hdu 5536 Chip Factory

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 208    Accepted Submission(s): 120


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

 
题目大意：
在一个数列中找出maxi,j,k(si+sj)⊕sk的
解题思路：
异或其实没有什么好办法，直接模拟也可以暴力过，但是当时没实现我们最后用字典树过的，在一个字典书中加入所有的数，再从树中拿出ai,aj，算完后在加入，计算最大值就好；
麻烦的就删除操作；

代码：

#include "iostream"
#include "cstdio"
#include "string.h"
#include "math.h"
using namespace std;
const int maxn=1000010;
struct Trie
{
int v,ch[2];
};
Trie node[maxn];
int a[1010];
int num=1,root=0;
void Insert(int x)
{
int u=root;
node[u].v++;
for(int i=30;i>=0;i--)
{
int c;
if(x&(1<<i))
c=1;
else
c=0;
if(node[u].ch[c]==NULL)
node[u].ch[c]=num++;
u=node[u].ch[c];
node[u].v++;
}
}
void Delete(int x)
{
int u=root;
node[u].v--;
for(int i=30;i>=0;i--)
{
int c;
if(x&(1<<i))
c=1;
else
c=0;
u=node[u].ch[c];
node[u].v--;
}
}
int query(int x)
{
int u=root;
for(int i=30;i>=0;i--)
{
int c;
if(x&(1<<i))
c=1;
else
c=0;
if(c==1)
{
if(node[u].ch[0]&&node[node[u].ch[0]].v)
u=node[u].ch[0];
else
{
u=node[u].ch[1];
x=x^(1<<i);
}
}
else
{
if(node[u].ch[1]&&node[node[u].ch[1]].v)
{
u=node[u].ch[1];
x=x^(1<<i);
}
else
u=node[u].ch[0];
}
}
return x;
}
int main(int argc, char* argv[])
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(node,0,sizeof(node));
num=1;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
Insert(a[i]);
}
int ans=-1;
for(int i=1;i<=n;i++)
{
Delete(a[i]);
for(int j=i+1;j<=n;j++)
{
Delete(a[j]);
ans=max(ans,query(a[i]+a[j]));
Insert(a[j]);
}
Insert(a[i]);
}
printf("%d\n",ans);
}
return 0;
}