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: Wrong Answer)

#include<bits/stdc++.h>
#include <math.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define lper(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9;
const ll INF=4e18;
const ll dy[4]={1,0,-1,0};
const ll dx[4]={0,-1,0,1};
int main() {
int N;
  cin >>N;
  if(N<4) {
    cout<<"No" << endl;
  }
else if (N>=4) {
  bool answer=false;
   repl(i,1,N/4) {
    if(((N-4*i)%4==0) || ((N-7*i)%4==0)){
      answer=true;
      break;
    }
  }
 
  repl(i,1,N/7) {
    if(((N-4*i)%7==0) || ((N-7*i)%7==0)) {
      answer=true;
      break;
    }
  }
  
    if(answer) {
      cout<<"Yes"<< endl;
    }
  else {
    cout<<"No"<< endl;
  }
}
}