Let us consider how we solve the problem for a fixed sequence $A$.

Assume we obtain $p_1,p_2,\cdots,p_N$ by using just Operation $2$. Let us consider the minimum cost needed here. First, for each $i$ ($1 \leq i \leq N$), the cost of $|A_i-p_i|$ will be incurred. Then, let us consider the cost needed to make the sequence $p$ using just Operation $2$. By focusing on the change in the difference between adjacent elements, we can see this cost is $M \times \sum_{i=0}^N |p_i - p_{i+1}|/2$, where $p_0=p_{N+1}=0$. We can rewrite the above formula as $M \times \sum_{i=0}^N \max(p_{i+1}-p_i,0)$. Now, the problem is to set $p_i$ to minimize the total cost above.

To solve this naively with DP, we can consider the following: $dp[i][j]=($the minimum cost when the first $i$ elements have been set and $p_i=j)$.

Here, let us observe $dp[i]$ for a fixed $i$, and it forms a convex polyline. Then, let us think of maintaining the set of points where the angle of the polyline changes, and the problem boils down to the following simple greedy strategy:

• Initialize $ans=\sum A_i$.
• Let $S$ be a multiset, which initially contains $C$ instances of $0$.
• For each $i$ ($1 \leq i \leq N$), do the following:
  • Add two instances of $A_i$ to $S$.
  • Take out the smallest element in $S$ and subtract this value from $ans$.
  • Delete the greatest element in $S$.

(By the way, this is the intended solution for the original problem.)

Let us turn this into a counting problem. For each element in $B_{i,j}$, we want to find the number of times it is used. We can find it with DP where the state is the number of instances smaller than $B_{i,j}$ in $S$. Eventually, we have a solution with the time complexity of $O(N^2K(M+K))$.