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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<list>
#include<vector>
#include<cmath>

#define rep(i,a,b) for(int i=a;i<b;i++)

using namespace std;


int main(){

    int n;
    cin >> n;

    bool flg = false;
    if(n>=7 && n%4==3)flg = true;
    else if(n>=14 && n%4==2)flg = true;
    else if(n>=21 && n%4==1)flg = true;
    else if(n>=4 && n%4==0)flg = true;

    if(flg)cout << "Yes" << endl;
    else cout << "No" << endl;
}