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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>

#define rep(i, n) for(int i=0; i<(n); ++i)
#define max  200001

//qsort(str, n, sizeof(int), up_comp);
int up_comp(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_comp(const void *a, const void *b){return *(int*)b - *(int*)a;}
int int_sort( const void * a , const void * b ) {return strcmp(( char * )a , ( char * )b );}
int count_digit(int a){int i=0;while(a!=0){a=a/10;i++;}return i;}
int euclid( int a, int b ){int temp;if(a<b){temp=a;a=b;b=temp;}if(b<1)return -1;if(a%b==0) return b;return euclid(b,a%b);}

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