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 main() {
    let mut n = String::new();
    let mut res = "No";

    io::stdin().read_line(&mut n).unwrap();

    let n: usize= n.trim().parse().unwrap();

    if n%4==0 || n%7==0 || !n<4{
        res = "Yes";
    }else {
        let mut i = 4;
        while i<n {
            if (n-i)%7==0 {
                res = "Yes";
                break;
            }
            i = i+4;
        }
    } 
    println!("{}", res);
}