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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main(){
    int n, m, x, y;
    cin >> n >> m >> x >> y;
    vector<int>X(n);
    vector<int>Y(m);

    for(int i=0; i<n; i++){
        cin >> X[i];
    }

    for(int i=0; i<m; i++){
        cin >> Y[i];
    }

    sort(X.begin(),X.end());
    sort(Y.begin(),Y.end());

    if(abs(X[n-1]-Y[0])<=1){
        cout << "War" << endl;
        return 0;
    }

    if(Y[0]<=x){
        cout << "War" << endl;
        return 0;
    }

    if(X[n-1]>=y){
        cout << "War" << endl;
        return 0;
    }

    cout << "No War" << endl;

    return 0;
}