Problem:
Score : 200 points

Problem Statement

La Confiserie d'ABC sells cakes at 4 dollars each and doughnuts at 7 dollars each.
Determine if there is a way to buy some of them for exactly N dollars. You can buy two or more doughnuts and two or more cakes, and you can also choose to buy zero doughnuts or zero cakes.

Constraints

N is an integer between 1 and 100, inclusive.

Input

Input is given from Standard Input in the following format:

N

Output

If there is a way to buy some cakes and some doughnuts for exactly N dollars, print Yes; otherwise, print No.

Sample Input 1

11

Sample Output 1

Yes

If you buy one cake and one doughnut, the total will be 4 + 7 = 11 dollars.

Sample Input 2

40

Sample Output 2

Yes

If you buy ten cakes, the total will be 4 \times 10 = 40 dollars.

Sample Input 3

3

Sample Output 3

No

The prices of cakes (4 dollars) and doughnuts (7 dollars) are both higher than 3 dollars, so there is no such way.

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

use std::io;

fn read_line() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s
}

macro_rules! from_line {
    ($($a:ident : $t:ty),+) => {
        $(let $a: $t;)+
        {
            let _line = read_line();
            let mut _it = _line.trim().split_whitespace();
            $($a = _it.next().unwrap().parse().unwrap();)+
            assert!(_it.next().is_none());
        }
    };
}

fn main() {
    from_line!(n: isize);
    if n % 7 == 0 || n % 4 == 0 {
        println!("Yes");
        return;
    }
    for i in 1..n/7 + 1 {
        let k = n - 7*i;
        if k % 4 == 0 {
            println!("Yes");
            return;
        }
    }
    println!("No");
}