HDU5536 Chip Factory(字典树)

题解：

给n个数字从中选择三个数字a[i],a[j],a[k]，使得a[i]+a[j]异或a[k]最大

建一个字典树，枚举a[i]+a[j]，删除之后查找然后再插回到字典树中，往相反的方向走，没有则同方向走

代码

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int max(int a,int b)
{
   if (a>=b)
       return a;
   else
       return b;
}
const int maxn = 500050;
int Next[maxn][2];
int word[maxn];
struct Tree
{

   int L,root;

   void init()
   {
      L = 0;
      root = newnode();
   }

   int newnode()
   {
      for (int i = 0;i<2;i++)
          Next[L][i] = -1;
      word[L++] = 0;
      return L-1;
   }

   void insert(int val)
   {
      int u = root,v;
      for (int i = 31;i>=0;i--)
      {
          v = (val & (1<<i))?1:0;
          if (Next[u][v] == -1)
              Next[u][v] = newnode();
          u = Next[u][v];
          word[u]++;
      }
   }

   void Delete(int val)
   {
      int u = root,v;
      for (int i = 31;i>=0;i--)
      {
          v = (val & (1<<i))?1:0;
          u = Next[u][v];
          word[u]--;
      }
   }

   int query(int val)
   {
      int u = root,v;
      for (int i = 31;i>=0;i--)
      {
          v = (val & (1<<i))?1:0;
          if (v==1)
          {
             if (Next[u][0] != -1 && word[Next[u][0]])
                 u = Next[u][0];
             else
             {
                 u = Next[u][1];
                 val^=(1<<i);
             }
          }
          else
          {
             if (Next[u][1] != -1 && word[Next[u][1]])
             {
                u = Next[u][1];
                val ^= (1<<i);
             }
             else
                 u = Next[u][0];
          }
      }
      return val;
   }
};
int a[1005];
int main()
{
   int T;
   scanf("%d",&T);

   while (T--)
   {   
      Tree tree;
      tree.init();
      int n;
      scanf("%d",&n);
      for (int i = 0;i<n;i++)
      {
          scanf("%d",&a[i]);
          tree.insert(a[i]);
      }

      int ans = 0;
      for ( int i = 0;i<n;i++)
      {
         tree.Delete(a[i]);
         for (int j = i+1;j<n;j++)
         {
            tree.Delete(a[j]);
            ans = max(tree.query(a[i]+a[j]),ans);
            tree.insert(a[j]);
         }
         tree.insert(a[i]);
      }
      printf("%d\n",ans);
      memset(a,0,sizeof(a));
   }
}

题目

Chip Factory

Time Limit:9000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

HDU 5536

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