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)

using namespace std;
#include<bits/stdc++.h>
#define BEGIN ios_base::sync_with_stdio(0);cin.tie(0);
#define END return EXIT_SUCCESS;
#define FOR(I,A,B) for((I)=(A);(I)<(B);(I)++)
#define REP(I,N) FOR((I),0,(N))
#define UP(I,A,B) for((I)=(A);(I)<=(B);(I)++)
#define DW(I,A,B) for((I)=(A);(I)>=(B);(I)--)
#define IN(P) cin>>(P)
#define IN2(P1,P2) cin>>(P1)>>(P2)
#define IN3(P1,P2,P3) cin>>(P1)>>(P2)>>(P3)
#define INS(I,N,V) REP((I),(N)) cin>>(V[I])
#define INS2(I,N,V1,V2) REP((I),(N)) cin>>(V1[I])>>(V2[I]);
#define INS3(I,N,V1,V2,V3) REP((I),(N)) cin>>(V1[I])>>(V2[I])>>(V3[I]);
#define OUT(P) cout<<(P)<<endl
#define OUT2(P1,P2) cout<<(P1)<<" "<<(P2)<<endl
#define OUT3(P1,P2,P3) cout<<(P1)<<" "<<(P2)<<" "<<(P3)<<endl
#define OUTS(I,N,V) REP((I),(N)) cout<<(V[i])<<endl
#define ALL(C) (C).begin(),(C).end()
#define RALL(C) (C).rbegin(),(C).rend()
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
template<class T>inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}

int N,i,j;
inline void solve(){
  IN(N);
  for(i=0;i*4<=N;i++){
    for(j=0;j*7<=N;j++){
      if(i*4+j*7==N){
        OUT("Yes");
        return;
      }else if(i*4+j*7>N)break;
    }
  }
  OUT("No");
}

int main(int argc,char**argv){
  BEGIN
  solve();
  END
}