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)

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

fn main() {
    let n: u32 = input();
    let mut flag: bool = false;
    for i in 0..n / 7 + 1 {
        for j in 0..(n - 7 * i) / 4 + 1 {
            if n == 7 * i + 4 * j {
                flag = true;
            }
        }
    }
    println!("{}", if flag {"Yes"} else {"No"});
}