Codeforces 1385B

B. Restore the Permutation by Merger

time limit per test1 second
memory limit per test256 megabytes
input standard input
output standard output
A permutation of length n is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4,3,5,1,2], [3,2,1] are permutations, and [1,1], [0,1], [2,2,1,4] are not.

There was a permutation p[1…n]. It was merged with itself. In other words, let’s take two instances of p and insert elements of the second p into the first maintaining relative order of elements. The result is a sequence of the length 2n.

For example, if p=[3,1,2] some possible results are: [3,1,2,3,1,2], [3,3,1,1,2,2], [3,1,3,1,2,2]. The following sequences are not possible results of a merging: [1,3,2,1,2,3], [3,1,2,3,2,1], [3,3,1,2,2,1].

For example, if p=[2,1] the possible results are: [2,2,1,1], [2,1,2,1]. The following sequences are not possible results of a merging: [1,1,2,2], [2,1,1,2], [1,2,2,1].

Your task is to restore the permutation p by the given resulting sequence a. It is guaranteed that the answer exists and is unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤400) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1≤n≤50) — the length of permutation. The second line of the test case contains 2n integers a1,a2,…,a2n (1≤ai≤n), where ai is the i-th element of a. It is guaranteed that the array a represents the result of merging of some permutation p with the same permutation p.

Output

For each test case, print the answer: n integers p1,p2,…,pn (1≤pi≤n), representing the initial permutation. It is guaranteed that the answer exists and is unique.

题目大意：

给出2*n长度的数组，里面只包含了1 - n之间的数，输出1 - n相应的顺序。

解题思路：

大水题，看样例就知道了，遍历数组，如果这个数是第一次出现那么输出就行了。做个标记后面再出现就不用输出了。AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <vector>
using namespace std;
const int N = 100;
typedef long long ll;
int a[N];
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		memset(a, 0 ,sizeof a);
		int n;
		cin >> n;
		for (int i = 1; i <= 2 * n; i ++)
		{
			int s;
			cin >> s;
			if (!a[s])
			{
				cout << s << " ";
				a[s] = 1;
			}
		}
		cout << endl;
	}
	return 0;
}