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)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
mod vig {
    /// prime_factor
    /// usage: for (div, exp) in prime_factor(n) {...}
    pub fn prime_factor(mut n: u64) -> Vec<(u64, u64)> {
        let mut ans: Vec<(u64,u64)> = vec![];
        for i in 2..((n as f64).sqrt() as u64 + 2) {
            if n%i==0 {
                let div = i;
                let mut exp = 0;
                loop {
                    if n%i!=0 {
                        break;
                    }
                    n /=div;
                    exp+=1;
                }
                ans.push((div, exp));
            }
        }
        if n>1 {
            ans.push((n, 1));
        }
        ans
    }
}
fn main() {
    input!{
        n: i64
    }
    // 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47 => 15
    let tg = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47];
    let mut primes = vec![0;50];
    for i in 1..n+1 {
        let p = vig::prime_factor(i as u64);
        for &(a,m) in &p {
            for &i in &tg {
                if a==i {
                    primes[a as usize]+=m;
                }
            }
        }
    }
    // println!("{:?}", primes);
    let mut v = vec![];
    for &ind in &tg {
        // primes[ind]
        v.push(primes[ind as usize]);
    }
    v.sort();
    v.reverse();
    // println!("{:?}", v);
    let mut n75 = 0;
    let mut n25 = 0;
    let mut n15 = 0;
    let mut n5 = 0;
    let mut n3 = 0;
    for &i in &v{
        if i>=74 {
            n75 += 1;
        }
        if i>=24 {
            n25 += 1;
        }
        if i>=14 {
            n15 += 1;
        }
        if i>=4 {
            n5 += 1;
        }
        if i>=2 {
            n3 += 1;
        }
    }
    // println!("{} {} {} {} {}", n75, n25, n15, n5, n3);
    let ans = n75 + n25*(n3-1) + n15*(n5-1) + n5*(n5-1)*(n3-2)/2;
    println!("{}", ans);

}