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(void) {

  int n;
  scanf("%d", &n);
  int prime_list[100];
  for (int i = 2; i < 100; i++) {
    prime_list[i] = 1;
  }
  for (int i = 2; i < 100; i++) {
    if (prime_list[i] == 1) {
      for (int j = 2; i*j < 100; j++) {
        prime_list[i*j] = 0;
      }
    }
  }
  int factors[100];
  for (int i = 0; i < 100; i++) {
    factors[i] = 0;
  }
  int number;
  for (int i = 2; i <= n; i++) {
    number = i;
    for (int j = 2; j <= n; j++) {
      if (prime_list[j] == 1) {
        while (number%j == 0) {
          factors[j]++;
          number /= j;
        }
        if (number == 1) {
          break;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 2; i < 100; i++) {
    if (prime_list[i] == 1 && factors[i] >= 2) {
      for (int j = 2; j < 100; j++) {
        if (j != i && prime_list[j] == 1 && factors[j] >= 4) {
          for (int k = j+1; k < 100; k++) {
            if (k != i && prime_list[k] == 1 && factors[k] >= 4) {
              ans++;
            }
          }
        }
      }
    }
  }
  for (int i = 2; i < 100; i++) {
    if (prime_list[i] == 1 && factors[i] >= 2) {
      for (int j = 2; j < 100; j++) {
        if (j != i && prime_list[j] == 1 && factors[j] >= 24) {
          ans++;
        }
      }
    }
  }
  for (int i = 2; i < 100; i++) {
    if (prime_list[i] == 1 && factors[i] >= 4) {
      for (int j = 2; j < 100; j++) {
        if (j != i && prime_list[j] == 1 && factors[j] >= 14) {
          ans++;
        }
      }
    }
  }
  for (int i = 2; i < 100; i++) {
    if (prime_list[i] == 1 && factors[i] >= 74) {
      ans++;
    }
  }
  printf("%d\n", ans);

  return 0;
}