時間限制: 5 Sec  記憶體限制: 128 MB                                                                          提交: 260  解決: 57                                                            [提交] [狀態] [討論版] [命題人:admin]

題目描述

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:

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?                                                   

輸入

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≤s i≤109
• There are at most 10 testcases with n > 100

輸出

For each test case, please output an integer indicating the checksum number in a line.

樣例輸入

2
3
1 2 3
3
100 200 300

 樣例輸出

6
400

                                                                                     [提交] [狀態]

題目大意：

給 N 個數，在這 N 個數裡找到三個值 i， j，k 使得（i+j）⊕ k 最大，輸出這個最大值。

解題思路：

每次在01字典樹裡刪除 i 和 j 然後 （i+j）和01字典樹裡的匹配求最大異或值。 

#include <bits/stdc++.h>
#define ll long long int
#define INF 0x3f3f3f3f
using namespace std;

const int MAXN = 1e3+10;
int ch[MAXN*32][2];
ll value[MAXN*32];
ll b[MAXN];
int num[MAXN*32];
int node_cnt;

void init()
{
    memset(ch[0], 0, sizeof(ch[0]));
    node_cnt = 1;
}

void Insert(ll x)
{
    int cur = 0;
    for(int i = 32; i >= 0; i--){
        int index = (x>>i)&1;
        if(!ch[cur][index]){
            memset(ch[node_cnt], 0, sizeof(ch[node_cnt]));
            ch[cur][index] = node_cnt;
            value[node_cnt] = 0;
            num[node_cnt++] = 0;
        }
        cur = ch[cur][index];
        num[cur]++;
    }
    value[cur] = x;
}

void update(ll x, int d)
{
    int cur = 0;
    for(int i =32; i >= 0; i--){
        int index = (x>>i)&1;
        cur = ch[cur][index];
        num[cur]+=d;
    }
}

ll query(ll x)
{
    int cur = 0;
    for(int i = 32; i >= 0; i--)
    {
        int index = (x>>i)&1;
        if(ch[cur][index^1] && num[ch[cur][index^1]]) cur = ch[cur][index^1];
        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 = 1; i <= N; i++)
        {
            scanf("%lld", &b[i]);
            Insert(b[i]);
        }
        ll ans = 0;
        for(int i = 1; i <= N; i++){
            for(int j = 1; j <= N; j++){
                if(i == j) continue;
                update(b[i], -1);
                update(b[j], -1);
                ans = max(ans, query(b[i]+b[j]));
                update(b[i], 1);
                update(b[j], 1);
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}