結果

問題 No.273 回文分解
ユーザー Coki628
提出日時 2020-05-27 23:34:49
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
AC  
実行時間 34 ms / 2,000 ms
コード長 2,096 bytes
コンパイル時間 239 ms
コンパイル使用メモリ 12,928 KB
実行使用メモリ 11,008 KB
最終ジャッジ日時 2024-10-13 03:54:07
合計ジャッジ時間 2,277 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 32
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 19
MOD = 10 ** 9 + 7
EPS = 10 ** -10
def Manacher(S):
    """ Manacher(文字iを中心とする最長の回文の半径を返す)：O(N) """
    N = len(S)
    R = [0] * N
    i = j = 0
    while i < N:
        while i-j >= 0 and i+j < N and S[i-j] == S[i+j]:
            j += 1
        R[i] = j
        k = 1
        while i-k >= 0 and i+k < N and k+R[i-k] < j:
            R[i+k] = R[i-k]
            k += 1
        i += k
        j -= k
    return R
def Manacher_even(_S):
    """ Manacher偶数長(文字iとi+1の間を中心とする最長の回文の半径を返す)：O(N) """
    _N = len(_S)
    S = []
    for i in range(_N-1):
        S.append(_S[i])
        S.append('$')
    S.append(_S[-1])
    N = len(S)
    R = [0] * N
    i = j = 0
    while i < N:
        while i-j >= 0 and i+j < N and S[i-j] == S[i+j]:
            j += 1
        R[i] = j
        k = 1
        while i-k >= 0 and i+k < N and k+R[i-k] < j:
            R[i+k] = R[i-k]
            k += 1
        i += k
        j -= k
    res = [0] * (_N-1)
    j = 0
    for i in range(1, N, 2):
        res[j] = R[i] // 2
        j += 1
    return res
def get_max(S):
    res1 = Manacher(S)
    res2 = Manacher_even(S)
    res = 0
    for a in res1:
        a = a*2 - 1
        res = max(res, a)
    for a in res2:
        a *= 2
        res = max(res, a)
    return res
S = input()
N = len(S)
ans = 0
for i in range(1, N):
    ans = max(ans, get_max(S[:i]), get_max(S[i:]))
print(ans)
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