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 = 100010;
struct Trie {
    int ch[maxn][2],cnt[maxn],tot;
    int newnode() {
        cnt[tot] = ch[tot][0] = ch[tot][1] = 0;
        return tot++;
    }
    void init() {
        tot = 0;
        newnode();
    }
    void update(int v,int d,int rt = 0) {
        for(int i = 30; i >= 0; --i) {
            int c = (v>>i)&1;
            if(!ch[rt][c]) ch[rt][c] = newnode();
            rt = ch[rt][c];
            cnt[rt] += d;
        }
    }
    int match(int v,int rt = 0,int ret = 0) {
        for(int i = 30; i >= 0; --i) {
            int c = (v>>i)&1;
            if(ch[rt][c^1] && cnt[ch[rt][c^1]]) {
                ret |= 1<<i;
                rt = ch[rt][c^1];
            } 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 = 0; i < n; ++i){
            scanf("%d",a + i);
            T.update(a[i],1);
        }
        for(int i = ret = 0; i < n; ++i){
            T.update(a[i],-1);
            for(int j = i + 1; j < n; ++j){
                T.update(a[j],-1);
                ret = max(ret,T.match(a[i] + a[j]));
                T.update(a[j],1);
            }
            T.update(a[i],1);
        }
        printf("%d
",ret);
    }
    return 0;
}