The key insight to solving this problem is the observation that for any ~K~ rice fields located at ~r_0 \le r_1 \le \dots \le r_{K-1}~, the transportation cost from all these ~K~ fields is minimized by placing the rice hub at a median. For example, when ~K = 1~, the hub should be at ~r_0~, and when ~K = 2~, placing it between ~r_0~ and ~r_1~ is optimal. In this problem, we will place the rice hub at ~r_{\lfloor K/2 \rfloor}~ for simplicity. Following this observation, we denote a solution by a sequence ~S \subseteq \langle r_0, \dots, r_{R-1} \rangle~ and let ~|S|~ denote the length of ~S~, which is the solution's value (the number of rice fields whose rice will be transported to the hub). The cost of ~S~ is ~cost(S) = \sum_{r_j \in S} |r_j-h(S)|~, where ~h(S)~ is the ~\left\lfloor \frac{|S|}{2} \right\rfloor~-th element of ~S~.

An ~\mathcal O(R^3)~ solution

Armed with this, we can solve the task by a guess-and-verify algorithm. We try all possible lengths of ~S~ (ranging between ~1~ and ~R~). Next observe that in any optimal solution ~S^*~, the rice fields involved must be contiguous; that is, ~S^*~ is necessarily ~\langle r_s, r_{s+1}, \dots, r_t \rangle~ for some ~0 \le s \le t \le R-1~. Therefore, there are ~R-K+1~ solutions of length ~K~. For each choice of ~S~, we compute ~h(S)~ and the transportation cost in ~\mathcal O(|S|)~ time and check if it is within the budget ~B~. This leads to an ~\mathcal O(R^3)~ algorithm, which suffices to solve subtask 2.

An ~\mathcal O(R^2)~ solution

To improve it to ~\mathcal O(R^2)~, we will speed up the computation of ~cost(S)~. Notice that we are only dealing with consecutive rice fields. Thus, for each ~S~, the cost ~cost(S)~ can be computed in ~\mathcal O(1)~ after precomputing certain prefix sums. Specifically, let ~T[i]~ be the sum of all coordinates to the left of rice field ~i~, i.e., ~T[0] = 0~ and ~T[i] = \sum_{j=0}^{i-1} X[j]~. Then, if ~S = \langle r_s, \dots, r_t \rangle~, ~cost(S)~ is given by ~(p-s)r_p-(T[p]-T[s])+(T[t+1]-T[p+1])-(t-p)r_p~, where ~p = \lfloor \frac{s+t}{2} \rfloor~. This ~\mathcal O(R^2)~ algorithm suffices to solve subtask 3.

An ~\mathcal O(R \log R)~ solution

Applying a binary search to find the right length in place of a linear search improves the running time to ~\mathcal O(R \log R)~ and suffices to solve all subtasks.

An ~\mathcal O(R)~ solution

We replace binary search with a variant of linear search carefully designed to take advantage of the feedback obtained each time we examine a combination of rice fields. In particular, imagine adding in the rice fields one by one. In iteration ~i~, we add ~r_i~ and find (1) ~S^*_i~, the best solution that uses only (a subsequence of) the first ~i~ rice fields (i.e., ~S^*_i \subseteq \langle r_0, \dots, r_i-1 \rangle~), and (2) ~S_i~, the best solution that uses only (a subsequence of) the first ~i~ rice fields and contains ~r_i-1~. This can be computed inductively as follows. As a base case, when ~i = 0~, both ~S_i~ and ~S^*_i~ are just ~\langle r_0 \rangle~ and cost ~0~, which is within the budget ~B \ge 0~. For the inductive case, assume that ~S^*_i~ and ~S_i~ are known. Now consider that ~S_{i+1}~ is ~S_i~ appended with ~r_i~, denoted by ~S_i \cdot r_i~, if the cost ~cost(S_i \cdot r_i)~ is at most ~B~, or otherwise it is the longest prefix of ~S_i \cdot r_i~ that costs at most ~B~. Furthermore, ~S^*_{i+1}~ is the better of ~S^*_i~ and ~S_{i+1}~. To implement this, we represent each ~S_i~ by its starting point ~s~ and ending point ~t~; thus, each iteration involves incrementing ~t~ and possibly ~s~, but ~s~ is always at most ~t~. Since ~cost(\langle r_s, \dots, r_t \rangle)~ takes ~\mathcal O(1)~ to compute, the running time of this algorithm is ~\mathcal O(R)~ and suffices to solve all subtasks.