HDU 6627 2017ACM/ICPC亚洲区域赛沈阳站 Rabbits-优快云博客

本文介绍了一道关于N个兔子在数轴上移动的游戏问题，目标是让最外侧的兔子移动到其他任意两个兔子之间，以实现尽可能长的游戏时间。文章提供了具体的解题思路与C++代码实现。

Problem 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

题意：

有N个兔子在数轴上，最外面的兔子可以移动到其他任意两个兔子之间的位置上，一个位置只能待一次。问最大移动的次数。

题解：

其实想想很简单，答案就是相邻兔子的距离和减去第一段距离和最后一段距离中的较小值。

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=505;
int a[maxn];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>a[i];
        int sum=0;
        for(int i=1;i<n;i++)
        {
            int x=a[i]-a[i-1]-1;
            sum+=x;
        }
        int x=a[1]-a[0]<a[n-1]-a[n-2]?a[1]-a[0]:a[n-1]-a[n-2];
        cout<<sum-x+1<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/orion7/p/7783554.html