Розбір

Johnny English and Bof reach the final point simultaneously in time $t$. Johnny English covers a distance $s_1=p\cdot t$, while Bof covers a distance $s_2=q\cdot t$. The ratio of the distances traveled is equal to the ratio of the speeds: $s_1/s_2=p/q$.

Let's construct such paths constructively. For example, consider a route consisting of $4$ points: $(0,0)\to (2q,0)\to (2q,p-q)\to (2q,0)$. Here, the coordinate $p-q$ is chosen to ensure that its value is non-negative (according to the problem's conditions, $q\le p$).

• The length of Johnny English's path is $2q+2\ast (p-q)=2p$.

• The length of Bof's path is $2q$.

If the speeds of Johnny English and Bof are the same, then the points $(2q,0)$ and $(2q,p-q)$ coincide, and according to the problem's conditions, there cannot be two identical points in a row on the route. In this case, as a solution, one could choose, for example, the following route: $(0,0)\to (1,0)$.

Algorithm realization

Read the input data.

scanf("%d %d", &q, &p);

If the speeds of Johnny English and Bof are the same, then they move from point $(0,0)$ to point $(1,1)$.

if (q == p)
{
  printf("2\n");
  printf("%d %d\n", 0, 0);
  printf("%d %d\n", 1, 0);
  return 0;
}

If $p\ne q$, then the route consists of four points.

printf("4\n");
printf("%d %d\n", 0, 0);
printf("%d %d\n", 2 * q, 0);
printf("%d %d\n", 2 * q, p - q);
printf("%d %d\n", 2 * q, 0);