HDU 5536/ 2015长春区域 J.Chip Factory Trie

Chip Factory

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

 

题意：给你n个数，为你不同的三个下标i,j,k，其中两个之和异或第三个的最大值是多少

题解：我们枚举其中两个得和，去不同下标最大，显然就裸字典树了。

///meek

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){
        if(ch=='-')f=-;ch=getchar();
    }
    while(ch>=''&&ch<=''){
        x=x*+ch-'';ch=getchar();
    }return x*f;
}
//****************************************
const int N=;
#define mod 10000007

#define inf 10000007
#define maxn 10000

struct Trie{
    int ch[N*][],siz,sum[N],word[N*];
    void init(){mem(ch),siz=;mem(word);}
    void insertt(int x) {
         int u=,k=,W=x,each[];mem(each);
         while(x) each[k++]=x%,x/=;
         for(int i=;i>=;i--) {
            int c=each[i];
            if(!ch[u][c]) {
                ch[u][c]=siz++;
                word[ch[u][c]]++;
            }
            else {
                word[ch[u][c]]++;
            }
            u=ch[u][c];
            if(i==)sum[u]=W;
         }
    }
    int ask(int x) {
        int u=,each[],k=,WW=x,g;mem(each);
        while(x) each[k++]=x%,x/=;
        for(int i=;i>=;i--) {
            int c=each[i];if(c==)g=;else g=;
            if(ch[u][g]&&word[ch[u][g]]) {
                u=ch[u][g];
            }
            else u=ch[u][c];
            if(i==) return WW^sum[u];
        }
    }
    void dele(int x) {
         int u=,each[],k=,g;mem(each);
        while(x) each[k++]=x%,x/=;
        for(int i=;i>=;i--) {
            int c=each[i];
            u=ch[u][c];
            word[u]--;
        }
    }
}trie;

int main(){
    int n,a[N],T=read();
    while(T--) {
        scanf("%d",&n);trie.init();
        for(int i=;i<=n;i++) {
            scanf("%d",&a[i]);
            trie.insertt(a[i]);
        }int ans=;
        for(int i=;i<=n;i++) {
            trie.dele(a[i]);
            for(int j=i+;j<=n;j++) {
                trie.dele(a[j]);
                ans=max(ans,trie.ask(a[i]+a[j]));
                trie.insertt(a[j]);
            }
            trie.insertt(a[i]);
        }
        printf("%d\n",ans);
    }
    return ;
}

代码