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 asumisu[128];

int memo[128][128];

int search(int pos, int kazu) {
	int answer = 0;
	int i;
	if (pos >= 120) return kazu == 75;
	if (kazu > 75) return 0;
	if (memo[pos][kazu]) return ~memo[pos][kazu];
	for (i = 0; i <= asumisu[pos]; i++) {
		answer += search(pos + 1, kazu * (i + 1));
	}
	return ~(memo[pos][kazu] = ~answer);
}

int main(void) {
	int N;
	int i;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 2; i <= N; i++) {
		int mimorinn = i;
		int j;
		for (j = 2; j * j <= mimorinn; j++) {
			while (mimorinn % j == 0) {
				asumisu[j]++;
				mimorinn /= j;
			}
		}
		if (mimorinn > 1) asumisu[mimorinn]++;
	}
	printf("%d\n", search(2, 1));
	return 0;
}