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
思路：这是我做过最水的C题。二分寻找符合条件的最小值就可以了。如果找不到，那么就不对。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=5e2+100;
vector<int> a,b,c;
int vis[maxx];
int n;
inline void init()
{
a.clear();
b.clear();
c.clear();
memset(vis,0,sizeof(vis));
}
int main()
{
int t,x,y;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
init();
for(int i=1;i<=n;i++) scanf("%d",&x),vis[x]=1,a.push_back(x);
for(int i=1;i<=2*n;i++) if(!vis[i]) b.push_back(i);
int flag=1;
for(int i=0;i<n;i++)
{
x=a[i];
y=lower_bound(b.begin(),b.end(),x)-b.begin();
if(y==b.size())
{
flag=0;
break;
}
c.push_back(x);
c.push_back(b[y]);
b.erase(b.begin()+y);
}
if(flag==0) cout<<"-1"<<endl;
else for(int i=0;i<2*n;i++) cout<<c[i]<<" ";cout<<endl;
}
return 0;
}
努力加油a啊，(o)/~
