hdu 5536 01字典树的删减

Chip Factory
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3409 Accepted Submission(s): 1494

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）

给你n个数，找出i,j,k满足 （ai+aj） ^ak最大
枚举i j即可
进行增加删减即可

#include <bits/stdc++.h>using namespace std;const int N = 101000;const int BB = 32;struct node{    int nxt[2],val;}T[N*BB+1];int sz;int a[N+10];void insert(int d,int val){    int p=0;    for(int i=31;i>=0;i--)    {        int dd=(d>>i)&1;        if(!T[p].nxt[dd]) T[p].nxt[dd]=++sz;        p=T[p].nxt[dd];        T[p].val+=val;    }}int ask(int x){    int ans=0;    int p=0;    for(int i=31;i>=0;i--)    {        int dd=(x>>i)&1;        if(T[T[p].nxt[1-dd]].val>0) p=T[p].nxt[1-dd],ans|=(1<<i);        else  if(T[T[p].nxt[dd]].val>0) p=T[p].nxt[dd];        else break;    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--){        sz=0;        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            insert(a[i],1);        }        int maxx=0;        for(int i=1;i<=n;i++)        {            insert(a[i],-1);            for(int j=i+1;j<=n;j++)            {                insert(a[j],-1);                int res=a[i]+a[j];                insert(a[j],1);                maxx=max(maxx,ask(res) );                cout<<maxx<<endl;            }            insert(a[i],1);        }        printf("%d\n",maxx );    }}