题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=5536

Chip Factory

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 4915 Accepted Submission(s): 2205

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

1 2 3

100 200 300

Sample Output

400

题目大意：

给 N 个数，在这 N 个数里找到三个值 i， j，k 使得（i+j）⊕ k 最大，输出这个最大值。

解题思路：

每次在01字典树里删除 i 和 j 然后（i+j）和01字典树里的匹配求最大异或值。

AC code：

 #include <bits/stdc++.h>

 #define ll long long int

 #define INF 0x3f3f3f3f

 using namespace std;

 const int MAXN = 1e3+;

 int ch[MAXN*][];

 ll value[MAXN*];

 ll b[MAXN];

 int num[MAXN*];

 int node_cnt;

 void init()

 {

     memset(ch[], , sizeof(ch[]));

     node_cnt = ;

 }

 void Insert(ll x)

 {

     int cur = ;

     for(int i = ; i >= ; i--){

         int index = (x>>i)&;

         if(!ch[cur][index]){

             memset(ch[node_cnt], , sizeof(ch[node_cnt]));

             ch[cur][index] = node_cnt;

             value[node_cnt] = ;

             num[node_cnt++] = ;

         }

         cur = ch[cur][index];

         num[cur]++;

     }

     value[cur] = x;

 }

 void update(ll x, int d)

 {

     int cur = ;

     for(int i =; i >= ; i--){

         int index = (x>>i)&;

         cur = ch[cur][index];

         num[cur]+=d;

     }

 }

 ll query(ll x)

 {

     int cur = ;

     for(int i = ; i >= ; i--)

     {

         int index = (x>>i)&;

         if(ch[cur][index^] && num[ch[cur][index^]]) cur = ch[cur][index^];

         else cur = ch[cur][index];

     }

     return x^value[cur];

 }

 int main()

 {

     int T_case, N;

     scanf("%d", &T_case);

     while(T_case--)

     {

         scanf("%d", &N);

         init();

         for(int i = ; i <= N; i++)

         {

             scanf("%lld", &b[i]);

             Insert(b[i]);

         }

         ll ans = ;

         for(int i = ; i <= N; i++){

             for(int j = ; j <= N; j++){

                 if(i == j) continue;

                 update(b[i], -);

                 update(b[j], -);

                 ans = max(ans, query(b[i]+b[j]));

                 update(b[i], );

                 update(b[j], );

             }

         }

         printf("%lld\n", ans);

     }

     return ;

 }

巴特西

HDU 5536 Chip Factory 【01字典树删除】