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 <iomanip> //! for setprecision(10)
#include <math.h>
#include <algorithm>
#include <functional> 
#include <string>
#include <vector>

#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <utility>

using namespace std;

typedef long long LL;

#define dump(c) { for (auto it = c.begin(); it != c.end(); ++it) if (it == c.begin()) cout << *it; else cout << ' ' << *it; cout << endl; } 
#define rep(i,n) for (int i = 0; i < (n); ++i)

const int MOD = 1000000007;

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(0);

    int N;
    cin >> N;

    int ans = 0;
    for (int i = 0; i*4 <= N; ++i) {
        if ((N-i*4)%7 == 0) {
            ++ans;
        }
    }

    cout << (ans ? "Yes" : "No") << endl;

    return 0;
}