题目：
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.
题目大意：不同身高的人围成一圈，问怎样排列才能使身高差异最小
解决方法：本题是想求一种能将两数之间差值最小化的排列方式，因为是环形排列，所以仅需先将原数组从小到大排序，在依照左边一个a1，右边一个a2，依次即可。
AC代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
int a[120];
int b[120];
int main()
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);int n;cin>>n;rep(i,1,n) cin>>a[i];sort(a+1,a+1+n);for(int i=1,j=1;i<=n;i+=2,j++) {b[j]=a[i];b[n-j+1]=a[i+1];}if(n%2!=0) b[n/2+1]=a[n];rep(i,1,n-1) cout<<b[i]<<" ";cout<<b[n]<<endl;return 0;
}
