結果
問題 |
No.186 中華風 (Easy)
|
ユーザー |
![]() |
提出日時 | 2022-12-13 13:33:17 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 81 ms / 2,000 ms |
コード長 | 1,879 bytes |
コンパイル時間 | 174 ms |
コンパイル使用メモリ | 82,176 KB |
実行使用メモリ | 68,608 KB |
最終ジャッジ日時 | 2024-11-07 09:06:31 |
合計ジャッジ時間 | 2,918 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
other | AC * 23 |
ソースコード
from typing import List, Tuple, Callable, TypeVar, Optional import sys import itertools import heapq import bisect import math from collections import deque, defaultdict from functools import lru_cache, cmp_to_key import string input = sys.stdin.readline if __file__ != 'prog.py': sys.setrecursionlimit(10 ** 6) def readints(): return map(int, input().split()) def readlist(): return list(readints()) def readstr(): return input()[:-1] def extgcd(a: int, b: int) -> Tuple[int, int, int]: """solve ax + by = gcd(a, b)""" x, y, u, v = 1, 0, 0, 1 while b: q, r = divmod(a, b) x -= q * u y -= q * v x, u = u, x y, v = v, y a, b = b, r return a, x, y def crt(P: Tuple[int, int]) -> Tuple[int, int]: """return the smallest x that satisfies x ≡ a (mod m) for all (a, m) pairs and lcm(M) # reference: https://qiita.com/drken/items/ae02240cd1f8edfc86fd """ """ if there exist x s.t. x ≡ a1 (mod m1) and x ≡ a2 (mod m2), let g = gcd(m1, m2). since m1 and m2 are multiple of g, x ≡ a1 (mod g) and x ≡ a2 (mod g). therefore a1 ≡ a2 (mod g). there exist (p, q) s.t. m1 * p + m2 * q = gcd(m1, m2). (this can be obtained by extgcd) let g = gcd(m1, m2) and s = (a2 - a1) / g then m1 * p + m2 * q = g <=> m1 * p + m2 * q = (a2 - a1) / s <=> s * m1 * p + s * m2 * q = a2 - a1 <=> a1 + s + m1 + p = a2 - s * m2 * q let x = a1 + s * m1 * p (= a2 - s * m2 * q) then x ≡ a1 (mod m1), x ≡ a2 (mod m2). """ r = 0 M = 1 for a, m in P: g, p, q = extgcd(M, m) if (a - r) % g != 0: return (0, -1) s = (a - r) // g tmp = s * p % (m // g) r += M * tmp M *= m // g return (r, M) P = [tuple(readints()) for _ in range(3)] ans, lcm = crt(P) print(ans if ans > 0 else lcm)