Problem:
Score: 200 points

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors.
Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the count.

Sample Input 1

105

Sample Output 1

1

Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

7

Sample Output 2

0

1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    token.parse().ok().unwrap()
}

fn main() {
    let n = input::<i64>();

    let mut ans = 0;
    for i in (1..n + 1).filter(|&i| i % 2 == 1) {
        let pf = prime_factorize(i);
        let cnt = pf.iter().map(|&(_, p)| p + 1).product::<i64>();
        if cnt == 8 {
            ans += 1;
        }
    }
    println!("{}", ans);
}

fn is_prime(n: i64) -> bool {
    let mut p = true;
    for i in 2..n {
        if i * i > n {
            break;
        }
        if n % i == 0 {
            p = false;
            break;
        }
    }
    p
}

fn prime_factorize(n: i64) -> Vec<(i64, i64)> {
    let mut n = n;
    let mut pf = vec![];
    for i in (2..n + 1).filter(|&n| is_prime(n)) {
        if n % i == 0 {
            let mut pow = 0;
            while n % i == 0 {
                pow += 1;
                n /= i;
            }
            pf.push((i, pow));
        }
    }
    pf
}