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::Read;

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<u32> = vec![0; 101];

    for mut i in 2..n + 1 {
        // iを祖神う分解
        let mut j = 2;
        while j * j <= i {
            while i % j == 0 {
                div[j] += 1;
                i /= j;
            }
            j += 1;
        }
        if i != 1 {
            div[i] += 1;
        }
    }
    let mut cnt75 = 0;
    let mut cnt25 = 0;
    let mut cnt15 = 0;
    let mut cnt5 = 0;
    let mut cnt3 = 0;
    for i in 2..n + 1 {
        if div[i] >= 2 {
            cnt3 += 1;
        }
        if div[i] >= 4 {
            cnt5 += 1;
        }
        if div[i] >= 14 {
            cnt15 += 1;
        }
        if div[i] >= 24 {
            cnt25 += 1;
        }
        if div[i] >= 74 {
            cnt75 += 1;
        }
    }
    let mut ans = cnt75;
    ans += cnt25 * (cnt3 - 1);
    ans += cnt15 * (cnt5 - 1);
    ans += cnt5 * (cnt5 - 1) * (cnt3 - 2) / 2;
    println!("{}", ans);
}