HDU 5805 NanoApe Loves Sequence (BC #86)

NanoApe Loves Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 1089    Accepted Submission(s): 449

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

题意：给出n个数字，求每个数字删除之后相邻数字差的绝对值的最大值的期望。

分析：因为要求输出期望*n，所以只需要把删除每个数字后，相邻数字差的绝对值的最大值（记为d）加起来即可。

删除一个数字最多关系两个d，所以，预先记录删除前的最大值、次大值、第三大值。判断每个删除的数字是否是d两边的数字。不是就加上最大值，依次类推。

代码如下：

#include <stdio.h>
#include <math.h>
int a[100010];
int main()
{
	int T;
	int n,i,j;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++)
			scanf("%d",a+i);
		long long d=0,max1=-1,max2=-1,max3=-1,index1,index2,index3;
		for(i=0;i<n-1;i++)
		{
			d=fabs(a[i]-a[i+1]);
			if(d>max1)
			{
				max3=max2;
				index3=index2;
				max2=max1;
				index2=index1;
				max1=d;
				index1=i;
			}
			else if(d>max2)
			{
				max3=max2;
				index3=index2;
				max2=d;
				index2=i;
			}
			else if(d>max3)
			{
				max3=d;
				index3=i;
			}
		}
		long long sum=0;
		for(i=1;i<n-1;i++)
		{
			d=fabs(a[i-1]-a[i+1]);
			if(i!=index1 && i!=index1+1 && d<max1)
				sum+=max1;
			else if(i!=index2 && i!=index2+1 && d<max2)
				sum+=max2;
			else if(i!=index3 && i!=index3+1 && d<max3)
				sum+=max3;
			else
				sum+=d;
		}
		if(index1!=0)
			sum+=max1;
		else if(index2!=0)
			sum+=max2;
		
		if(index1+1!=n-1)
			sum+=max1;
		else if(index2+1!=n-1)
			sum+=max2;
		
		printf("%I64d\n",sum);
	}
	
	return 0;
}