結果
問題 | No.3075 Mex Recurrence Formula |
ユーザー |
👑 |
提出日時 | 2025-03-18 01:27:56 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 428 ms / 2,000 ms |
コード長 | 5,444 bytes |
コンパイル時間 | 263 ms |
コンパイル使用メモリ | 82,548 KB |
実行使用メモリ | 122,016 KB |
最終ジャッジ日時 | 2025-03-27 12:53:22 |
合計ジャッジ時間 | 10,837 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 46 |
ソースコード
# https://github.com/tatyam-prime/SortedSet/blob/main/SortedSet.pyimport mathfrom bisect import bisect_left, bisect_rightfrom typing import Generic, Iterable, Iterator, TypeVarT = TypeVar('T')class SortedSet(Generic[T]):BUCKET_RATIO = 16SPLIT_RATIO = 24def __init__(self, a: Iterable[T] = []) -> None:"Make a new SortedSet from iterable. / O(N) if sorted and unique / O(N log N)"a = list(a)n = len(a)if any(a[i] > a[i + 1] for i in range(n - 1)):a.sort()if any(a[i] >= a[i + 1] for i in range(n - 1)):a, b = [], afor x in b:if not a or a[-1] != x:a.append(x)n = self.size = len(a)num_bucket = int(math.ceil(math.sqrt(n / self.BUCKET_RATIO)))self.a = [a[n * i // num_bucket : n * (i + 1) // num_bucket] for i in range(num_bucket)]def __iter__(self) -> Iterator[T]:for i in self.a:for j in i: yield jdef __reversed__(self) -> Iterator[T]:for i in reversed(self.a):for j in reversed(i): yield jdef __eq__(self, other) -> bool:return list(self) == list(other)def __len__(self) -> int:return self.sizedef __repr__(self) -> str:return "SortedSet" + str(self.a)def __str__(self) -> str:s = str(list(self))return "{" + s[1 : len(s) - 1] + "}"def _position(self, x: T) -> tuple[list[T], int, int]:"return the bucket, index of the bucket and position in which x should be. self must not be empty."for i, a in enumerate(self.a):if x <= a[-1]: breakreturn (a, i, bisect_left(a, x))def __contains__(self, x: T) -> bool:if self.size == 0: return Falsea, _, i = self._position(x)return i != len(a) and a[i] == xdef add(self, x: T) -> bool:"Add an element and return True if added. / O(√N)"if self.size == 0:self.a = [[x]]self.size = 1return Truea, b, i = self._position(x)if i != len(a) and a[i] == x: return Falsea.insert(i, x)self.size += 1if len(a) > len(self.a) * self.SPLIT_RATIO:mid = len(a) >> 1self.a[b:b+1] = [a[:mid], a[mid:]]return Truedef _pop(self, a: list[T], b: int, i: int) -> T:ans = a.pop(i)self.size -= 1if not a: del self.a[b]return ansdef discard(self, x: T) -> bool:"Remove an element and return True if removed. / O(√N)"if self.size == 0: return Falsea, b, i = self._position(x)if i == len(a) or a[i] != x: return Falseself._pop(a, b, i)return Truedef lt(self, x: T) -> T | None:"Find the largest element < x, or None if it doesn't exist."for a in reversed(self.a):if a[0] < x:return a[bisect_left(a, x) - 1]def le(self, x: T) -> T | None:"Find the largest element <= x, or None if it doesn't exist."for a in reversed(self.a):if a[0] <= x:return a[bisect_right(a, x) - 1]def gt(self, x: T) -> T | None:"Find the smallest element > x, or None if it doesn't exist."for a in self.a:if a[-1] > x:return a[bisect_right(a, x)]def ge(self, x: T) -> T | None:"Find the smallest element >= x, or None if it doesn't exist."for a in self.a:if a[-1] >= x:return a[bisect_left(a, x)]def __getitem__(self, i: int) -> T:"Return the i-th element."if i < 0:for a in reversed(self.a):i += len(a)if i >= 0: return a[i]else:for a in self.a:if i < len(a): return a[i]i -= len(a)raise IndexErrordef pop(self, i: int = -1) -> T:"Pop and return the i-th element."if i < 0:for b, a in enumerate(reversed(self.a)):i += len(a)if i >= 0: return self._pop(a, ~b, i)else:for b, a in enumerate(self.a):if i < len(a): return self._pop(a, b, i)i -= len(a)raise IndexErrordef index(self, x: T) -> int:"Count the number of elements < x."ans = 0for a in self.a:if a[-1] >= x:return ans + bisect_left(a, x)ans += len(a)return ansdef index_right(self, x: T) -> int:"Count the number of elements <= x."ans = 0for a in self.a:if a[-1] > x:return ans + bisect_right(a, x)ans += len(a)return ansdef main():N, X = map(int, input().split())A = list(map(int, input().split()))if X <= N:print(A[X - 1]);returnF = [0] * (N + 1)S = SortedSet(range(N + 1))def add(val):if val > N:returnif F[val] == 0:S.discard(val)F[val] += 1returndef discard(val):if val > N:returnF[val] -= 1if F[val] == 0:S.add(val)returnfor i in range(N):add(A[i])B = [0] * (N + 1)for i in range(N + 1):B[i] = S[0]add(B[i])if i < N:discard(A[i])print(B[(X - N - 1) % (N + 1)])returnmain()