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：這是我做過最水的C題。二分尋找符合條件的最小值就可以了。如果找不到，那么就不對。
代碼如下：

#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)/~

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