HDU 5536--Chip Factory(暴力)

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 5394    Accepted Submission(s): 2422

Problem Description

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:

maxi,j,k(si+sj)⊕sk

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

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

时间充裕，暴力一发0.0

 

 #include <iostream>
 #include<cstring>
 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 using namespace std;
 int a[];
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         int n,m=-;
         scanf("%d",&n);
         for(int i=;i<n;i++)
             scanf("%d",&a[i]);
         for(int i=;i<n;i++){
             for(int j=i+;j<n;j++){
                 for(int k=j+;k<n;k++){
                     m=max(m,(a[i]+a[j])^a[k]);
                     m=max(m,(a[j]+a[k])^a[i]);
                     m=max(m,(a[i]+a[k])^a[j]);
                 }
             }
         }
         printf("%d\n",m);
     }
     return ;
 }