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 <array>

using ll = long long;
using namespace std;

int main() {
    array<int, 101> A = {};
    fill(A.begin()+1, A.end(), 10000);
    int n;
    cin >> n;
    int B[2] = {4, 7};
    for (int i = 1; i < 101; ++i) {
        for (auto &&b : B) {
            if(b <= i && A[i-b] != 10000){
                A[i] = min(A[i-b] + 1, A[i]);
            }
        }
    }

    cout << (A[n] < 10000 ? "Yes" : "No") << "\n";
    return 0;
}