Difference Row

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;
}