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 primes(n: usize) -> Vec<usize> {
    let mut ret = Vec::new();
    for i in 2..n {
        let mut is_prime = true;
        for j in 2..i {
            if i % j == 0 {
                is_prime = false;
                break;
            }
        }
        if is_prime {
            ret.push(i);
        }
    }
    ret
}

fn factor_num(n: usize, p: usize) -> usize {
    let mut ret: usize = 0;
    let mut x = p;
    while x <= n {
        ret += n / x;
        x *= p;
    }
    ret
}

fn main() {
    let mut buf = String::new();

    std::io::stdin().read_to_string(&mut buf).unwrap();

    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();

    let prime = primes(100);

    let mut dp = [0i64; 100];
    dp[1] = 1;

    for i in 0..prime.len() {
        let p = prime[i];
        let f = factor_num(n, p);
        if f == 0 {
            continue;
        }
        for j in (0..76).rev() {
            for k in 1..f + 1 {
                let t = j * (k + 1);
                if t < dp.len() {
                    dp[t] += dp[j];
                }
            }
        }
    }

    println!("{}", dp[75]);
}