C-Rabbits【思维题·暑假训练赛1】

最新推荐文章于 2020-02-16 16:55:54 发布

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最新推荐文章于 2020-02-16 16:55:54 发布

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本文链接：https://blog.youkuaiyun.com/qq_43611144/article/details/98980480

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本文介绍了一种关于兔子在线上跳跃的游戏算法。游戏的目标是使N只兔子尽可能长时间地玩耍，通过让位于两端的兔子跳到其他兔子之间的空位。文章详细解释了输入输出格式，包括测试案例的数量、兔子初始位置的设定以及最大可能的移动次数。

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Description

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 a1 < a2 < a3 < · · · < aN which are the initial positions of the rabbits.
For each rabbit, its initial position ai satisfies 1 ≤ ai ≤ 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

Code

/*
题意：站在line中最左端和最右端的兔子向line中跳入任意两只中间存在空位的兔子的空位中,
*/
#include <iostream>
#include <cstdio>
using namespace std;
int num[510];

int main(){
    int t;
    cin>>t;
    while(t--){
       int n;
       cin>>n;
       for(int i=0;i<n;i++){
          scanf("%d",&num[i]);
          getchar();
       }
       int sum=0;
       for(int i=0;i<n-1;i++){
          sum=sum+num[i+1]-num[i]-1;
       }
       //舍去最左端、最右端中较小的一个
       if((num[1]-num[0])>(num[n-1]-num[n-2])) sum=sum-(num[n-1]-num[n-2]-1);
       else sum=sum-(num[1]-num[0]-1);
       printf("%d\n",sum);
    }
    return 0;
}