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)*}
    };
}
 
#[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")
    };
}









use std::cmp::min;
use std::collections::BTreeMap;

fn prime_factor(mut n:usize) -> BTreeMap<usize, usize>{
    let mut res = BTreeMap::new();
    let t = ((n as f64).sqrt() as usize) + 1;
    for i in 2..t{
        while n % i == 0{
            *res.entry(i).or_insert(0) += 1;
            
            n = n / i;    
        }
    }
    if n != 1{
            res.insert(n,1);
        }
    res
}
fn main() {
    input! {
        n: usize,
    }
    let mut res = BTreeMap::new();
    for i in 1..n+1{
        let tmp = prime_factor(i);

        for(k, v) in tmp.into_iter(){
            *res.entry(k).or_insert(0) += v;
        }
    }
    let mut vec = Vec::new();
    for (k, v) in res.into_iter(){
        vec.push((v, k));
    }
    vec.sort();
    let mut ans = 0;
    for s in 0..vec.len(){
        if vec[s].0 >= 74{ans = ans + 1;}
        for p in s+1..vec.len(){
            if s == p{continue;}
            
            if vec[s].0 >= 2 && vec[p].0>=24{
                ans = ans + 1;  
            }
            if vec[s].0 >= 4 && vec[p].0>=14{
                ans = ans + 1;    
            }
            if vec[s].0 >= 24 && vec[p].0>=2{
                ans = ans + 1;  
            }
            if vec[s].0 >= 14 && vec[p].0>=4{
                ans = ans + 1;    
            }
            for k in p+1..vec.len(){
                if p == k || s == k {continue;}
                if vec[s].0 >= 2 && vec[p].0 >= 4 && vec[k].0 >= 4{
                    ans += 1;    
                }
                if vec[s].0 >= 4 && vec[p].0 >= 2 && vec[k].0 >= 4{
                    ans += 1;    
                }
                if vec[s].0 >= 4 && vec[p].0 >= 4 && vec[k].0 >= 2{
                    ans += 1;    
                }
            }
            
        }

    }

    println!("{}", ans);
}