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Problem solution · Python

ABC313 C — Approximate Equalization 2

ABC313 C — Approximate Equalization 2: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Sorting and greedy selection
Source
KATO-Hiro AtCoder Solutions
Length
47 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

Sorting and greedy selection

For ABC313 C — Approximate Equalization 2, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 47 lines of Python from the credited upstream file abc313_c.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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 codeABC313 C — Approximate Equalization 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())    a = list(map(int, input().split()))     # 考察のステップ    # 1. 数列の最終的な値: x, x, ..., x, x + 1, x + 1, ..., x + 1    # xがc個、x + 1がd個あるとする（c + d = n）    # x + 1が0個のケースもありうる     # 2. 不変量に着目。aの総和sは、操作の前後・順番に関わらず一定。    # s = x * c + (x + 1) * d    #   = x * (c + d) + d    #   = x * n + d     # 3. xを求める    s = sum(a)    x, q = divmod(s, n)     # 4. 操作後の数列をbとすると、求めたい値は|bi - ai|を最小化したもの    # 数列a、bを昇順にソートする部分の正当性を十分理解できていない    a = sorted(a)    b = [x] * n     for i in range(q):        b[i] += 1     b = b[::-1]     ans = 0     for ai, bi in zip(a, b):        ans += abs(bi - ai)     # print(a, b)    print(ans // 2)  if __name__ == "__main__":    main()

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