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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int main()
{
	int N;
	scanf("%d", &N);
	
	int i, j, k, l, p[25] = {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}, m[100] = {}, n;
	for (n = 2; n <= N; n++) {
		for (i = 0, k = n; k > 1; i++) {
			if (k % p[i] != 0) continue;
			for (; k % p[i] == 0; m[i]++, k /= p[i]);
		}
	}
	
	int ans = 0;
	for (i = 0; i < 25; i++) {
		if (m[i] < 2) break;
		if (m[i] >= 74) ans++;
		for (j = 0; j < 25; j++) {
			if (j == i || m[j] < 4) continue;
			if (m[j] >= 24) ans++;
			if (m[i] >= 4 && m[j] >= 14) ans++;
			for (l = j + 1; l < 25; l++) {
				if (l == i || m[l] < 4) continue;
				ans++;
			}
		}
	}
	
	printf("%d\n", ans);
	fflush(stdout);
	return 0;
}