Super Brain

最新推荐文章于 2019-02-18 16:58:29 发布

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本文介绍了一档科学真人秀节目SuperBrain中的一项挑战任务——参赛者需在短时间内记忆两组整数序列，并找出唯一一个在这两组序列中都出现的整数。文章提供了用于解决此问题的C++程序示例，该程序能够处理多个测试用例并输出正确答案。

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Super Brain is a famous scientific reality and talent show aiming to find people with exceptional brainpower.

In one of the challenges, two integer sequences and of length are given to the contestant. It's guaranteed that and hold for all , and there is exactly one integer in the first sequence which also appears in the second sequence. The contestant has to memorize the two sequences in a very short time, and find the integer which appears in both sequences correctly.

As a technical staff of the show, you are required to write a program and find out the correct integer.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains an integer (), indicating the length of the sequence.

The second line contains integers (), indicating the first sequence.

The third line contains integers (), indicating the second sequence.

It's guaranteed that and hold for all , and there is exactly one integer in the first sequence which also appears in the second sequence.

It's also guaranteed that the sum of over all test cases will not exceed .

<h4< dd="">Output

For each test case output one line containing one integer, indicating the integer which appears in both sequences.

<h4< dd="">Sample Input

3
3
3 1 2
5 3 4
2
38324 14122
38323 14122
1
180310
180310

<h4< dd="">Sample Output

3
14122
180310

这道题我能说我wa了好几次嘛w(ﾟДﾟ)w，老是超时......

#include<cstring>
#include<cstdio>
#include<iostream>
#include<cstdlib>
using namespace std;
int hash[1000001];
int main()
{
    int t,n;
    int x;
    scanf("%d",&t);
    while(t--)
    {
        memset(hash,0,sizeof(hash));
        scanf("%d",&n);
        for(int i=0;i<2*n;i++)
          {
              scanf("%d",&x);
              hash[x]++;
              if(hash[x]==2)
                  printf("%d\n",x);
          }
    }
    return 0;
}