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[1005];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m=-999;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&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++){
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 0;
}