結果
問題 | No.1433 Two color sequence |
ユーザー |
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提出日時 | 2022-06-05 11:40:14 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 343 ms / 2,000 ms |
コード長 | 2,347 bytes |
コンパイル時間 | 225 ms |
コンパイル使用メモリ | 82,404 KB |
実行使用メモリ | 116,464 KB |
最終ジャッジ日時 | 2024-09-21 04:08:17 |
合計ジャッジ時間 | 8,321 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 21 |
ソースコード
import sysimport pypyjitimport itertoolsimport heapqimport mathfrom collections import deque, defaultdictimport bisectinput = sys.stdin.readlinesys.setrecursionlimit(10 ** 6)pypyjit.set_param('max_unroll_recursion=-1')def index_lt(a, x):'return largest index s.t. A[i] < x or -1 if it does not exist'return bisect.bisect_left(a, x) - 1def index_le(a, x):'return largest index s.t. A[i] <= x or -1 if it does not exist'return bisect.bisect_right(a, x) - 1def index_gt(a, x):'return smallest index s.t. A[i] > x or len(a) if it does not exist'return bisect.bisect_right(a, x)def index_ge(a, x):'return smallest index s.t. A[i] >= x or len(a) if it does not exist'return bisect.bisect_left(a, x)class SegTree:def __init__(self, N, func, e):self.N = Nself.func = funcself.X = [e] * (N << 1)self.e = edef build(self, seq):for i in range(self.N):self.X[self.N + i] = seq[i]for i in range(self.N)[::-1]:self.X[i] = self.func(self.X[i << 1], self.X[i << 1 | 1])def add(self, i, x):i += self.Nself.X[i] += xwhile i > 1:i >>= 1self.X[i] = self.func(self.X[i << 1], self.X[i << 1 | 1])def update(self, i, x):i += self.Nself.X[i] = xwhile i > 1:i >>= 1self.X[i] = self.func(self.X[i << 1], self.X[i << 1 | 1])def query(self, L, R):L += self.NR += self.NvL = self.evR = self.ewhile L < R:if L & 1:vL = self.func(vL, self.X[L])L += 1if R & 1:R -= 1vR = self.func(self.X[R], vR)L >>= 1R >>= 1return self.func(vL, vR)N = int(input())T = input()[:-1]A = list(map(int, input().split()))A = [A[i] if T[i] == 'R' else -A[i] for i in range(N)]S = [0] * NS[0] = A[0]for i in range(1, N):S[i] = S[i - 1] + A[i]INF = 1 << 30stmin = SegTree(N, min, INF)stmax = SegTree(N, max, -INF)stmin.build(S)stmax.build(S)ans = -INFfor i in range(N):ma = stmax.query(i, N)mi = stmin.query(i, N)s = S[i - 1] if i > 0 else 0ans = max(ans, ma - s)ans = max(ans, -(mi - s))print(ans)