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.