Rabbits (2017ACM/ICPC亚洲区沈阳站)

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

题意：

有n只兔子在不同的位置，任意一只最外面的兔子可以跳到其余任意两只兔子中间空位里，问所有兔子最多可移动的总次数。 

思路：

只要最外面的两只兔子中，有一只兔和其他兔相邻，即它们之间没有空格，那么最大可移动次数就是剩下的空格数之和。

所以就判断一下位于端点的两只兔子与它们最近兔子之间的距离，取相对距离小的一组，用总空格数-该组之间的空格数就可以了。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#include<map>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
const int mod=1e9+7;
const int M=1e4 + 10;
int a[M];
int main()
{
    int T, n, sum,minn;
    cin >> T;
    while(T--){
        cin >> n;
        cin >> a[0];
        sum = 0;
        for (int i = 1; i < n; i++){
           
            cin >> a[i];
            sum += a[i] - a[i - 1] - 1;
        }
        minn = inf;
        minn = min(a[1] - a[0] - 1, a[n - 1] - a[n - 2] - 1);
        cout << sum - minn << endl;
    }

    return 0;
}