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:

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

大致題意：給你n個數，讓你從中選擇三個不同的數，將其中兩個數相加後異或上第三個數，使得結果最大，問最大結果為多少。

思路：先將這n個數建立成一顆01字典樹，然後O（n^2）的去列舉兩個不同的數，將其從字典樹中刪除，然後再查詢此時異或最大值，然後再向字典樹中加入這兩個數。

程式碼如下

1.陣列

#include<iostream>
#include<cstdio>
#include<algorithm>  
#include<cstring>
using namespace std;
#define LL long long 
#define ULL unsigned long long 
const int MAXN=32*1005;
struct Trie
{
    int ch[2];
    int size;
}T[MAXN];

int root=1,tot=1;
void add(int x)
{
    int p=root;
    T[p].size++;
    for(int i=31;i>=0;i--)
    {
        int k=(x>>i)&1;
        if(!T[p].ch[k]) T[p].ch[k]=++tot;

        p=T[p].ch[k];
        T[p].size++;    
    }
}
void Delete(int x)
{
    int p=root;
    T[p].size--;
    for(int i=31;i>=0;i--)
    {
        int k=(x>>i)&1;
        p=T[p].ch[k];
        T[p].size--;
    }
}

int query(int x)
{
    int ans=0;
    int p=root;
    for(int i=31;i>=0;i--)
    {
        int k=(x>>i)&1;
        if(T[p].ch[k^1]&&T[T[p].ch[k^1]].size)
            ans^=(1<<i),p=T[p].ch[k^1];     
        else 
            p=T[p].ch[k];
    }
    return ans;
}
void init()
{
    for(int i=1;i<=tot;i++)
    T[i].ch[0]=T[i].ch[1]=T[i].size=0;

    tot=1;
}
int a[1005];
int main() 
{
    int T;
    scanf("%d",&T);
    int n,num;
    for(int cas=1;cas<=T;cas++)
    {
        init();
        int ans=-1;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            add(a[i]);
        }
        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]));
                add(a[j]);
            }
            add(a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}

2.指標

#include<iostream>
#include<cstdio>
#include<algorithm>  
#include<cstring>
using namespace std;
#define LL long long 
#define ULL unsigned long long 

struct node
{
    int size;
    node *Next[2];
};

void add(node *head,int num)
{
    node *p=head;
    p->size++;
    for(int i=31;i>=0;i--)
    {
        int k=(num>>i)&1;
        if(p->Next[k]==NULL)
        {
            node *q=new node();
            p->Next[k]=q;
        }
        p=p->Next[k];
        p->size++;
    }
}
void Delete(node *head,int num)
{
    node *p=head;
    p->size--;
    for(int i=31;i>=0;i--)
    {
        int k=(num>>i)&1;
        p=p->Next[k];
        p->size--;
    }
}
int query(node *head,int num)
{
    node *p=head;
    int x=0;
    for(int i=31;i>=0;i--)
    {
        int k=(num>>i)&1;
        if(p->Next[k^1]&&p->Next[k^1]->size)
            p=p->Next[k^1],x^=(1<<i);
        else 
        p=p->Next[k];
    }
    return x;
}
int a[1005];
int main() 
{
    int T;
    scanf("%d",&T);
    int n,num;
    for(int cas=1;cas<=T;cas++)
    {
        node *head=new node();
        int ans=-1;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            add(head,a[i]);
        }
        for(int i=1;i<=n;i++)
        {
            Delete(head,a[i]);
            for(int j=i+1;j<=n;j++)
            {
                Delete(head,a[j]);
                ans=max(ans,query(head,a[i]+a[j]));
                add(head,a[j]);
            }
            add(head,a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}