結果
問題 |
No.3075 Mex Recurrence Formula
|
ユーザー |
![]() |
提出日時 | 2025-03-28 21:25:43 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 837 ms / 2,000 ms |
コード長 | 5,101 bytes |
コンパイル時間 | 233 ms |
コンパイル使用メモリ | 82,912 KB |
実行使用メモリ | 187,428 KB |
最終ジャッジ日時 | 2025-03-28 21:26:07 |
合計ジャッジ時間 | 22,682 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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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.py import math from bisect import bisect_left, bisect_right class SortedSet(): BUCKET_RATIO = 16 SPLIT_RATIO = 24 def __init__(self, a) -> 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 = [], a for 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): for i in self.a: for j in i: yield j def __reversed__(self): for i in reversed(self.a): for j in reversed(i): yield j def __eq__(self, other) -> bool: return list(self) == list(other) def __len__(self) -> int: return self.size def __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): "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]: break return (a, i, bisect_left(a, x)) def __contains__(self, x) -> bool: if self.size == 0: return False a, _, i = self._position(x) return i != len(a) and a[i] == x def add(self, x) -> bool: "Add an element and return True if added. / O(√N)" if self.size == 0: self.a = [[x]] self.size = 1 return True a, b, i = self._position(x) if i != len(a) and a[i] == x: return False a.insert(i, x) self.size += 1 if len(a) > len(self.a) * self.SPLIT_RATIO: mid = len(a) >> 1 self.a[b:b+1] = [a[:mid], a[mid:]] return True def _pop(self, a, b: int, i: int): ans = a.pop(i) self.size -= 1 if not a: del self.a[b] return ans def discard(self, x) -> bool: "Remove an element and return True if removed. / O(√N)" if self.size == 0: return False a, b, i = self._position(x) if i == len(a) or a[i] != x: return False self._pop(a, b, i) return True def lt(self, x): "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) : "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) : "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) : "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) : "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 IndexError def pop(self, i: int = -1) : "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 IndexError def index(self, x) -> int: "Count the number of elements < x." ans = 0 for a in self.a: if a[-1] >= x: return ans + bisect_left(a, x) ans += len(a) return ans def index_right(self, x) -> int: "Count the number of elements <= x." ans = 0 for a in self.a: if a[-1] > x: return ans + bisect_right(a, x) ans += len(a) return ans from collections import defaultdict N,X=map(int,input().split()) A=list(map(int,input().split())) ss = SortedSet([i for i in range(N+2)]) dic=defaultdict(int) for i in range(N): dic[A[i]]+=1 if A[i] in ss: ss.discard(A[i]) for i in range(N+1): l=ss[0] A.append(l) dic[l]+=1 if dic[l]==1: ss.discard(l) dic[A[i]]-=1 if dic[A[i]]==0: ss.add(A[i]) if X<=N*2+1: print(A[X-1]) exit() X-=N+1 X%=(N+1) print(A[N+X])