The key insight to solving this problem is the observation that for any $K$ rice fields located at $r_0\le r_1\le \cdots \le r_{K-1}$, 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 $S\subseteq ⟨r_0,\dots ,r_{R-1}⟩$ 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 $\lfloor \frac{|S|}{2}\rfloor$-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 $S^{\ast }$, the rice fields involved must be contiguous; that is, $S^{\ast }$ is necessarily $⟨r_s,r_{s+1},\dots ,r_t⟩$ 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 $S=⟨r_s,\dots ,r_t⟩$, $cost(S)$ 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) $S_i^{\ast }$, the best solution that uses only (a subsequence of) the first $i$ rice fields (i.e., $S_i^{\ast }\subseteq ⟨r_0,\dots ,r_i-1⟩$), 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 $S_i^{\ast }$ are just $⟨r_0⟩$ and cost $0$, which is within the budget $B\ge 0$. For the inductive case, assume that $S_i^{\ast }$ 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, $S_{i+1}^{\ast }$ is the better of $S_i^{\ast }$ 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 $cost(⟨r_s,\dots ,r_t⟩)$ takes $\mathcal{O}(1)$ to compute, the running time of this algorithm is $\mathcal{O}(R)$ and suffices to solve all subtasks.