You are given a sequence b1,b2,…,bn. Find the lexicographically minimal permutation a1,a2,…,a2n such that bi=min(a2i−1,a2i), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1≤t≤100).
The first line of each test case consists of one integer n — the number of elements in the sequence b (1≤n≤100).
The second line of each test case consists of n different integers b1,…,bn — elements of the sequence b (1≤bi≤2n).
It is guaranteed that the sum of n by all test cases doesn’t exceed 100.
Output
For each test case, if there is no appropriate permutation, print one number −1.
Otherwise, print 2n integers a1,…,a2n — required lexicographically minimal permutation of numbers from 1 to 2n.
Example
input
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
output
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10
题意：给你n个数的数组b，问是否存在1到2*n的排列a满足b[i]=min(a2i-1,a2i)
思路：首先由题意可知这个数列里必须有1，然后把每个出现的数都做标记，然后对数组b依次找对应的组合，1到2*n必须只出现一次
AC代码：
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int a[100010],b[100010],v[100010],c[100010];
int main()
{
int t;
cin>>t;
while(t--)
{
int n,f=0;
cin>>n;
memset(v,0,sizeof(v));
for(int i=1;i<=n;i++)
{
cin>>a[i];
v[a[i]]=1;
if(v[a[i]]>1)f=1;
}
if(v[1]==0)f=1;
int tot=1;
for(int i=1;i<=n;i++)
{
c[tot++]=a[i];
while(v[a[i]])
{
a[i]++;
if(a[i]>2*n)
f=1;
}
c[tot++]=a[i];
v[a[i]]=1;
}
if(f)
{
cout<<-1<<endl;
continue;
}
for(int i=1;i<=2*n;i++)
cout<<c[i]<<" ";
cout<<endl;
}
}