結果
| 問題 | 
                            No.1368 サイクルの中に眠る門松列
                             | 
                    
| コンテスト | |
| ユーザー | 
                             GER_chen
                         | 
                    
| 提出日時 | 2021-01-29 21:48:45 | 
| 言語 | PyPy3  (7.3.15)  | 
                    
| 結果 | 
                             
                                AC
                                 
                             
                            
                         | 
                    
| 実行時間 | 306 ms / 2,000 ms | 
| コード長 | 1,410 bytes | 
| コンパイル時間 | 157 ms | 
| コンパイル使用メモリ | 81,864 KB | 
| 実行使用メモリ | 107,872 KB | 
| 最終ジャッジ日時 | 2024-06-27 08:02:28 | 
| 合計ジャッジ時間 | 3,593 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 * 1 | 
| other | AC * 15 | 
ソースコード
#import sys;input = lambda: sys.stdin.readline().rstrip()
#import heapq
#from collections import deque
#import numpy as np
#from collections import Counter as cnt
#from collections import defaultdict as ddc
#from math import factorial as fct
#from math import gcd
#from bisect import bisect_left as bsl
#from bisect import bisect_right as bsr
#from itertools import accumulate as acc
#from itertools import combinations as cmb
#from itertools import permutations as pmt
#from itertools import product as prd
#from functools import reduce as red
#import sys
#sys.setrecursionlimit(10**9)  #再帰を多く使う(デフォルトは1000)
for _ in range(int(input())):
    n = int(input())
    A = list(map(int,input().split()))
    A += A[:2]
    K = [0]*n
    for i in range(n):
        S = A[i:i+3]
        if len(set(S)) == 3 and S[1] in {min(S), max(S)}:
            K[i] = S[0]
    DP0 = [0]*n
    DP0[0], DP0[1], DP0[2] = K[0], K[0], K[0]
    for i in range(3, n):
        DP0[i] = max(DP0[i-1], DP0[i-2], DP0[i-3]+K[i])
    cand0 = DP0[-3]
    
    DP1 = [0]*n
    DP1[0], DP1[1], DP1[2] = 0, K[1], K[1]
    for i in range(3, n):
        DP1[i] = max(DP1[i-1], DP1[i-2], DP1[i-3]+K[i])
    cand1 = DP1[-2]
    
    DP2 = [0]*n
    DP2[0], DP2[1], DP2[2] = 0, 0, K[2]
    for i in range(3, n):
        DP2[i] = max(DP2[i-1], DP2[i-2], DP2[i-3]+K[i])
    cand2 = DP2[-1]
    print(max(cand0, cand1, cand2))
            
            
            
        
            
GER_chen