CF1968G1 Division + LCP (easy version)

This is the easy version of the problem. In this version $ l=r $ . You are given a string $ s $ . For a fixed $ k $ , consider a division of $ s $ into exactly $ k $ continuous substrings $ w_1,\dots,w_k $ . Let $ f_k $ be the maximal possible $ LCP(w_1,\dots,w_k) $ among all divisions. $ LCP(w_1,\dots,w_m) $ is the length of the Longest Common Prefix of the strings $ w_1,\dots,w_m $ . For example, if $ s=abababcab $ and $ k=4 $ , a possible division is $ \color{red}{ab}\color{blue}{ab}\color{orange}{abc}\color{green}{ab} $ . The $ LCP(\color{red}{ab},\color{blue}{ab},\color{orange}{abc},\color{green}{ab}) $ is $ 2 $ , since $ ab $ is the Longest Common Prefix of those four strings. Note that each substring consists of a continuous segment of characters and each character belongs to exactly one substring. Your task is to find $ f_l,f_{l+1},\dots,f_r $ . In this version $ l=r $ .

The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. The first line of each test case contains two integers $ n $ , $ l $ , $ r $ ( $ 1 \le l = r \le n \le 2 \cdot 10^5 $ ) — the length of the string and the given range. The second line of each test case contains string $ s $ of length $ n $ , all characters are lowercase English letters. It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 2\cdot 10^5 $ .

For each test case, output $ r-l+1 $ values: $ f_l,\dots,f_r $ .

In the first sample $ n=k $ , so the only division of $ aba $ is $ \color{red}a\color{blue}b\color{orange}a $ . The answer is zero, because those strings do not have a common prefix. In the second sample, the only division is $ \color{red}a\color{blue}a\color{orange}a $ . Their longest common prefix is one.