Problem 1375

1. Understanding the Problem:
- We have a board of size $n \times k$.
- Pasha and Vova take turns placing a border of length 1 between two adjacent cells.
- The player loses if after their move there is no way from the bottom left cell to the top right cell without crossing any border.

2. Analyzing the Path:
- Any path from the bottom left cell to the top right cell will cross $n + k - 2$ borders. This is because the path must move $n-1$ steps up and $k-1$ steps to the right, totaling $n + k - 2$ steps.

3. Total Number of Borders:
- The total number of borders that can be placed on the board is $n(k-1) + k(n-1)$. This is because:
- There are $n$ rows, each with $k-1$ horizontal borders.
- There are $k$ columns, each with $n-1$ vertical borders.
- Simplifying, we get:
$$n(k-1) + k(n-1) = nk - n + kn - k = 2nk - n - k$$

4. Parity Consideration:
- We need to consider the parity (odd or even nature) of the number of borders.
- The total number of borders modulo 2 is:
$$(n(k-1) + k(n-1)) \mod 2 = (nk - n + kn - k) \mod 2 = (2nk - n - k) \mod 2$$
- Since $2nk$ is always even, we have:
$$(2nk - n - k) \mod 2 = (n + k) \mod 2$$

5. Determining the Winning Strategy:
- If $n + k$ is even, then the total number of borders is even.
- If $n + k$ is odd, then the total number of borders is odd.
- The game ends when there are no more valid moves, i.e., when the last border is placed.
- If $n + k$ is even, the first player (Pasha) will place the last border.
- If $n + k$ is odd, the second player (Vova) will place the last border.

6. Conclusion:
- The player who places the last border loses because they will block the path from the bottom left to the top right cell.
- Therefore, if $n + k$ is even, the second player (Vova) wins.
- If $n + k$ is odd, the first player (Pasha) wins.

The final answer is the second player (Vova) will win if $\boxed{ n + k }$ is even, and the first player (Pasha) will win if $n + k$ is odd.