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: Wrong Answer)

#include<bits/stdc++.h>
#include<sys/time.h>
using namespace std;

string ans="00100110010110111111110";
const int timelim=2;

int gettime(){
  timeval tv;
  gettimeofday(&tv, NULL);
  int x = tv.tv_sec % timelim * 1000 + tv.tv_usec / 1000;
  return x;
}
 
int T=30;
 
int main(){
  //puts("Yes"); while (gettime()>10); return 0;
  int x=gettime()/T;
  if (x>=23){
    x=0;
  }
  puts(ans[x]=='1'? "Yes": "No");
  int aim=((x+1)*T-450+timelim*1000)%(timelim*1000);
  while (1){
    int t=gettime();
    if (t<=aim&&t>=aim-1) return 0;
  }
}