Secular reception#9061

Secular reception (Editorial)

Johnny English and Bof reach the final point simultaneously in time $t$ . Johnny English covers a distance $s_{1} = p \cdot t$ s_1 = p \cdot t, while Bof covers a distance $s_{2} = q \cdot t$ s_2 = q \cdot t. The ratio of the distances traveled is equal to the ratio of the speeds: $s_{1} / s_{2} = p / q$ s_1 / s_2 = p / q.

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

The length of Johnny English's path is $2 q + 2 * (p - q) = 2 p$ 2q + 2 * (p - q) = 2p.
The length of Bof's path is $2 q$ .

If the speeds of Johnny English and Bof are the same, then the points $(2 q, 0)$ (2q, 0) and $(2 q, p - q)$ (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: $(0, 0) \to (1, 0)$ (0, 0) → (1, 0).

Algorithm implementation

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;
}

C++

If $p \neq = q$ 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);