hdu 6227 Rabbits题解

最新推荐文章于 2018-10-22 20:53:52 发布

原创最新推荐文章于 2018-10-22 20:53:52 发布 · 587 阅读

1 ·

CC 4.0 BY-SA版权

C++原创程序专栏收录该内容

15 篇文章

订阅专栏

本文介绍了一道编程题，题目要求帮助N只兔子通过跳跃的方式尽可能长时间地玩耍。文章提供了完整的代码实现，采用贪心策略来计算最大可能的游戏回合数。

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

Sample Output

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<iostream>
#include<cstdio>
#include<algorithm>
#define N 510
using namespace std;
int a[N];
int b[N];
int main(){
	int n,k;
	scanf("%d",&k);
	while(k--){
		scanf("%d",&n);
		int sum=0;
		for(int i=0;i<n;++i)
			scanf("%d",&a[i]);
		for(int i=0;i<n-1;++i){
			b[i]=a[i+1]-a[i]-1;
			sum+=b[i]; 
		}
		int ans=sum-min(b[0],b[n-2]);
		printf("%d\n",ans);			 
	}
	return 0;
}