Chip Factory
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 6090 Accepted Submission(s): 2745
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）
Recommend
hujie
题意：n个数，求任取两个数的加和，与剩下的数异或的最大值
分析：
01字典树求异或的最大值，需要删除操作,因为01字典树，我们可以对每一个节点进行标记，num[u]表示插入字符的01串经过该节点的次数，删除的时候，遍历数字二进制所对应的节点01，使其num【u】-1就可以了。（其实num跟统计前缀差不多作用）
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=1005;
const int maxnode=1000000+100;
int num[maxnode];
const int sigma_size=3;
struct Trie
{
int ch[maxnode][sigma_size];
int sz;
void init()
{
sz=1;
memset(ch[0],0,sizeof(ch[0]));
}
void insert(ll x)
{
int u=0;
for(int i=32;i>=0;i--)
{
int id=(x>>i)&1;//取出x二进制第i位
if(ch[u][id]==0)
{
ch[u][id]=sz;
memset(ch[sz],0,sizeof(ch[sz]));
num[sz]=0;
sz++;
}
u=ch[u][id];
num[u]++;
}
}
void update(ll x, int d)
{
int u = 0;
for(int i =32; i >= 0; i--){
int index = (x>>i)&1;
u = ch[u][index];
num[u]+=d;
}
}
ll find(ll x)
{
ll ans=0;
int u=0;
for(int i=32;i>=0;i--)
{
int id=(x>>i)&1;
int y=id^1; //找与id不同的
if(ch[u][y]&&num[ch[u][y]])
{
u=ch[u][y];
ans=ans<<1|1; //不同为1,ans=ans*2+1
}
else
{
u=ch[u][id];
ans<<=1; //相同为0,ans=ans*2
}
}
return ans;
}
};
Trie trie;
ll a[N];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
ll maxx=0;
int n;
trie.init();
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
trie.insert(a[i]);
}
for(int i=1;i<n;i++)
for(int j=i+1;j<=n;j++)
{
trie.update(a[i],-1);
trie.update(a[j],-1);
maxx=max(maxx,trie.find(a[i]+a[j]));
trie.update(a[i],1);
trie.update(a[j],1);
}
cout<<maxx<<endl;
}
return 0;
}
