codeforces 1385B - Restore the Permutation by Merger(肯定用vector啦)
B. Restore the Permutation by Merger
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.
Example
input
5
2
1 1 2 2
4
1 3 1 4 3 4 2 2
5
1 2 1 2 3 4 3 5 4 5
3
1 2 3 1 2 3
4
2 3 2 4 1 3 4 1
output
1 2
1 3 4 2
1 2 3 4 5
1 2 3
2 3 4 1
链接：https://codeforces.ml/problemset/problem/1385/B
题意：给定一个排列数的数组，让同样的排列数数组，与其混合，要求不改变元素间的相对位置，请根据混合后的数组，找寻原来的排列数数组
思路：原数组是没有重复的，所以可以直接根据哪个先出现就先放到vector里面，最后输出就好了
ac代码：
#include
#include
#include
using namespace std;
int main()
{
int t,a;
cin>>t;
while(t--)
{
int n;
vector s;
cin>>n;
n = 2*n;
while(n--)
{
cin>>a;
if(find(s.begin(),s.end(),a)==s.end())
{
s.push_back(a);
}
}
vector::iterator it;
for(it=s.begin();it!=s.end();it++)
{
cout<< *it <<" ";
}
cout<