HDU 5536 Chip Factory

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 620    Accepted Submission(s): 318

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）

 

解题：很明显这种题目用字典树，今年的多校有一道类似的但是难度更大的题目

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 struct Trie {
     int ch[maxn][],cnt[maxn],tot;
     int newnode() {
         cnt[tot] = ch[tot][] = ch[tot][] = ;
         return tot++;
     }
     void init() {
         tot = ;
         newnode();
     }
     void update(int v,int d,int rt = ) {
         for(int i = ; i >= ; --i) {
             int c = (v>>i)&;
             if(!ch[rt][c]) ch[rt][c] = newnode();
             rt = ch[rt][c];
             cnt[rt] += d;
         }
     }
     int match(int v,int rt = ,int ret = ) {
         for(int i = ; i >= ; --i) {
             int c = (v>>i)&;
             if(ch[rt][c^] && cnt[ch[rt][c^]]) {
                 ret |= <<i;
                 rt = ch[rt][c^];
             } else rt = ch[rt][c];
         }
         return ret;
     }
 }T;
 int a[maxn];
 int main() {
     int kase,n,ret;
     scanf("%d",&kase);
     while(kase--){
         scanf("%d",&n);
         T.init();
         for(int i = ; i < n; ++i){
             scanf("%d",a + i);
             T.update(a[i],);
         }
         for(int i = ret = ; i < n; ++i){
             T.update(a[i],-);
             for(int j = i + ; j < n; ++j){
                 T.update(a[j],-);
                 ret = max(ret,T.match(a[i] + a[j]));
                 T.update(a[j],);
             }
             T.update(a[i],);
         }
         printf("%d\n",ret);
     }
     return ;
 }