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::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

use std::collections::HashSet;

fn divisors(n: i64) -> HashSet<i64> {
    let mut ds = vec![];
    let mut d = 1;
    while d * d <= n {
        if n % d == 0 {
            ds.push(d);
            ds.push(n / d);
        }
        d += 1;
    }
    let uniq: HashSet<i64> = ds.into_iter().collect();
    uniq
}

fn main() {
    let n: i64 = read();

    let mut ans = 0;
    for i in 0..n+1 {
        if i % 2 == 0 {
            continue;
        }
        let a = divisors(i);
        if a.len() == 8 {
            ans += 1;
        }
    }
    println!("{}", ans);
}