Oops, I forgot to make it public. Thanks for the heads up.

Let S denote the set of squares with no guards installed among the squares of k*k that are completely contained within the plate.

If there is an intersection of the squares belonging to S, the game is over by placing a guard in that place, so you should not hand over such a situation.

Imagine that there is no current intersection of S. At this point, we can unconditionally choose two squares that do not overlap each other.

All you have to do is drop the guard in the remaining space of the two squares which are not overlap each other than your opponent will never win unless I win

The number of such cells on the board is nm-2*k*k, so if nm is odd, Yuto can always execute the above strategy, and if the number is even, same for Platina.

Separately, you have to consider the fact that the first player finishes the first turn.

I'm sorry for my English skills, and thank you for interest in my problem

Divide and conquer DP optimisation.

For convenience, add a dummy element to the start and end of the array that we must both miss. Denote by $$$DP[b][k]$$$ the maximum score we can get from the first $$$b$$$ notes, such that we miss $$$k$$$ of them. Denote by $$$Add[a][b]$$$ the amount of score we get if we miss notes $$$a$$$ and $$$b$$$, and hit all notes in between. Then, \begin{equation*} DP[b][k] = \max_{a < b} DP[a][k-1] + Add[a][b]. \end{equation*} Define \begin{equation*} A[b][k] = arg\,max_{a < b — 1} DP[a][k-1] + Add[a][b]. \end{equation*} Then \begin{equation*} DP[b][k] = \max(DP[b-1][k-1], DP[A[b][k]][k-1] + Add[A[b][k]][b] \end{equation*} Thus, we can use divide & conquer optimisation, since $$$A[b][k] \leq A[b+1][k]$$$.

Code: 110039963

F (I, j) means the number of non decreasing subarrays of [i, j].

Observation 1: Platina naturally calls L or R.

Observation 2: Also on this, Yuto will call X where max(f(L, X), f(X, R)) is the minimum.

Observation 3: For L,R, f(L,X) will be smaller up to a specific position, and f(X,R) will be smaller from a specific position.

Observation 4: The number called by Yuto by Observation 2 and Observation 3, the smaller one calls one of the two boundaries.

So our goal is to find that boundary.

Therefore, it is possible to find the boundary using binary search. To find this boundary once, we have to do it in O(logN), so we have to find F(i,j) in O(1).

NDSA means non-decreaing subarray. If K consecutively non-decreasing, the number of NDSAs in the zone is k(k+1)/2. Therefore, it can be grouped by these zones as a pretreatment.

Using the prefix sum, you can get the NDSA in a few areas in the middle, and you only need to add x(x+1)/2 only for parts that do not completely include the area.

Therefore, it can be solved in O(1), and the whole process can be processed in O(QlgN). I don't know the solution, but there are some people who have solved O(N+Q).

I'm sorry because I'm late, and thank you for interest in my problem