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Problem solution · Python

ABC117 D — XXOR

ABC117 D — XXOR: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Dynamic programming
Source
KATO-Hiro AtCoder Solutions
Length
50 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For ABC117 D — XXOR, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 50 lines of Python from the credited upstream file abc117_d.py.
  • The implementation visibly relies on sequence storage, ordered lookup, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from KATO-Hiro AtCoder Solutions by KATO-Hiro and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeABC117 D — XXOR · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys     input = sys.stdin.readline     n, k = map(int, input().split())    a = list(map(int, input().split()))     # See:    # https://drken1215.hatenablog.com/entry/2019/02/04/013700    # dp[digit][is_smaller]    dp = [[-1 for _ in range(100)] for _ in range(2)]    dp[0][0] = 0    digit_max = 50     for digit in range(digit_max):        mask = 1 << (digit_max - digit - 1)        count = 0         for ai in a:            if ai & mask:                count += 1         cost0 = mask * count        cost1 = mask * (n - count)         # smaller to smaller        if dp[1][digit] != -1:            dp[1][digit + 1] = max(dp[1][digit + 1], dp[1][digit] + max(cost0, cost1))         # exact to smaller        if dp[0][digit] != -1 and (k & mask):            dp[1][digit + 1] = max(dp[1][digit + 1], dp[0][digit] + cost0)         # exact to exact        if dp[0][digit] != -1:            if k & mask:                dp[0][digit + 1] = max(dp[0][digit + 1], dp[0][digit] + cost1)            else:                dp[0][digit + 1] = max(dp[0][digit + 1], dp[0][digit] + cost0)     print(max(dp[0][digit_max], dp[1][digit_max]))  if __name__ == "__main__":    main()

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