Rabbits

Here N (N ≥ 3) rabbits are playing by the river. They are playing on a number line, each occupying a different integer. In a single move, one of the outer rabbits jumps into a space between any other two. At no point may two rabbits occupy the same position.
Help them play as long as possible
Input
The input has several test cases. The first line of input contains an integer t (1 ≤ t ≤ 500) indicating the number of test cases.
For each case the first line contains the integer N (3 ≤ N ≤ 500) described as above. The second line contains n integers a1a1 < a2a2 < a3a3 < … < aNaN which are the initial positions of the rabbits. For each rabbit, its initial position
aiai satisfies 1 ≤ aiai ≤ 10000.
Output
For each case, output the largest number of moves the rabbits can make.
Sample Input
5
3
3 4 6
3
2 3 5
3
3 5 9
4
1 2 3 4
4
1 2 4 5
Sample Output
1
1
3
0
1
这道题主要是审题问题，有两个容易误解的地方。第一个是题目中所说的两边其实是指第一只兔子和最后一只兔子，也就是队伍的最左最右两边，而不是相对于某一只特定的兔子的左边和右边。也就是说不是最左最右的兔子是不能动的。第二个是两边其实就是大范围的两边，而不一定是相邻的两边，因为题中强调了是any other two。接下来就是一次贪心。

#include"cstdio"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
#define N 550
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int n;
		int a[N],b[N];
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
		}
		int sum=0;
		for(int i=0;i<n-1;i++)
		{
			b[i]=a[i+1]-a[i]-1;
			sum+=b[i];
		}
		sum-=min(b[0],b[n-2]);
		printf("%d\n",sum);
	}
	return 0;
}