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，
代码语言：javascript复制#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");
else
solve();
}
}
