2015ACM/ICPC亚洲区长春站 J hdu 5536 Chip Factory

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 368    Accepted Submission(s): 202

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）

 

题意：给出一组n的数，问max((a[i]+a[j])^a[k]) i != j != k

分析：

有几种方法

1、现将a[i]+a[j]存起来，枚举k，在数据结构中查询。。。注意要去除ak的影响，我是用了主席树，其实可以用trie，这是最蠢的一种做法。。。就是我的做法。。。

2、将ak存起来，枚举ai,aj，注意到时间9s，同上的方法。比第一种快

3、暴力。。。。实在太猥琐了。。。居然没卡掉。。。。n^3暴力居然过了。。。

4、cheat做法。。。枚举前80%

 

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <iostream>
 #include <map>
 #include <set>
 #include <algorithm>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define MLL (1000000000000000001LL)
 #define INF (1000000001)
 #define For(i, s, t) for(int i = (s); i <= (t); i ++)
 #define Ford(i, s, t) for(int i = (s); i >= (t); i --)
 #define Rep(i, n) for(int i = (0); i < (n); i ++)
 #define Repn(i, n) for(int i = (n)-1; i >= (0); i --)
 #define mk make_pair
 #define ft first
 #define sd second
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define sz(x) ((int) (x).size())
 #define clr(x, y) (memset(x, y, sizeof(x)))
 inline void SetIO(string Name)
 {
     string Input = Name + ".in";
     string Output = Name + ".out";
     freopen(Input.c_str(), "r", stdin);
     freopen(Output.c_str(), "w", stdout);
 }

 inline int Getint()
 {
     char ch = ' ';
     int Ret = ;
     bool Flag = ;
     while(!(ch >= '' && ch <= ''))
     {
         if(ch == '-') Flag ^= ;
         ch = getchar();
     }
     while(ch >= '' && ch <= '')
     {
         Ret = Ret *  + ch - '';
         ch = getchar();
     }
     return Ret;
 }

 const int N = , M = , Dep = ;
 struct SegmentType
 {
     int Child[], Sum;
     #define Lc(x) (Seg[x].Child[0])
     #define Rc(x) (Seg[x].Child[1])
     #define Child(x, y) (Seg[x].Child[y])
     #define Sum(x) (Seg[x].Sum)
 } Seg[(M+N)*Dep];
 int Tot;
 int n, Arr[N];
 int Cnt[M], Length;
 int Ans;

 inline void Solve();

 inline void Input()
 {
     int TestNumber = Getint();
     while(TestNumber--)
     {
         n = Getint();
         For(i, , n) Arr[i] = Getint();
         Solve();
     }
 }

 inline void Init(int x) {
     clr(Seg[x].Child, ), Sum(x) = ;
 }

 inline void Insert(int Val, int x, int Depth)
 {
     Sum(x)++;
     if(Depth < ) return;
     int Type = (Val & ( << Depth)) > ;
     if(!Child(x, Type)) {
         Child(x, Type) = ++Tot;
         Init(Child(x, Type));
     }
     Insert(Val, Child(x, Type), Depth-);
 }

 inline int Work(int Val, int x, int Depth)
 {
     if(!x) return ;
     if(Depth < ) return Sum(x);
     int Type = (Val & ( << Depth)) > ;
     if(Type) return Sum(Lc(x)) + Work(Val, Rc(x), Depth - );
     return Work(Val, Lc(x), Depth - );
 }

 inline void Query(int Val, int x, int Depth, int &Ret)
 {
     if(Depth < ) return;
     int Type = (Val & ( << Depth)) > ;
     if(Child(x, Type ^ ))
     {
         int p1 = Ret;
         p1 |= ( << Depth) * (Type ^ );
         p1 = p1 - Val - ;
         int A = Work(p1, , );
         int p2 = Ret;
         p2 |= ( << Depth) * (Type ^ );
         p2 |= ( << Depth) - ;
         p2 = p2 - Val;
         int B = Work(p2, , );
         int p = B - A;
         if(Val > p1 && Val <= p2) p--;
         if(Sum(Child(x, Type ^ )) - p > )
         {
             Ret |= ( << Depth) * (Type ^ );
             Query(Val, Child(x, Type ^ ), Depth - , Ret);
             return;
         }
     }
     Ret |= ( << Depth) * Type;
     Query(Val, Child(x, Type), Depth - , Ret);
 }

 inline void Solve()
 {
     Tot = , Length = ;
     Init(), Init(), Init();
     For(i, , n)
     {
         For(j, i + , n)
         {
             Cnt[++Length] = Arr[i] + Arr[j];
             Insert(Cnt[Length], , );
         }
         Insert(Arr[i], , );
     }

     Ans = ;
     for(int i = ; i <= n; i++)
     {
         int x = ;
         Query(Arr[i], , , x);
         Ans = max(Ans, x ^ Arr[i]);
     }
     printf("%d\n", Ans);
 }

 int main()
 {
     Input();
     //Solve();
     return ;
 }