Codeforces Round 778 (Div. 1 + Div. 2, based on Technocup 2022 Final Round)
A
Maximum Cake Tastiness
Codeforces
800
原题
←B →
翻译中…

There are $n$ pieces of cake on a line. The $i$-th piece of cake has weight $a_i$ ($1 \leq i \leq n$).

翻译中…

The tastiness of the cake is the maximum total weight of two adjacent pieces of cake (i. e., $\max(a_1+a_2,\, a_2+a_3,\, \ldots,\, a_{n-1} + a_{n})$).

翻译中…

You want to maximize the tastiness of the cake. You are allowed to do the following operation at most once (doing more operations would ruin the cake):

翻译中…
  • Choose a contiguous subsegment $a[l, r]$ of pieces of cake ($1 \leq l \leq r \leq n$), and reverse it.
翻译中…

The subsegment $a[l, r]$ of the array $a$ is the sequence $a_l, a_{l+1}, \dots, a_r$.

翻译中…

If you reverse it, the array will become $a_1, a_2, \dots, a_{l-2}, a_{l-1}, \underline{a_r}, \underline{a_{r-1}}, \underline{\dots}, \underline{a_{l+1}}, \underline{a_l}, a_{r+1}, a_{r+2}, \dots, a_{n-1}, a_n$.

翻译中…

For example, if the weights are initially $[5, 2, 1, 4, 7, 3]$, you can reverse the subsegment $a[2, 5]$, getting $[5, \underline{7}, \underline{4}, \underline{1}, \underline{2}, 3]$. The tastiness of the cake is now $5 + 7 = 12$ (while before the operation the tastiness was $4+7=11$).

翻译中…

Find the maximum tastiness of the cake after doing the operation at most once.

翻译中…

Input

翻译中…

The first line contains a single integer $t$ ($1 \le t \le 50$) — the number of test cases.

翻译中…

The first line of each test case contains a single integer $n$ ($2 \le n \le 1000$) — the number of pieces of cake.

翻译中…

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — $a_i$ is the weight of the $i$-th piece of cake.

翻译中…

Output

翻译中…

For each test case, print a single integer: the maximum tastiness of the cake after doing the operation at most once.

翻译中…

Note

翻译中…

In the first test case, after reversing the subsegment $a[2, 5]$, you get a cake with weights $[5, \underline{7}, \underline{4}, \underline{1}, \underline{2}, 3]$. The tastiness of the cake is now $\max(5+7, 7+4, 4+1, 1+2, 2+3) = 12$. This is the maximum possible tastiness of the cake one can obtain by performing the operation at most once.

翻译中…

In the second test case, it's optimal not to do any operation. The tastiness is $78+78 = 156$.

翻译中…

In the third test case, after reversing the subsegment $a[1, 2]$, you get a cake with weights $[\underline{54}, \underline{69}, 91]$. The tastiness of the cake is now $\max(54+69, 69+91) = 160$. There is no way to make the tastiness of the cake greater than $160$ by performing at most one operation.

翻译中…

样例

Input #1
5
6
5 2 1 4 7 3
3
32 78 78
3
69 54 91
8
999021 999021 999021 999021 999652 999021 999021 999021
2
1000000000 1000000000
Output #1
12
156
160
1998673
2000000000
InputCopy OutputCopy
题目信息
题号
1654A
时间限制
2 s
内存限制
256 MB
标签
暴力
贪心
模拟
排序

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。