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 <stdio.h>
#include <string.h>
#define Max 100005
#define LL long long
using namespace std;
int data[200];
int main()
{
    int  n;
    scanf("%d",&n);
 /*   for(int i=4;i<=100;i+=4)
			data[i]=1;
	for(int i=7;i<=100;i+=7)
			data[i]=1;*/
	for(int i=0;i<=25;i++)
		for(int j=0;j<=14;j++)
		{
			if(i*4+j*7<=100)
				data[i*4+j*7]=1;
		}
	if(data[n])
		printf("Yes\n");
	else
		printf("No\n");
    return 0;
}