*hdu 5536（字典树的运用）

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

题意：求下面这个公式的最大值：

maxi,j,k(si+sj)⊕sk

思路：如果用普通方法你要分别枚举3个数，n^3感觉会超时的。

然而完全莫有想到能用字典树，你先把所有的数保存下来，然后删去要用的i和j，再在里面找出能和a[i]+a[j]异或

出的最大值。相当于值需要枚举i和j即可。 /*好机智

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <vector>
#include <algorithm>
#include <functional>
typedef long long ll;
using namespace std;

int a[1005];

struct node
{
int number;
int flag;
int next[2];
void ini()
{
next[0] = next[1] = 0;
flag = 0;
}
} pnode[1000005];
int root = 0;
int tot;

void inser(int x)
{
int tt = root;
for(int i = 30; i >= 0; i --)
{
int t;
if(x & (1<<i))t = 1;
else  t = 0;
if(!pnode[tt].next[t])
{
pnode[tt].next[t] = ++tot;
pnode[tot].number = t;
}
tt = pnode[tt].next[t];
pnode[tt].flag++;
}
}

void delet(int x)
{
int tt = root;
for(int i = 30; i >= 0; i--)
{
int t;
if(x & (1<<i))t = 1;
else  t = 0;
tt = pnode[tt].next[t];
pnode[tt].flag --;
}
}

int query(int x)
{
int tt = root;
for(int i = 30; i >= 0; i--)
{
int t;
if(x & (1<<i)) t = 1;
else t = 0;
if(t == 1)
{
int nex = pnode[tt].next[0];
if(pnode[nex].flag > 0 && nex) tt = pnode[tt].next[0];
else
{
tt = pnode[tt].next[1];
x ^= (1<<i);
}
}
else
{
int nex = pnode[tt].next[1];
if(pnode[nex].flag > 0 && nex)
{
tt = pnode[tt].next[1];
x ^= (1<<i);
}
else tt = pnode[tt].next[0];
}
}
return x;
}

int main()
{
int T,n;
scanf("%d",&T);
while(T--)
{
int ans = 0;
scanf("%d",&n);
tot = 0;
for(int i = 0; i < n; i++)   scanf("%d",a+i);
for(int i = 0; i < n; i++)   inser(a[i]);

for(int i = 0; i < n; i++)
{
delet(a[i]);
for(int j = i+1; j < n; j++)
{
delet(a[j]);
ans = max(ans,query(a[i] + a[j]));
inser(a[j]);
}
inser(a[i]);
}
for(int i = 0;i < tot;i++)
{
pnode[i].ini();
}
printf("%d\n",ans);
}
return 0;
}