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Problem solution · Python

ABC114 D — 756

ABC114 D — 756: 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
84 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 ABC114 D — 756, 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

  • 84 lines of Python from the credited upstream file abc114_d.py.
  • The implementation visibly relies on hash 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 codeABC114 D — 756 · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def run_prime_factorization(max_number: int) -> dict:    """Run prime factorization.    Args:        max_number: Int of number (greater than 1).    Returns:        A dictionary's items ((base, exponent) pairs).    Landau notation: O(log n)    """     from math import sqrt     ans = dict()    remain = max_number     for base in range(2, int(sqrt(max_number)) + 1):        if remain % base == 0:            exponent_count = 0             while remain % base == 0:                exponent_count += 1                remain //= base             ans[base] = exponent_count     if remain != 1:        ans[remain] = 1     return ans  def main():    import sys     input = sys.stdin.readline     n = int(input())    ans = 0    numbers = dict()     for i in range(2, n + 1):        r = run_prime_factorization(i)         for key, value in r.items():            if key not in numbers.keys():                numbers[key] = value + 1            else:                numbers[key] += value     count_75 = sum([1 for value in numbers.values() if value >= 75])    count_25 = sum([1 for value in numbers.values() if value >= 25])    count_15 = sum([1 for value in numbers.values() if value >= 15])    count_5 = sum([1 for value in numbers.values() if value >= 5])    count_3 = sum([1 for value in numbers.values() if value >= 3])     # [75, ∞]が1つ以上    ans += count_75     # [25, ∞]が1つ以上, [3, 25)が1つ以上（[25, ∞]が1つのとき）    count_3_24 = count_3 - count_25     if count_25 > 0 and (count_25 - 1 + count_3_24) > 0:        ans += count_25 * (count_25 - 1 + count_3_24)     # [15, ∞]が1つ以上, [5, 15)が1つ以上（[15, ∞]が1つのとき）    count_5_14 = count_5 - count_15     if count_15 > 0 and (count_15 - 1 + count_5_14) > 0:        ans += count_15 * (count_15 - 1 + count_5_14)     # [5, ∞]が2つ以上, [3, 5)が1つ以上（[5, ∞]が2つのとき）    count_3_4 = count_3 - count_5     if count_5 > 1 and (count_5 - 2 + count_3_4) > 0:        ans += (count_5 * (count_5 - 1) // 2) * (count_5 - 2 + count_3_4)     print(ans)  if __name__ == "__main__":    main()

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