill(prime, 1);
		for(int k=2;k<=N;k++) {
			int tmp = k;
			for(int i=2;i*i<=k;i++) {
				while(tmp%i==0) {
					tmp /= i;
					prime[i]++;
				}
			}
			if(tmp!=1)prime[tmp]++;
			
		}
		
		int _3up = 0;
		int _5up = 0;
		int _15up = 0;
		int _25up = 0;
		int _75up = 0;
		
		for(int i=0;i<N+1;i++) {
			
			if(prime[i]>=75) _75up ++;
			else if(prime[i]>=25) _25up ++;
			else if(prime[i]>=15) _15up ++;
			else if(prime[i]>=5)  _5up ++;
			else if(prime[i]>=3)  _3up ++;
		}
		
		int sum5 = _75up + _25up + _15up + _5up + _3up;
		int sum4 = _75up + _25up + _15up + _5up;
		int sum3 = _75up + _25up + _15up;
		int sum2 = _75up + _25up;
		int sum1 = _75up;
		
		int ans = sum1 + sum2 * (sum5 - 1) + sum3 * (sum4 - 1) + sum4 * (sum4 - 1)/2 * (sum5 - 2);
				
		
		System.out.println(ans);
	}
	
	
}