Editorial

First key fact:

• Knights always attack cells of the opposite color to the one they are on.

Second key fact:

• For $n\ne 2$, it is advantageous to place all the knights on cells of the same color.

To obtain the maximum number of knights that we can place on one board for $n\ne 2$, let's calculate the number of cells of each color:

• Black cells: $\lfloor \frac{n^2}{2}\rfloor$

• White cells: $\lceil \frac{n^2}{2}\rceil$

When $n=2$, we can place knights on all $n^2$ cells, because no knight can reach another.

Now, having the maximum number of knights, $max$, that we can place on one board, the number of boards needed to accommodate $K$ knights is found by the formula:

$re=\lceil \frac{K}{max}\rceil$

Asymptotic complexity: $O(1)$