Codeforces Round #201 (Div. 2) 347A Difference Row(脑洞)

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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 test(s)

input
5
100 -100 50 0 -50

output
100 -50 0 50 -100 

Note

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 thatx1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1.

给一个长度为n的数组，输出使得(x1 - x2) + (x2 - x3) + ... + (xn - 1 - xn)最大的数组顺序。

先排序，输出最大，输出中间，最后输出最小。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int maxn = 1005;
int main(int argc, char const *argv[])
{
    int n;
    while(scanf("%d", &n) != EOF) {
        int a[maxn];
        for(int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        sort(a, a + n);
        printf("%d", a[n - 1]);
        for(int i = 1; i < n - 1; ++i)
            printf(" %d", a[i]);
        printf(" %d\n", a[0]);
    }
    return 0;
}

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