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Problem solution · Python

ABC122 D — We Like AGC

ABC122 D — We Like AGC: 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
51 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 ABC122 D — We Like AGC, 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

  • 51 lines of Python from the credited upstream file abc122_d.py.
  • The implementation visibly relies on 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 codeABC122 D — We Like AGC · 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())    mod = 10 ** 9 + 7    # dp[size][i][j][k]: 文字列の長さ、文字列の直前の1, 2, 3番目のとき、条件を満たす組み合わせ    dp = [[[[0 for _ in range(4)] for _ in range(4)] for _ in range(4)] for _ in range(n + 2)]    dp[0][3][3][3] = 1 # 初期化     for size in range(n + 1):        for i in range(4):            for j in range(4):                for k in range(4):                    if dp[size][i][j][k] == 0:                        continue                     for new_s in range(4):                        # 条件を満たさない場合を弾く                        if new_s == 0 and i == 1 and j == 2:                            continue                        if new_s == 0 and i == 2 and j == 1:                            continue                        if new_s == 1 and i == 0 and j == 2:                            continue                        if new_s == 0 and i == 1 and k == 2:                            continue                        if new_s == 0 and j == 1 and k == 2:                            continue                         dp[size + 1][new_s][i][j] += dp[size][i][j][k]                        dp[size + 1][new_s][i][j] %= mod            ans = 0     for i in range(4):        for j in range(4):            for k in range(4):                ans += dp[n][i][j][k]                ans %= mod        print(ans)  if __name__ == "__main__":    main()

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