CPMagical CP Library

Problem solution · Python

ABC247 C — 1 2 1 3 1 2 1

ABC247 C — 1 2 1 3 1 2 1: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Depth-first search
Source
KATO-Hiro AtCoder Solutions
Length
22 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

Depth-first search

For ABC247 C — 1 2 1 3 1 2 1, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 22 lines of Python from the credited upstream file abc247_c.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeABC247 C — 1 2 1 3 1 2 1 · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys     input = sys.stdin.readline     n = int(input())     def dfs(count, numbers):        if count == n:            return numbers                return dfs(count + 1, numbers + [count + 1] + numbers)        print(*dfs(1, [1]))  if __name__ == "__main__":    main()

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