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 <bits/stdc++.h>

int main()
{
    int n,panduan=0;
    scanf("%d",&n);

   
        if(n%4==0) panduan=1;
    else if(n%7==0) panduan=1;
    else if(n%11==0) panduan=1;
    else if((n%7)%4==0)panduan=1;
    else if((n%11)%7==0)panduan=1;
    else if((n%11)%4==0)panduan=1;
    else if(n/4>0&&(n%4)%3==0) panduan=1;
    else if(n%15==0)panduan=1;
    else if((n%15)%4==0)panduan=1;
    else if((n%15)%7==0)panduan=1;
    else if(n%18==0)panduan=1;
    else if((n%18)%4==0)panduan=1;
    else if((n%18)%7==0)panduan=1;
    else if(n>=20) panduan=1;
    else panduan=0;
        if(panduan) printf("Yes\n");
        else printf("No\n");

    return 0;
}