D-D. Gluttony（构造）

D. Gluttony
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array a with n distinct integers. Construct an array b by permuting a such that for every non-empty subset of indices S = {x1, x2, …, xk} (1 ≤ xi ≤ n, 0 < k < n) the sums of elements on that positions in a and b are different, i. e.

Input
The first line contains one integer n (1 ≤ n ≤ 22) — the size of the array.

The second line contains n space-separated distinct integers a1, a2, …, an (0 ≤ ai ≤ 109) — the elements of the array.

Output
If there is no such array b, print -1.

Otherwise in the only line print n space-separated integers b1, b2, …, bn. Note that b must be a permutation of a.

If there are multiple answers, print any of them.

Examples
inputCopy
2
1 2
outputCopy
2 1
inputCopy
4
1000 100 10 1
outputCopy
100 1 1000 10
Note
An array x is a permutation of y, if we can shuffle elements of y such that it will coincide with x.

Note that the empty subset and the subset containing all indices are not counted.

题意：要求每段区间的数字和都不相同的排列。
思路：需要有一个满足题意排列策略，一个可行的办法是将每个数字映射为它排序后位置的后一位（最后一位映射为第一位）。证明：每次选到一个区间，num[i] ~ num[i + k],排列后得到 num[i + x]（排序后下一位,最大则映射为最小） + num[i + 1 + y]…没有选到最大的数时，两者做差可得结果大于0。选到最大的数时，假设全部选上（实际不可能全部选上），作差得0，去掉任意一个原数中非最大值的数和映射后的结果，作差结果必然小于0,（可以自己手算模拟一下）。结论：区间映射后两者作差不为0，故不相等。

#include <bits/stdc++.h>

using namespace std;

struct node
{
    int x;
    int id;
}num[30];

int cmp(node a,node b)
{
    return a.x < b.x;
}

int main()
{
    int n,_hash[30];
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&num[i].x);
        num[i].id = i;
    }
    sort(num + 1,num + 1 + n,cmp);
    for(int i = 1;i <= n;i++)
    {
        _hash[num[i].id] = i;
    }
    for(int i  =1;i <= n;i++)
    {
        int t = (_hash[i] + 1) % n;
        if(t == 0)
            t = n;
        printf("%d ",num[t].x);
    }
    return 0;
}