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::{stdin, Read, StdinLock};
use std::str::FromStr;

#[allow(dead_code)]
struct Scanner<'a> {
    cin: StdinLock<'a>,
}
#[allow(dead_code)]
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }

    fn read<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }

    fn input<T: FromStr>(&mut self) -> T {
        self.read().unwrap()
    }
}

fn enumerate_divisors(n: i64) -> Vec<i64> {
    let mut ret = Vec::new();
    let mut i = 1;
    while i * i <= n {
        if n % i == 0 {
            ret.push(i);
            if i * i != n {
                ret.push(n / i);
            }
        }
        i += 1;
    }
    ret
}

fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);

    let n: i64 = sc.input();

    let mut count = 0;
    for i in 1..n + 1 {
        if i % 2 == 1 && enumerate_divisors(i).len() == 8 {
            count += 1;
        }
    }
    println!("{}", count);
}