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 <string>
#include <vector>
#include <climits>
#include <set>
#include <algorithm>
#include <unordered_map>
#include <queue>
#include <iomanip>
#include <map>
#include <utility>
#include <numeric>

#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define ll long long
#define P pair<int, int>

using namespace std;

int N;
int main() {
  cin >> N;
  bool ans[101] = {false};
  ans[0] = true;
  rep(i, 101) {
    if (i+4 < 101) {
      ans[i + 4] = ans[i + 4] || ans[i];
    }
  }
  rep(i, 101) {
    if (i+7 < 101) {
      ans[i+7] = ans[i+7] || ans[i];
    }
  }
  cout << (ans[N] ? "Yes" : "No") << endl;
}