1408191944-hd-shǎ崽 OrOrOrOrz.cpp_c语言兔子生崽问题,成熟期为一个月,第二个月开始生兔子假设一对小兔的成熟期是一-优快云博客

本文链接：https://blog.youkuaiyun.com/wangluoershixiong/article/details/38687189

本文介绍了一个特殊的序列选择算法，该算法要求从一个不重复的整数序列中按特定顺序选择数字，即先选最大值，再选最小值，依此类推。文章提供了完整的C++实现代码，并针对输入序列长度为奇数的情况进行了特别处理。

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shǎ崽 OrOrOrOrz

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3853 Accepted Submission(s): 1193

Problem Description

Acmer in HDU-ACM team are ambitious, especially shǎ崽, he can spend time in Internet bar doing problems overnight. So many girls want to meet and Orz him. But Orz him is not that easy.You must solve this problem first.
The problem is :
Give you a sequence of distinct integers, choose numbers as following : first choose the biggest, then smallest, then second biggest, second smallest etc. Until all the numbers was chosen .
For example, give you 1 2 3 4 5, you should output 5 1 4 2 3

Input

There are multiple test cases, each case begins with one integer N(1 <= N <= 10000), following N distinct integers.

Output

Output a sequence of distinct integers described above.

Sample Input

51 2 3 4 5

Sample Output

5 1 4 2 3

题目大意

给你一个不同的整数序列，依次输出号码如下：首先选择最大的，然后第二的最小，最大，最小二等选择所有的号码。

注意事项

要注意给出的长度是奇数还是偶数，分别考虑。

代码

#include<stdio.h>
#include<algorithm>
using namespace std;
int s[11000];
int main()
{
	int n;
	int i,j,k;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
		    scanf("%d",&s[i]);
		sort(s,s+n);
		k=n/2;
		for(i=0,j=n-1;i<k;i++,j--)
		{
			printf("%d %d",s[j],s[i]);
			if(i!=k-1)
				printf(" ");
		}
		if(n%2==1)//长度是奇数的时候要另外考虑 
		    printf(" %d",s[k]);
		printf("\n");
	}
	return 0;
}