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>
using namespace std;
#define _GLIBCXX_DEBUG
#define rep(i, from, to) for (int i = from; i < (to); ++i)
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
using ll = long long;
using vin=vector<int>;
using vll=vector<ll>;
using P = pair<int, int>;
const int inf=1e9+7;
const ll INF=1e18;
template <typename T> void chmin(T &a, T b) { a = min(a, b); }
template <typename T> void chmax(T &a, T b) { a = max(a, b); }

int main() {
  int n;
  cin>>n;
  bool ok=false;
  rep(i,0,26){
   rep(j,0,16){
     if(i*4+j*7==n){
       ok=true;
       break;
     }
   }
    
  }
  if(ok) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
}






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