ZOJ3860-Find the Spy

Find the Spy

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Whoooa! There is a spy in Marjar University. All we know is that the spy has a special ID card. Please find him out!

Input

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

The first line contains a integer N (3 <= N <= 100), which describes that there are N students need to be checked.

The second line contains N integers indicating the ID card number of N students. All ID card numbers are 32-bit integers.

Output

For each test case, output the ID card number which is different from others.

Sample Input

3
10
1 1 1 1 1 1 1 1 6 1
3
9 9 8
5
90016 90016 90016 2009 90016

Sample Output

6
8
2009

很显然的一道签到题，学长们看第一句话然后看看样例就分分钟把代码敲好了，而我这种菜鸟虽然不相信题目这么简单但还是敲好了代码，测试过，学长看到32-bit integers以为要用long long,之前在FZU上已经吃过亏了，所以果断用I64d输出，第一遍

CE了，显示什么头文件的问题，改改还是CE，才发现交题要用c++，但又PE了，哎，，，换了种思路却wa了两遍，于是又回到第一种思路代码上，仔细看看题，原来它指定的是lld输出，醉了，，改了一交，过了。无言以对。。。

在ZOJ上测试发现每个oj的测评系统都有点不同，下次还是要细心一点；

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int  N=100+10;
int a[N];
int main()
{
   int t,n,i;//居然int也能过；
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
      sort(a,a+n);//搜了下代码，原来思路和标准代码一模一样；
       if(a[0]==a[1])
           printf("%d\n",a[n-1]);
       else
        printf("%d\n",a[0]);
    }
    return 0;
}