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Problem solution · Python

ABC257 E — Addition and Multiplication 2

ABC257 E — Addition and Multiplication 2: 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
38 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 ABC257 E — Addition and Multiplication 2, 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

  • 38 lines of Python from the credited upstream file abc257_e.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 codeABC257 E — Addition and Multiplication 2 · 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())    c = list(map(int, input().split()))[::-1]     # xの最大値 → 桁数をなるべく多く、先頭の桁ほど値が大きくする    # 桁数の上限    c_min = min(c)    digit_max = n // c_min    digit = digit_max     ans = list()     # 先頭の桁から条件を満たすように構築    # コストciを払ったときに、最大の桁数が維持されるかどうか?    for d in range(digit_max):        for i, ci in enumerate(c):            j = 9 - i             if (n - ci) >= (digit - 1) * c_min:                ans.append(str(j))                n -= ci                break                digit -= 1        print(''.join(ans))  if __name__ == "__main__":    main()

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