codeforces 891 B. Gluttony（构造）

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本篇博客介绍了一种算法挑战，任务是通过排列给定的唯一整数数组A来构造数组B，确保所有非空子集索引上的元素总和各不相同。提供了问题的输入输出格式、示例及解决方案。

B. Gluttony

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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.

$\text{[math]}$

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

input

2
1 2

output

2 1

input

4
1000 100 10 1

output

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.

思路：

每次b对应位置的元素是a对应位置刚好比a小一点的那个元素。

AC code

#include<bits/stdc++.h>
using namespace std;

int a[25],b[25],sub[25];

bool cmp(int x,int y)
{
    return a[x]<a[y];
}

int main()
{
    int n;cin>>n;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sub[i]=i;
    }
    sort(sub+1,sub+n+1,cmp);
    for(int i=1;i<n;i++)
        b[sub[i+1]]=a[sub[i]];
    b[sub[1]]=a[sub[n]];
    for(int i=1;i<n;i++)
        cout<<b[i]<<" ";
    cout<<b[n]<<endl;
    return 0;
}