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$n$ 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$F$.

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

Input

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

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

The second line of the input contains $n$ 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$n$.

Sample Input

141 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 longusing 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);    }}