Gym 100947C——Rotate It !!（水题水题）

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本文介绍了一个关于序列旋转的问题，通过合理的旋转策略来最大化特定玩家在取数游戏中获得的分数总和。文章详细解释了游戏规则及输入输出要求，并提供了一个示例帮助理解。

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Dzy and Fox have a sequence A consisting of N numbers [A1...AN]. Dzy starts by taking the first number, then Fox takes the second number, then Dzy takes the third number and so on, they continue taking turns until all of the N numbers are taken. The player with the highest sum of numbers wins.

Since Dzy is your dear friend, you decided to rotate the sequence (you may rotate it as many times as you like) in order to maximize Dzy's sum of numbers.

Rotation is defined as removing the first element from the beginning of the sequence and adding it to the end of the sequence.

So given the sequence A , you have to help Dzy and let him achieve the maximum possible sum.

Input

The first line containts a single integer T, the number of test cases.

Then T testcases are given as follows :

The first line of each testcase contains a single integer N (1 ≤ n ≤ 104).

The second line of each testcase contains N space-separated integers [A1...AN],the elements of the sequence A (1 ≤ i ≤ n) ( - 109 ≤ Ai ≤ 109).

Output

Output T lines , The answer for each testcase which is the maximum achievable sum by Dzy if you help him.

Examples

input
1
5
1 5 3 2 4

output
12

Note

Consider all 5 rotations of the sequence:

1 5 3 2 4 (Dzy score = 1 + 3 + 4 = 8)

5 3 2 4 1 (Dzy score = 8)

3 2 4 1 5 (Dzy score = 12)

2 4 1 5 3 (Dzy score = 6)

4 1 5 3 2 (Dzy score = 11)

水题一道随便看看就当凑数。

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <iostream>
using namespace std;

#define MAXN 10100
#define INF 10e9+7

long long a[MAXN];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",a+i);
        }
        int x=n/2;
        if(n%2==1)
            x++;
        long long maxn=-10e15;
        for(int i=1;i<=n;i++)
        {
            int cnt=1;
            long long sum=a[i];
            int j=i;
            while(cnt<x)
            {
                if((j+2)==n+2)
                {
                    j=2;
                }
                else if((j+2)==n+1)
                {
                    j=1;
                }
                else
                {
                    j=j+2;
                }
                sum+=a[j];
                cnt++;
            }
            maxn=max(sum,maxn);
        }
        printf("%I64d\n",maxn);
    }
}