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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();

    let mut div = vec![0; 101];
    for mut i in 2..n + 1 {
        let mut j = 2;
        while j * j <= i {
            while i % j == 0 {
                i /= j;
                div[j] += 1;
            }
            j += 1;
        }
        if i != 1 {
            div[i] += 1;
        }
    }

    // 75 = 3 * 5 * 5
    // 75 = 3 * 5 * 5, 5 * 15, 3 * 25, 75

    let mut c3 = 0i32;
    let mut c5 = 0i32;
    let mut c15 = 0i32;
    let mut c25 = 0i32;
    let mut c75 = 0i32;

    for i in 2..101 {
        div[i] += 1;
        if div[i] >= 3 {
            c3 += 1;
        }
        if div[i] >= 5 {
            c5 += 1;
        }
        if div[i] >= 15 {
            c15 += 1;
        }
        if div[i] >= 25 {
            c25 += 1;
        }
        if div[i] >= 75 {
            c75 += 1;
        }
    }
    let mut ans = 0;
    ans += c5 * (c5 - 1) / 2 * (c3 - 2);
    ans += c15 * (c5 - 1);
    ans += c25 * (c3 - 1);
    ans += c75;
    println!("{}", ans);
}