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")
}

pub fn read_n<T: FromStr>(n: usize) -> Vec<T> {
    read_n_logic::<T>(n, vec![])
}

pub fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
    match n {
        0 => a,
        _ => {
            a.push(read::<T>());
            read_n_logic(n - 1, a)
        }
    }
}
use std::cmp::{max, min};
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 cnt: i64 = 0;
    for i in 1..n+1 {
        if i %2 == 1 && divisors(i).len() == 8 {
            cnt += 1;
        }
    }
    println!("{}", cnt);
}