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 nn chips today, the ii-th chip produced this day has a serial number sis_i.

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\max_{i,j,k} (s_i+s_j) \oplus s_k

which i,j,ki,j,k are three different integers between 11 and nn. And ⊕\oplus is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

Input

The first line of input contains an integer TT indicating the total number of test cases.

The first line of each test case is an integer nn, indicating the number of chips produced today. The next line has nn integers s1,s2,..,sns_1, s_2, .., s_n, separated with single space, indicating serial number of each chip.

1≤T≤10001 \le T \le 1000

3≤n≤10003 \le n \le 1000

0≤si≤1090 \le s_i \le 10^9

There are at most 1010 testcases with n>100n > 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