HDU 5805 NanoApe Loves Sequence（水模拟）

http://acm.hdu.edu.cn/showproblem.php?pid=5805
NanoApe Loves Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 48 Accepted Submission(s): 23

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

很水的题，少考虑了一种 情况，赛后改了一点东西就AC了，真是日了狗了。
先把每两个之间的差求出来，然后sort一下，然后每删除一个值的话，会出现两个新的值，消失两个旧的值。
然后开始和最后特判一下就OK了。
下面是AC代码：

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<queue>#include<stack>using namespace std;#define ll long longint a[100005];int b[300004];int main(){    int n,m;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]);        }        int num=0;        for(int i=2; i<=n; i++)        {            b[num++]=abs(a[i]-a[i-1]);        }        ll sss=0;        sort(b,b+num);        if(abs(a[2]-a[1])==b[num-1])        {            sss+=b[num-2];        }        else        {            sss+=b[num-1];        }        for(int i=2; i<n; i++)        {            if(abs(a[i+1]-a[i-1])>b[num-1])            {                sss+=abs(a[i+1]-a[i-1]);            }            else if((abs(a[i]-a[i-1])==b[num-1]&&abs(a[i+1]-a[i])==b[num-2])||(abs(a[i]-a[i-1])==b[num-2]&&abs(a[i+1]-a[i])==b[num-1]))            {                sss+=b[num-3];            }            else if(abs(a[i]-a[i-1])==b[num-1]||abs(a[i+1]-a[i])==b[num-1])            {                sss+=b[num-2];            }            else            {                sss+=b[num-1];            }        }        if(abs(a[n]-a[n-1])==b[num-1])        {            sss+=b[num-2];        }        else        {            sss+=b[num-1];        }        printf("%I64d\n",sss);    }    return 0;}