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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <math.h>
#include <iomanip>
#include <bitset>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <utility>
#include <set>
#include <map>
using namespace std;
typedef long long int ll;

#define MOD 998244353


ll gcd(ll a,ll b){
    if (a%b==0)return(b);
    else return(gcd(b,a%b));
}

bool primeNumber(ll n){
    if(n < 2) return false;
    else{
        for(ll i = 2; i * i <= n; i++){
            if(n % i == 0) return false;
        }
        return true;
    }
}
  
  
int main(){
    ll n;
    cin>>n;
    if(n%11 == 0 || n%4== 0 || n%7 == 0)puts("Yes");
    else puts("No");
    return 0;
}