The key idea is that when watering a subarray, do not consider the entire subarray. Instead, consider the number of times a certain plant is watered as a result of these subarrays. Let ~dp[i][j]~ be the minimum cost to water the first ~i~ plants and such that the ~i^\text{th}~ plant has been watered by ~j~ subarrays. Then we add ~B~ for each new subarray at this step and ~A~ for each subarray at this step. So we obtain the following recurrence:

$$\displaystyle dp[i][j] = \min\left(\min_{k \le j} \left(dp[i-1][k] + B(j-k) + Aj + C(v_i-j)^2\right), \min_{k \ge j} \left(dp[i-1][k] + Aj + C(v_i-j)^2\right) \right)$$

Call ~V~ the largest value of ~v_i~. It is not hard to see that we only need to consider ~j~ up to ~V~. The constraints for the first subtask allow this recurrence to pass in time.

Time Complexity: ~\mathcal O(NV^2)~

For the second subtask, we will perform an optimization so that the transition becomes ~\mathcal O(1)~ instead of the original ~\mathcal O(V)~. Rewrite the recurrence by rearranging the terms which do not depend on ~k~:

$$\displaystyle dp[i][j] = \min\left(Bj + Aj + C(v_i-j)^2 + \min_{k \le j} (dp[i-1][k]-Bk), Aj + C(v_i-j)^2 + \min_{k \ge j} dp[i-1][k]\right)$$

We can precompute $$\displaystyle \min_{k \le j} (dp[i-1][k]-Bk)$$ and $$\displaystyle \min_{k \ge j} dp[i-1][k]$$ in ~\mathcal O(V)~ for all ~j~ and for each ~i~ by doing a running minimum. So we can now do the transition in ~\mathcal O(1)~.

Time Complexity: ~\mathcal O(NV)~