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Problem solution · Python

ABC150 D — Semi Common Multiple

ABC150 D — Semi Common Multiple: 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
82 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 ABC150 D — Semi Common Multiple, 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

  • 82 lines of Python from the credited upstream file abc150_d.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 codeABC150 D — Semi Common Multiple · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def lcm(a: int, b: int) -> int:    '''Compute least common multiple of a and b.    Args:        a: Int of number (greater than 0).        b: Int of number (greater than 0).    Returns:        least common multiple.    Landau notation: O(log n)    See:    https://gist.github.com/endolith/114336/eff2dc13535f139d0d6a2db68597fad2826b53c3    https://docs.python.org/3/library/sys.html#sys.version_info    '''     from math import gcd     return a * b // gcd(a, b)  # See:# https://blog.hamayanhamayan.com/entry/2020/01/11/145024def naive(n, m, b):    ans = 0        for i in range(m):        ok = True         for j in range(n):            # 各数字が(ak / 2)で割り切れない            if i % b[j] != 0:                ok = False            # 2p + 1が偶数になる            if (i // b[j]) % 2 == 0:                ok = False                        if ok:                print(i)                ans += 1        return ans  def main():    from math import ceil    import sys     input = sys.stdin.readline     n, m = map(int, input().split())    a = list(map(int, input().split()))    b = [0 for _ in range(n)]     for index, ai in enumerate(a):        if ai % 2 == 0:            b[index] = ai // 2        else:            print(0)            exit()        l = b[0]     for bi in b[1:]:        l = lcm(l, bi)        # 2p + 1が偶数になるかどうかをチェック    for bi in b:        if (l // bi) % 2 == 0:            print(0)            exit()     count = m // l    print(ceil(count / 2))      # print(naive(n, m, b))  if __name__ == "__main__":    main()

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