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Problem solution · Python

AGC001 A — BBQ Easy

AGC001 A — BBQ Easy: 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
27 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 AGC001 A — BBQ Easy, 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

  • 27 lines of Python from the credited upstream file agc001_a.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 codeAGC001 A — BBQ Easy · PythonPython
Use this to learn the idea, then write your own version.
'''input5100 1 2 3 14 15 58 58 58 29135 21 3 1 23 ''' # -*- coding: utf-8 -*- # AtCoder Grand Contest# Problem A  if __name__ == '__main__':    number = int(input())    l = sorted(list(map(int, input().split())), reverse=True)    count = 0     for i in range(number):        count += min(l[2 * i], l[2 * i + 1])     print(count)

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