Rotate It !!(思维)

Rotate It !!

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice Gym 100947C

Description

standard input/output  Announcement

 

• Statements

  Dzy and Fox have a sequence A consisting of N numbers [A₁...A_N]. 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...A_N],the elements of the sequence A (1 ≤ i ≤ n) ( - 10⁹ ≤ A_i ≤ 10⁹).

Output

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

Sample Input

Input

1 5 1 5 3 2 4

Output

12

Hint

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)

题解：一个环状数据，Dzy从第一个开始取，每隔一个取一个，取的数最大是多少，注意若数为奇数，Dzy取的数一定最多；

想法是先找到总数，然后枚举每个位置为最后一个位置；

见代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1e4 + ;
int num[MAXN];
typedef long long LL;
int main(){
    int T, N;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &N);
        LL sum = , cur;
        if(N == ){
            scanf("%lld", &cur);
            printf("%lld\n", cur);
            continue;
        }
        for(int i = ; i < N; i++){
            scanf("%d", num + i);
            sum += num[i];
        }
        if(N %  == ){
            cur = ;
            for(int i = ; i < N; i+= )
                cur += num[i];
            printf("%lld\n", max(cur, sum - cur));
        }
        else{
            LL sum1 = , sum2 = , cur1 = , cur2 = ;
            for(int i = ; i < N; i++){
                if(i %  == )
                    sum1 += num[i];
                else
                    sum2 += num[i];
            }
            LL ans = sum1;
            for(int i = ; i < N; i++){
                if(i % == )
                    cur1 += num[i];
                else
                    cur2 += num[i];
                if(i %  == ){
                    ans = max(ans, sum2 - cur2 + cur1);
                //    printf("%lld\n", sum2 - cur2 + cur1);
                }
                else{
                    ans = max(ans, sum1 - cur1 + cur2);
                //    printf("%lld\n", sum1 - cur1 + cur2);
                }

            }
            printf("%lld\n", ans);
        }
    }
    return ;
}