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Problem solution · Python

ABC346 D — Gomamayo Sequence

ABC346 D — Gomamayo Sequence: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Direct simulation
Source
KATO-Hiro AtCoder Solutions
Length
68 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

Direct simulation

For ABC346 D — Gomamayo Sequence, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 68 lines of Python from the credited upstream file abc346_d.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeABC346 D — Gomamayo Sequence · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from itertools import accumulate, pairwise     input = sys.stdin.readline     n = int(input())    s = input().rstrip()    c = list(map(int, input().split()))    inf = 10**18    ans = inf    one_starts = [0] * n    zero_starts = [0] * n     # 前計算: コストの累積和    for i, si in enumerate(s):        if (i % 2 == 0 and si == "0") or (i % 2 == 1 and si == "1"):            one_starts[i] = c[i]        if (i % 2 == 0 and si == "1") or (i % 2 == 1 and si == "0"):            zero_starts[i] = c[i]     one_starts = list(accumulate(one_starts, initial=0))    zero_starts = list(accumulate(zero_starts, initial=0))     for i, (si, sj) in enumerate(pairwise(s), 1):        # i, i + 1が'1'のパターン        candidate1 = 0         if si == "0":            candidate1 += c[i - 1]        if sj == "0":            candidate1 += c[i]         if i % 2 == 1:            candidate1 += zero_starts[n] - zero_starts[min(n, i + 1)]            candidate1 += one_starts[max(0, i - 1)]        else:            candidate1 += one_starts[n] - one_starts[min(n, i + 1)]            candidate1 += zero_starts[max(0, i - 1)]         ans = min(ans, candidate1)         # i, i + 1が'0'のパターン        candidate2 = 0         if si == "1":            candidate2 += c[i - 1]        if sj == "1":            candidate2 += c[i]         if i % 2 == 1:            candidate2 += one_starts[n] - one_starts[min(n, i + 1)]            candidate2 += zero_starts[max(0, i - 1)]        else:            candidate2 += zero_starts[n] - zero_starts[min(n, i + 1)]            candidate2 += one_starts[max(0, i - 1)]         ans = min(ans, candidate2)     print(ans)  if __name__ == "__main__":    main()

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