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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int main() {
	int n;
	int result = 0;
  int result2 = 0;
	int kai[130] = {};
	int num[130] = {};
	int sosu[26] = { 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101 };
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		kai[i] = i;
	}
	for (int i = 0; sosu[i] <= n; i++) {
		for (int j = 1; j <= n; j++) {
			for (int k = 1; k <= 6; k++) {
				if (kai[j] % sosu[i] == 0) {
					num[sosu[i]]++;
					kai[j] /= sosu[i];
				}
			}
		}
	}
	for (int i = 0; sosu[i] <= n; i++) {
		if (num[sosu[i]] >= 74) {
			result++;
		}
	}

	for (int i = 25;  i >= 0; i--) {
		if (num[sosu[i]] >= 2) {
			for (int j = 25; j >= 0; j--) {
				if (num[sosu[j]] >= 24) {
					if (sosu[i] != sosu[j]) {
						result++;
					}
				}
			}
		}
	}

	for (int i = 25;  i >= 0; i--) {
		if (num[sosu[i]] >= 4) {
			for (int j = 25;  j >= 0; j--) {
				if (num[sosu[j]] >= 14) {
					if (sosu[i] !=sosu[j]) {
						result++;
                      
					}
				}
			}
		}
	}

	for (int i = 25; i>=0; i--) {
		if (num[sosu[i]] >= 2) {
			for (int j = 25; j >= 0; j--) {
				if (num[sosu[j]] >= 4) {
					for (int k = 25;  k >= 0; k--) {
						if (num[sosu[k]] >= 4) {
                        
							if (sosu[i] != sosu[j] && sosu[i] !=sosu[k] && sosu[k] != sosu[j]) {
								result2++;
							}
						}
					}
				}
			}
		}
	}
  result+=result2/2;
	printf("%d", result);
  for (int i = 0; sosu[i] <= n; i++) {
	}
}