CustomCC0-1.0#cpq0081800
Treewise Bitwise Convolution
Summary
- •Phase 6 / tree/xor
- •Reasoning-first competitive programming drill
Problem Description
Given a tree of n nodes, each node assigned an integer, compute for every node the XOR of values in its subtree, then for every possible XOR value (≤2^{20}), count how many nodes have that XOR as their subtree value.
How to read this problem in plain language:
- This is a Phase 6 reasoning drill focused on tree/xor.
- Typical lenses to test first: tree, xor, subtree.
- Constraints reminder: 1 ≤ , 0 ≤
Mini examples for mental simulation:
1) Boundary example: Describe why this case is tricky. Explain expected behavior and why naive logic may fail.
2) Adversarial example: Adversarial case where naive greedy/local decision looks correct but fails globally.
Lite-mode writing target:
- Write 1~2 observations that shrink the search space.
- Name one final algorithm and state target complexity explicitly.
- Validate with at least 2 edge cases and one hand simulation.
Constraints
- •1 ≤ , 0 ≤
Analysis
Key Insight
Use this hint to refine your reasoning. This step should reduce search space or formalize correctness. State why this insight changes your algorithm choice.
treexorsubtreedfs