ZOJ3860-Find the Spy_whoooa! there is a spy in marjar university. all w-优快云博客

本文介绍了解决FindtheSpy编程题目的方法。通过输入学生ID卡号并找出特工的独特ID，使用C++实现了一个简单的算法。文章还分享了作者在解决此问题过程中遇到的一些常见错误及解决技巧。

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Find the Spy

Time Limit: 2 Seconds Memory Limit: 65536 KB

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;
}