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: Wrong Answer)

#include <stdio.h>
#include <string.h>
int main(void){
    int n;
    scanf("%d",&n);
    int a[110];
    memset(a,1,sizeof(a));
    for(int i = 1;i<=100;i++){
        if(i % 4 == 0 || i % 7 == 0){
            a[i] = 0;
        }else if(i - 4 < 0 && (a[i - 4] == 0 || a[i - 7] == 0)){
            a[i] = 0;
        }else{
            a[i] = 1;
        }
        // printf("a[%d]=%d\n",i,a[i]);
    }
    // printf("%d\n",a[2 % 4]);
    // printf("%d\n",a[2 % 7]);
    if(a[n] == 0){
        printf("Yes\n");
    }else{
        printf("No\n");
    }
    return 0;
}