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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define ll long long
#define rep(i,n) for(ll i=0;i<(n);i++)
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))
#define chmax(a,b) ((a)=(a)>(b)?(a):(b))
#define chmin(a,b) ((a)=(a)<(b)?(a):(b))
#define abs(p) ((p)>=(0)?(p):(-(p)))
#define MOD 1000000007
ll power(ll a,ll b){ll r=1;rep(i,b){r*=a;}return r;}
#define swap(a,b) do{ll w=(a);(a)=(b);(b)=w;}while(0)

//your code here!

int main(void){
    ll N;
	scanf("%lld", &N);
	ll i=0, j=0;
	for(ll i=0; i*4<=N; i++){
		for(ll j=0; i*4+j*7<=N; j++){
			if(i*4+j*7 == N){
				printf("Yes\n");
				return 0;
			}
		}
	}
	
	printf("No\n");
    return 0;
}