Chip Factory ---- UVALive - 7192

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5204

John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces n chips today, the i-th chip produced this day has a serial number si .

At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below:

which i, j, k are three different integers between 1 and n. And ⊕ is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

Input

The first line of input contains an integer T indicating the total number of test cases.

The first line of each test case is an integer n, indicating the number of chips produced today. The next line has n integers s1, s2, . . . , sn, separated with single space, indicating serial number of each chip.

• 1 ≤ T ≤ 1000

• 3 ≤ n ≤ 1000

• 0 ≤ si ≤ 109

• There are at most 10 testcases with n > 100

Output

For each test case, please output an integer indicating the checksum number in a line.

Sample Input

2

3

1 2 3

3

100 200 300

Sample Output

6

400

这道题没啥好说的，训练赛的时候脑子抽抽了觉的以下代码会T掉，就想别的招去了，WA到怀疑人生QWQ

#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <climits>
#include <string>
#include <cstring>
using namespace std;
typedef long long ll;


int main()
{
    ios::sync_with_stdio(false);
    int T, n;
    ll a[1010];
    ll maxn;
    cin>>T;
    while( T-- )
    {

        maxn = 0;
        cin>>n;
        for(int i=0; i<n; i++ )
        {
            cin>>a[i];
        }

        for(int i=0; i<n; i++ )
        {
            for(int j=i+1; j<n; j++ )
            {
                for(int k=0; k<n; k++ )
                {
                    if( i!=j && i!=k && j!=k )
                    {
                        if( ((a[i]+a[j])^a[k]) > maxn )
                            maxn = (a[i]+a[j])^a[k];
                    }
                }
            }
        }
        cout<<maxn<<endl;
    }
    return 0;
}

再快一点的：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <climits>
#include <string>
#include <cstring>
using namespace std;
typedef long long ll;


int main()
{
    ios::sync_with_stdio(false);
    int T, n;
    ll a[1010];
    ll maxn;
    cin>>T;
    while( T-- )
    {

        maxn = 0;
        cin>>n;
        for(int i=0; i<n; i++ )
        {
            cin>>a[i];
        }

        for(int i=0; i<n; i++ )
        {
            for(int j=i+1; j<n; j++ )
            {
                for(int k=j + 1; k<n; k++ )
                {
                    maxn = max((a[i]+a[j])^a[k], maxn);
                    maxn = max((a[i]+a[k])^a[j], maxn);
                    maxn = max((a[j]+a[k])^a[i], maxn);
                }
            }
        }
        cout<<maxn<<endl;
    }
    return 0;
}