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Problem solution · Python

ABC414 E — Count A%B=C

ABC414 E — Count A%B=C: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Sliding window or two pointers
Source
KATO-Hiro AtCoder Solutions
Length
57 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

Sliding window or two pointers

For ABC414 E — Count A%B=C, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 57 lines of Python from the credited upstream file abc414_e.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeABC414 E — Count A%B=C · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def floor_decomposition(number):    """    count = floor(number / pos), left < pos <= right     Returns a list of tuples (count, left, right) where:        - count is the number of times the number can be divided by right        - left is the next lower number that can be divided by count        - right is the current divisor     The process continues until right becomes zero.     Landau's O(√n) algorithm is used to decompose the number.     See:    https://atcoder.jp/contests/abc414/submissions/67527693    """     results = []    inf = 10**18    right = inf     while right:        count = number // right        left = number // (count + 1)         results.append((count, left, right))        right = left     return results  def main():    import sys     input = sys.stdin.readline     n = int(input())    mod = 998244353     ans = n * (n + 1) // 2    ans %= mod     results = floor_decomposition(n)     for count, left, right in results:        ans -= count * (right - left)        ans %= mod     print(ans)  if __name__ == "__main__":    main()

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