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Problem solution · Python

ABC262 D — I Hate Non-integer Number

ABC262 D — I Hate Non-integer Number: 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
39 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 ABC262 D — I Hate Non-integer Number, 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

  • 39 lines of Python from the credited upstream file abc262_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 codeABC262 D — I Hate Non-integer Number · 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())    a = list(map(int, input().split()))    mod = 998244353    ans = 0     # See:    # https://atcoder.jp/contests/abc262/submissions/33664314    for k in range(1, n + 1):        # 余りを状態として持つことをどうしたら思いつけるか?        dp = [[0] * k for _ in range(k + 1)]  # dp[i][j]: 数列からi個を選択したときに、その総和 % mod kがjとなる組合せ        dp[0][0] = 1         for ai in a:            # aiが巨大な数字である場合への対処            ai %= k             # ループを逆向きなのは、重複して数えるのを回避するためか?            for i in range(k - 1, -1, -1):                for j in range(k):                    dp[i + 1][(j + ai) % k] += dp[i][j]                    dp[i + 1][(j + ai) % k] %= mod                ans += dp[k][0]        ans %= mod     print(ans)  if __name__ == "__main__":    main()

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