HDU 5536 Chip Factory(字典树)-优快云博客

该问题描述了一个CPU芯片工厂的经理John如何计算每日生产的芯片的校验和。每个芯片有一个唯一的序列号，校验和是通过选取三个不同的芯片，计算它们序列号之和的异或值的最大值。输入包含芯片的数量和每个芯片的序列号，输出是计算得到的校验和。题目提示解决这类问题时要考虑异或运算的性质，并给出了一些样例输入和输出。

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Chip Factory

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 5258 Accepted Submission(s): 2358

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）

与HDU4825类似

https://blog.youkuaiyun.com/GYH0730/article/details/79821507

有关异或的题一定要先往2进制上想，这道题不同的地方是要先在字典树上删除a[i]与a[j]，在查询与(a[i]+a[j])的和异或的最大值

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 400005;
struct Trie_node
{
    int next[2];
    int flag;
    void init()
    {
        next[0] = next[1] = 0;
        flag = 0;
    }
}trie[MAXN];
int trie_n;
void Insert(int num)
{
    int root = 0,t,next;
    for(int i = 30; i >= 0; i--) {
        if(num & (1 << i)) t = 1;
        else t = 0;
        next = trie[root].next[t];
        if(next == 0) {
            next = trie[root].next[t] = ++trie_n;
        }
        root = next;
        trie[root].flag++;
    }
}
void Delete(int num)
{
    int root = 0,t;
    for(int i = 30; i >= 0; i--) {
        if(num & (1 << i)) t = 1;
        else t = 0;
        root = trie[root].next[t];
        trie[root].flag--;
    }
}
int Query(int num)
{
    int root = 0,t,next;
    for(int i = 30; i >=0; i--) {
        if(num & (1 << i)) t = 1;
        else t = 0;
        if(t == 1) {
            next = trie[root].next[0];
            if(trie[next].flag > 0 && next) {
                root = trie[root].next[0];
            }
            else {
                root = trie[root].next[1];
                num ^= (1 << i);
            }
        }
        else {
            next = trie[root].next[1];
            if(trie[next].flag > 0 && next) {
                root = trie[root].next[1];
                num ^= (1 << i);
            }
            else {
                root = trie[root].next[0];
            }
        }
    }
    return num;
}
int a[MAXN];
int main(void)
{
    int T,n,ans;
    scanf("%d",&T);
    while(T--) {
        for(int i = 0; i <= trie_n; i++) {
            trie[i].init();
        }
        trie_n = 0;
        scanf("%d",&n);
        for(int i = 0; i < n; i++) {
            scanf("%d",&a[i]);
            Insert(a[i]);
        }
        ans = 0;
        for(int i = 0; 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]));
                Insert(a[j]);
            }
            Insert(a[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}