Cowboy Vlad has a birthday today! There are n 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.web
Formally, let's number children from 1 to n in a circle order, that is, for every i child with number i will stand next to the child with number i+1, also the child with number 1 stands next to the child with number n. Then we will call the discomfort of the circle the maximum absolute difference of heights of the children, who stand next to each other.ide
Please help children to find out how they should reorder themselves, so that the resulting discomfort is smallest possible.ui
The first line contains a single integer n (2≤n≤100) — the number of the children who came to the cowboy Vlad's birthday.spa
The second line contains integers a1,a2,…,an (1≤ai≤109) denoting heights of every child.code
Print exactly n 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.orm
If there are multiple possible answers, print any of them.xml
5 2 1 1 3 2
1 2 3 2 1
3 30 10 20
10 20 30
In the first example, the discomfort of the circle is equal to 1, since the corresponding absolute differences are 1, 1, 1 and 0. Note, that sequences [2,3,2,1,1] and [3,2,1,1,2] form the same circles and differ only by the selection of the starting point.排序
In the second example, the discomfort of the circle is equal to 20, since the absolute difference of 10 and 30 is equal to 20.ip
有n我的围成一个圈,每一个人有本身一个高度,让你对这n我的进行从新的排列,使得知足任意两我的之间高度差的最大值最小。ci
咱们就能够很容易的得出规律来,对高度序列排序以后,从n开始先输出下标序列为偶数的,而后在从1开始输出下标序列为奇数的值。
#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;
}