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 ll long long
#define ull unsigned long long
#define PI acos(-1.0)
#define eps 1e-12
#define fi first
#define se second
#define pb push_back
#define MEM(a,b) memset((a),(b),sizeof(a))
#define mod(x) ((x)%MOD)
#define pii pair<int,int>
#define wz cout<<"-----"<<endl;
const int INF_INT = 2147483647;
const ll INF_LL = 9223372036854775807LL;
const ull INF_ULL = 18446744073709551615Ull;
const ll P = 92540646808111039LL;

const ll maxn = 1e5 + 10, MOD = 1e9 + 7;
const int Move[4][2] = {-1,0,1,0,0,1,0,-1};
const int Move_[8][2] = {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}


void init() {

}



int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
//    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);
//    ios_base::sync_with_stdio(0);
    init();
    int n; cin >>n;
    bool flg = 0;
    for (int i = 0; i <= n / 4; i++){
        if ((n - i * 4) % 7 == 0) {
            flg = 1;break;
        }
    }
    if (flg) puts("Yes");
    else puts("No");
    return 0;
}