Codeforces Round #623 (Div. 2, based on VK Cup 2019-2020 - Elimination Round, Engine) C. Restoring

C. Restoring Permutation
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
inputCopy
5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8
outputCopy
1 2
-1
4 5 1 2 3 6
-1
1 3 5 6 7 9 2 4 8 10
暴力暴力，

#include <bits/stdc++.h> using namespace std;int a[1000], b[1000], n, t; bool vis[1000]; void solve() {bool r=1;for (int i = 1; i <= n; ++i){b[2 * i - 1] = a[i];r = 1;for (int j = a[i]; j <= 2 * n; ++j)if (!vis[j]){b[2 * i] = j;vis[j] = 1;r = 0;break;}if (r == 1)break;}r = 1;for (int i = 1; i <= 2 * n; ++i)if (b[i] == 0){r = 0;break;}if (r == 0)puts("-1");else{for (int i = 1; i <= 2 * n; ++i)cout << b[i] << ' ';cout << endl;} } int main() {cin >> t;while (t--){memset(vis, 0, sizeof vis);memset(b, 0, sizeof b);cin >> n;bool r = 1;for (int i = 1; i <= n; ++i){cin >> a[i];vis[a[i]] = 1;if (a[i] == 2 * n)r = 0;}if (r == 0)puts("-1");elsesolve();} }

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