結果
問題 | No.2240 WAC |
ユーザー |
|
提出日時 | 2023-03-10 21:44:12 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 93 ms / 2,000 ms |
コード長 | 4,928 bytes |
コンパイル時間 | 148 ms |
コンパイル使用メモリ | 82,156 KB |
実行使用メモリ | 89,400 KB |
最終ジャッジ日時 | 2024-09-18 04:00:23 |
合計ジャッジ時間 | 4,349 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 43 |
ソースコード
import sysinput = lambda :sys.stdin.readline()[:-1]ni = lambda :int(input())na = lambda :list(map(int,input().split()))yes = lambda :print("yes");Yes = lambda :print("Yes");YES = lambda : print("YES")no = lambda :print("no");No = lambda :print("No");NO = lambda : print("NO")######################################################################## https://github.com/tatyam-prime/SortedSet/blob/main/SortedSet.pyimport mathfrom bisect import bisect_left, bisect_rightfrom typing import Generic, Iterable, Iterator, TypeVar, Union, ListT = TypeVar('T')class SortedSet(Generic[T]):BUCKET_RATIO = 50REBUILD_RATIO = 170def _build(self, a=None) -> None:"Evenly divide `a` into buckets."if a is None: a = list(self)size = self.size = len(a)bucket_size = int(math.ceil(math.sqrt(size / self.BUCKET_RATIO)))self.a = [a[size * i // bucket_size : size * (i + 1) // bucket_size] for i in range(bucket_size)]def __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)if not all(a[i] < a[i + 1] for i in range(len(a) - 1)):a = sorted(set(a))self._build(a)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 __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 _find_bucket(self, x: T) -> List[T]:"Find the bucket which should contain x. self must not be empty."for a in self.a:if x <= a[-1]: return areturn adef __contains__(self, x: T) -> bool:if self.size == 0: return Falsea = self._find_bucket(x)i = bisect_left(a, 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 = self._find_bucket(x)i = bisect_left(a, x)if i != len(a) and a[i] == x: return Falsea.insert(i, x)self.size += 1if len(a) > len(self.a) * self.REBUILD_RATIO:self._build()return Truedef discard(self, x: T) -> bool:"Remove an element and return True if removed. / O(√N)"if self.size == 0: return Falsea = self._find_bucket(x)i = bisect_left(a, x)if i == len(a) or a[i] != x: return Falsea.pop(i)self.size -= 1if len(a) == 0: self._build()return Truedef lt(self, x: T) -> Union[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) -> Union[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) -> Union[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) -> Union[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, x: int) -> T:"Return the x-th element, or IndexError if it doesn't exist."if x < 0: x += self.sizeif x < 0: raise IndexErrorfor a in self.a:if x < len(a): return a[x]x -= 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 ansn,m = na()s = input()a = []for i in range(n*2+m*2):if s[i] == "A":a.append(i)r = 0for i in range(n*2+m*2):if s[i] == "C":if a[r] > i:No()exit()r += 1r = 0for i in range(n*2+m*2-1,-1,-1):if s[i] == "W":if a[-1-r] < i:No()exit()r += 1Yes()