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Problem solution · Python

ABC184 F — Programming Contest

ABC184 F — Programming Contest: a Python solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Binary search
Source
KATO-Hiro AtCoder Solutions
Length
60 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

Binary search

For ABC184 F — Programming Contest, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 60 lines of Python from the credited upstream file abc184_f.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 1 loop block detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 codeABC184 F — Programming Contest · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def generate_patterns(elements, bound):    results = [0]    size = len(elements)     for bit in range(1, 1 << size):        summed = 0         for i in range(size):            if bit & (1 << i):                summed += elements[i]         if summed <= bound:            results.append(summed)     return results  def main():    import sys     input = sys.stdin.readline     n, t = map(int, input().split())    a = list(map(int, input().split()))     mid = n // 2    first = a[:mid]    second = a[mid:]    ans = 0     first_patterns = generate_patterns(first, t)    second_patterns = sorted([0] + generate_patterns(second, t))     left_index = 0    right_index = len(second_patterns)     for first_pattern in first_patterns:        remain = t - first_pattern        left = left_index        right = right_index         while (right - left) > 1:            mid = (left + right) // 2             if second_patterns[mid] > remain:                right = mid            else:                left = mid         ans = max(ans, first_pattern + second_patterns[left])     print(ans)  if __name__ == "__main__":    main()

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