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

求删除每个数后的序列最大差值的和

记录第一 第二 第三大的和，每次删除点时删除掉两个差值，新加入一个差值  第一个的最后一个数只删除一个差值

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

#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);
    }
}