The XOR Largest Pair bis

Description
today with a small W less than 1s time to complete such a question:
at a given N integers A_1, A_2, ..., A_N for one of two exclusive-or operation, the result is the largest number?
As he complacency, L showed the teacher another problem:
you numbers a1 to 1000 a1000, which is selected from the three digital AI, AJ, AK
(. 1 <= I, J, K <= 1000, and i, j, k different from each other)
meet (ai + aj) xor ak maximum value of
the small W suddenly confused .........

the Input
first line gives digital N, N subsequent lines, each a digital line
Output
If that
the Sample the Input
. 4
. 1
2
. 3
. 4
the Sample Output
. 7
the HINT
(. 1 + 2) =. 7 XOR4

sol: Trie first join all numbers, then the two numbers i and j, and get it in the Trie and the third number k XOR before enumeration. Note that the three numbers to vary, so we went to the front Trie was XOR, from Trie was deleted i and j, after the end of the XOR, then i and j are added back. When deleted, the number of times the +1 each node i and j -1 appears, when inserted, each node i and j occurring.

code show as below:

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 int a[2005],ch[35*1005][2],val[35*1005];
 5 int tol;
 6   
 7 void insert(ll x,ll p)
 8 {
 9     int u=0;
10     for(int i=31;i>=0;i--)
11     {
12         int v=(x>>i)&1;
13         if(!ch[u][v])
14             ch[u][v]=tol++;
15         u=ch[u][v];
16         val[u]+=p;
17     }
18 }
19   
20 ll query(ll x)
21 {
22     int u=0;
23     ll ans=0;
24     for(int i=31;i>=0;i--)
25     {
26         int v=(x>>i)&1;
27         if(val[ch[u][v^1]])
28             u=ch[u][v^1],ans=ans<<1|1; 
29         else
30             u=ch[u][v],ans=ans<<1;
31     }
32     return ans;
33 }
34   
35 int main()
36 {
37     int T;
38     scanf("%d",&T);//T组数据 
39     while(T-- )
 40      {
 41 is          Tol = . 1 ;
 42 is          int n-;
 43 is          Scanf ( " % D " , & n-); // number of each n- 
44 is          for ( int I = 0 ; I <= n-* 32 ; I ++ ) // build the entire TRIE 
45              CH [I] [ 0 ] = CH [I] [ . 1 ] Val = [I] = 0 ;
 46 is          for ( int I = . 1 ; I <= n-; I ++) // each was added to the digital TRIE 
47          {
 48             Scanf ( " % D " , A & [I]);
 49              INSERT (A [I], . 1 ); // inserted, the number of occurrence of each node increases. 1 
50          }
 51 is          LL Maxx = 0 ;
 52 is          for ( int = I . 1 ; I <= n-; I ++) // enumeration first number 
53 is          {
 54 is              for ( int J = I + . 1 ; J <= n-; J ++) // enumeration second number of 
55              {
 56 is                  INSERT ( A [I], - . 1 ); // delete A [i], a [j ]
57 is                  INSERT (A [J], - . 1 ); // delete operation using the inserting operation equals the number of the node appears -1 
58                  Maxx = max (Maxx, Query (A [I] + A [J]));
 59                  // with a [i] + a [j ] in the Trie to XOR number, maximum value 
 60                  // because it is not as a [i] and a [j] themselves, to delete these two numbers, after XOR coupled back 
61 is                  INSERT (A [J], . 1 );
 62 is                  INSERT (A [I], . 1 );
 63 is              }
 64          }
 65          the printf ( " % LLD \ n- " , Maxx);
 66      }
 67      return  0 ;
 68 }

The XOR Largest Pair bis

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