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Problem solution · Python

ABC370 E — Avoid K Partition

ABC370 E — Avoid K Partition: 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
40 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 ABC370 E — Avoid K Partition, 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

  • 40 lines of Python from the credited upstream file abc370_e.py.
  • The implementation visibly relies on sequence storage, hash lookup, 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 codeABC370 E — Avoid K Partition · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from collections import defaultdict    from itertools import accumulate     input = sys.stdin.readline     n, k = map(int, input().split())    a = list(map(int, input().split()))    s = list(accumulate(a, initial=0))    mod = 998244353    dp = [0] * (n + 1)  # i番目まで決めたときの条件を満たす分割方法    dp[0] = 1    dp_total = 1    dp_sum = defaultdict(int)    dp_sum[0] = 1     # 累積和と余事象を活用して高速化    # dp[i]の和を変数で、dp_sum[si]を連想配列で持つ    for i in range(1, n + 1):        dp[i] += dp_total        sj = s[i] - k        dp[i] -= dp_sum[sj]        dp[i] %= mod         dp_total += dp[i]        dp_total %= mod        si = s[i]        dp_sum[si] += dp[i]        dp_sum[si] %= mod     print(dp[n])  if __name__ == "__main__":    main()

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