题目：
Description
You want to arrange n integers a1,?a2,?...,?an in some order in a row. Let's define the value of an arrangement as the sum of differences between all pairs of adjacent integers.More formally, let's denote some arrangement as a sequence of integers x1,?x2,?...,?xn, where sequence x is a permutation of sequence a. The value of such an arrangement is (x1?-?x2)?+?(x2?-?x3)?+?...?+?(xn?-?1?-?xn).
Find the largest possible value of an arrangement. Then, output the lexicographically smallest sequence x that corresponds to an arrangement of the largest possible value.
Input
The first line of the input contains integer n (2?≤?n?≤?100). The second line contains n space-separated integers a1, a2, ..., an (|ai|?≤?1000).
Output
Print the required sequence x1,?x2,?...,?xn. Sequence x should be the lexicographically smallest permutation of a that corresponds to an arrangement of the largest possible value.
Sample Input
Input
5
100 -100 50 0 -50
Output
100 -50 0 50 -100
Hint
In the sample test case, the value of the output arrangement is (100?-?(?-?50))?+?((?-?50)?-?0)?+?(0?-?50)?+?(50?-?(?-?100))?=?200. No other arrangement has a larger value, and among all arrangements with the value of 200, the output arrangement is the lexicographically smallest one.
Sequence x1,?x2,?... ,?xp is lexicographically smaller than sequence y1,?y2,?... ,?yp if there exists an integer r(0?≤?r?<?p) such that x1?=?y1,?x2?=?y2,?... ,?xr?=?yr and xr?+?1?<?yr?+?1.
解题思路：
观察题给公式，发现中间的数可以抵消，只剩下最两边的数，要想此值最多，只有将最大的数放在开头，最小的数放在末尾，就能得到的最大值，对于中间的数，
按从小到大排序便可以满足排列中lexicographically的最小
细节处理：
可以将输入的数字从小到大排序，再将头尾数交换
代码：
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,i,x;
vector<int> a;
cin>>n;
for(i=0;i<n;i++)
{cin>>x; a.push_back(x);}
sort(a.begin(),a.end());
int t;
t=a[0]; a[0]=a[n-1]; a[n-1]=t;
for(i=0;i<n;i++)
cout<<a[i]<<" ";
return 0;
}