C. Birthday
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Cowboy Vlad has a birthday today! There are nn children who came to the celebration. In order to greet Vlad, the children decided to form a circle around him. Among the children who came, there are both tall and low, so if they stand in a circle arbitrarily, it may turn out, that there is a tall and low child standing next to each other, and it will be difficult for them to hold hands. Therefore, children want to stand in a circle so that the maximum difference between the growth of two neighboring children would be minimal possible.

Formally, let's number children from 11 to nn in a circle order, that is, for every ii child with number ii will stand next to the child with number i+1i+1, also the child with number 11 stands next to the child with number nn. Then we will call the discomfort of the circle the maximum absolute difference of heights of the children, who stand next to each other.

Please help children to find out how they should reorder themselves, so that the resulting discomfort is smallest possible.

Input

The first line contains a single integer nn (2≤n≤1002≤n≤100) — the number of the children who came to the cowboy Vlad's birthday.

The second line contains integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) denoting heights of every child.

Output

Print exactly nn integers — heights of the children in the order in which they should stand in a circle. You can start printing a circle with any child.

If there are multiple possible answers, print any of them.

Examples
input

Copy
5
2 1 1 3 2
output

Copy
1 2 3 2 1
input

Copy
3
30 10 20
output

Copy
10 20 30
Note

In the first example, the discomfort of the circle is equal to 11, since the corresponding absolute differences are 11, 11, 11 and 00. Note, that sequences [2,3,2,1,1][2,3,2,1,1] and [3,2,1,1,2][3,2,1,1,2] form the same circles and differ only by the selection of the starting point.

In the second example, the discomfort of the circle is equal to 2020, since the absolute difference of 1010 and 3030 is equal to 2020.

【题意】

有n个人围成一个圈,每个人有自己一个高度,让你对这n个人进行重新的排列,使得满足任意两个人之间高度差的最大值最小。

【分析】

我们就可以很容易的得出规律来,对高度序列排序之后,从n开始先输出下标序列为偶数的,然后在从1开始输出下标序列为奇数的值。

【代码】

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=105;
int n,a[N];
inline void Init(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",a+i);
}
inline void Solve(){
sort(a+1,a+n+1);
for(int i=1;i<=n;i+=2) printf("%d ",a[i]);
for(int i=n&1?n-1:n;i;i-=2) printf("%d%s",a[i],i<2?"":" ");
}
int main(){
Init();
Solve();
return 0;
}

 

 

 

CF 1131C Birthday的更多相关文章

  1. CF 1131A,1131B,1131C,1131D,1131F(Round541 A,B,C,D,F)题解

    A. Sea Battle time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  2. ORA-00494: enqueue [CF] held for too long (more than 900 seconds) by 'inst 1, osid 5166'

    凌晨收到同事电话,反馈应用程序访问Oracle数据库时报错,当时现场现象确认: 1. 应用程序访问不了数据库,使用SQL Developer测试发现访问不了数据库.报ORA-12570 TNS:pac ...

  3. cf之路,1,Codeforces Round #345 (Div. 2)

     cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....   ...

  4. cf Round 613

    A.Peter and Snow Blower(计算几何) 给定一个点和一个多边形,求出这个多边形绕这个点旋转一圈后形成的面积.保证这个点不在多边形内. 画个图能明白 这个图形是一个圆环,那么就是这个 ...

  5. ARC下OC对象和CF对象之间的桥接(bridge)

    在开发iOS应用程序时我们有时会用到Core Foundation对象简称CF,例如Core Graphics.Core Text,并且我们可能需要将CF对象和OC对象进行互相转化,我们知道,ARC环 ...

  6. [Recommendation System] 推荐系统之协同过滤(CF)算法详解和实现

    1 集体智慧和协同过滤 1.1 什么是集体智慧(社会计算)? 集体智慧 (Collective Intelligence) 并不是 Web2.0 时代特有的,只是在 Web2.0 时代,大家在 Web ...

  7. CF memsql Start[c]UP 2.0 A

    CF memsql Start[c]UP 2.0 A A. Golden System time limit per test 1 second memory limit per test 256 m ...

  8. CF memsql Start[c]UP 2.0 B

    CF memsql Start[c]UP 2.0 B B. Distributed Join time limit per test 1 second memory limit per test 25 ...

  9. CF #376 (Div. 2) C. dfs

    1.CF #376 (Div. 2)    C. Socks       dfs 2.题意:给袜子上色,使n天左右脚袜子都同样颜色. 3.总结:一开始用链表存图,一直TLE test 6 (1)如果需 ...

随机推荐

  1. RAID各种级别详细介绍

    独立硬盘冗余阵列(RAID, Redundant Array of Independent Disks),旧称廉价磁盘冗余阵列(RAID, Redundant Array of Inexpensive ...

  2. javascript对内容的操作

    我在这里介绍innerHTML.innerText.innerContent 一,innerHTML(可以识别标签): 案例1:替换掉整个标签 <!--innerHTML--> <p ...

  3. 【容斥】Four-tuples @山东省第九届省赛 F

    时间限制: 10 Sec 内存限制: 128 MB 题目描述 Given l1,r1,l2,r2,l3,r3,l4,r4, please count the number of four-tuples ...

  4. 轻量对象存储服务——minio

    minio Minio是一个非常轻量的对象存储服务. Github: minio 它本身不支持文件的版本管理.如果有这个需求,可以用 s3git 搭配使用. Github: s3git 安装 mini ...

  5. PHP03

    PHP03 1.提交地址: action.用户点击提交后,发送请求的地址.一般为了便于维护,最常见的是提交给当前文件,然后在当前文件判断是否为表单提交请求,表单的处理逻辑放在Html之前,为了避免写死 ...

  6. Ubuntu 18.04 安装Virtual Box or VMWare workstation Pro 14

    Linux相关的知识:https://www.cnblogs.com/dunitian/p/4822808.html#linux Virtual Box:sudo apt-get install vi ...

  7. .Net转Java.07.IDEA和VS常用操作、快捷键对照表

      功能 IDEA 2017.1 快捷键   Visual Studio 2015 快捷键 文档 格式化整个文档 Ctrl+Alt+L   Ctrl+E,D 或者 Ctrl+K,D  文件 显示最近的 ...

  8. nvidia-smi命令输出详解

    nvidia-smi命令输出如下: +-----------------------------------------------------------------------------+ | ...

  9. 正則表達式 - C语言

    http://blog.csdn.net/pipisorry/article/details/37073843 sscanf/scanf正则使用方法 %[ ] 的使用方法:%[ ]表示要读入一个字符集 ...

  10. 保存一个经常用的Makefile

    ############################################################# # Generic Makefile for C/C++ Program # ...