NanoApe Loves Sequence-待解决

NanoApe Loves Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 440    Accepted Submission(s): 205

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,...,An , denoting the elements of the sequence.

1≤T≤10, 3≤n≤100000, 1≤Ai≤109

 

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

1 4 1 2 3 4

 

Sample Output

6

 

Source

BestCoder Round #86

 

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#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    int t;
    int n;
    int cha=0;
    int cha2=0;
    int a[100005]={0};
    int maxx1;
    int first;
    int maxx2;
    int second;
    int maxx3;
    int third;
    int sum=0;
    scanf("%d",&t);
    for(int z=0;z<t;z++){
        sum=0;
        maxx1=0;
        maxx2=0;
        maxx3=0;
        scanf("%d",&n);

        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            if(i!=0){
                cha=abs(a[i]-a[i-1]);
                if(maxx1<cha){
                    maxx1=cha;
                    first=i;
                }
            }
        }

        for(int i=1;i<n;i++){
            cha=abs(a[i]-a[i-1]);
            if(maxx2<cha){
                if(maxx2<=maxx1){
                    if(i==first){
                        continue;
                    }else{
                        maxx2=cha;
                        second=i;
                    }
                }
            }
        }

        for(int i=1;i<n;i++){
            cha=abs(a[i]-a[i-1]);
            if(maxx2<cha){
                if(maxx3<=maxx1&&maxx3<=maxx2){
                    if(i==first||i==second){
                        continue;
                    }else{
                        maxx3=cha;
                        third=i;
                    }
                }
            }
        }

        for(int i=1;i<n-1;i++){
            cha2=abs(a[i+1]-a[i-1]);
            if(maxx1<=cha2){
                sum+=cha2;
            }
            if(maxx1>cha2){
                if(i==first&&i+1==second||i==second&&i+1==first){
                    if(maxx3!=0){
                        if(maxx3>=cha2){
                            sum+=maxx3;
                        }else{
                            sum+=cha2;
                        }

                    }else{
                        sum+=cha2;
                    }

                }
                if(i==first||i==first-1){
                    if(cha2<=maxx2){
                        sum+=maxx2;
                    }else{
                        sum+=cha2;
                    }
                }else{
                    sum+=maxx1;
                }

            }
        }
        if(first==1){
            sum+=maxx2;
            sum+=maxx1;
        }
        if(first==n-1){
            sum+=maxx2;
            sum+=maxx1;
        }
        if(first!=1&&first!=n-1){
            sum+=(2*maxx1);
        }
        printf("%d
",sum);
    }
    return 0;
}