First, let's solve the problem as if the pizza was a line instead of a cycle. (This is subtask 3.) If the goal is to find the longest segment equal in length to or shorter than $K$, then the length of a segment can be quickly computed by building a prefix sum array on the line. If the sum of the range $[a,b]$ can be found by subtracting position $a-1$ of the prefix sum array from position $b$, then for every position $e$ in the line, the largest sum segment ending at $e$ can be computed as the minimum of all positions in the range $[e-k,e)$ in the prefix sum array, subtracted from position $e$. After building the prefix sum array, a set can be used to solve this (the sliding range minimum query problem) in $\mathcal{O}(N\log K)$, or alternatively, it can be solved in $\mathcal{O}(N)$ using a double ended queue.

To solve the problem for a cycle, just arbitrarily choose any starting point, take two periods, and treat it as a line.