hdu NanoApe Loves Sequence

NanoApe Loves Sequence

Accepts: 505 Submissions: 1646 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others) Problem Description

NanoApe, the Retired Dog, has returned back to prepare for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with nnn numbers on the paper and then randomly deleted a number in the sequence. After that, he calculated the maximum absolute value of the difference of each two adjacent remained numbers, denoted as FFF.

Now he wants to know the expected value of FFF, if he deleted each number with equal probability.

Input

The first line of the input contains an integer TTT, denoting the number of test cases.

In each test case, the first line of the input contains an integer nnn, denoting the length of the original sequence.

The second line of the input contains nnn integers A1,A2,...,AnA_1, A_2, ..., A_nA​1​​,A​2​​,...,A​n​​, denoting the elements of the sequence.

1≤T≤10, 3≤n≤100000, 1≤Ai≤1091 \le T \le 10,~3 \le n \le 100000,~1 \le A_i \le 10^91≤T≤10, 3≤n≤100000, 1≤A​i​​≤10​9​​

Output

For each test case, print a line with one integer, denoting the answer.

In order to prevent using float number, you should print the answer multiplied by nnn.

Sample Input

1
4
1 2 3 4      

Sample Output

6      

求删除每個數後的序列最大內插補點的和

記錄第一 第二 第三大的和，每次删除點時删除掉兩個內插補點，新加入一個內插補點  第一個的最後一個數隻删除一個內插補點

/************************************************
┆  ┏┓　　　┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　┻　　　┃ ┆
┆┗━┓　  　┏━┛ ┆
┆　　┃　  　┃　　┆　　　　　　
┆　　┃　  　┗━━━┓ ┆
┆　　┃　 AC代馬 　　┣┓┆
┆　　┃　        　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#define ll long long
using namespace std;

int a[100010],Max[100010];

int Abs(int i)
{
    return i>0?i:(-i);
}

bool cmp(int i,int j)
{
    return i>j;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(Max,-1,sizeof(Max));
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
        for(int i=0;i<n-1;i++)
        Max[i]=Abs(a[i]-a[i+1]);
        sort(Max,Max+n-1,cmp);
        ll ans=0;
        for(int i=0;i<n;i++)
        {
            if(i==0)
            {
                int num=Abs(a[1]-a[0]);
                if(num==Max[0])
                {
                    ans+=Max[1];
                }
                else ans+=Max[0];
            }
            else if(i==n-1)
            {
                int num=Abs(a[n-1]-a[n-2]);
                if(num==Max[0])
                {
                    ans+=Max[1];
                }
                else ans+=Max[0];
            }
            else
            {
                int num1=Abs(a[i]-a[i-1]);
                int num2=Abs(a[i]-a[i+1]);
                int num3=Abs(a[i+1]-a[i-1]);
                if(num1==Max[0]&&num2==Max[1])
                {
                    ans+=max(num3,Max[2]);
                }
                else if(num1==Max[1]&&num2==Max[0])
                {
                    ans+=max(num3,Max[2]);
                }
                else if(num1==Max[0])
                {
                    ans+=max(num3,Max[1]);
                }
                else if(num2==Max[0])
                {
                    ans+=max(num3,Max[1]);
                }
                else ans+=max(num3,Max[0]);
            }
        }
        printf("%lld\n",ans);
    }
}