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hdu5536Chip Factory

2015-11-12 13:48 471 查看
算机学院大学生程序设计竞赛（新生为主） 

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 640    Accepted Submission(s): 320


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）
 

#include<bits/stdc++.h>
const int maxn=51000;
const int ch_size=2;
using namespace std;
int n,a[maxn];
unsigned int ans;
struct Trie{
int a[maxn][ch_size],tot,sum[maxn];
void init(){sum[0]=tot=0;memset(a[0],0,sizeof(a[0]));}
void clear(int x){memset(a[x],0,sizeof(a[x]));sum[x]=0;}
void insert(unsigned int v){
int x=0;sum[0]++;
for(int i=31;i>=0;i--){
bool son=((v>>i)&1);
if(a[x][son]){
x=a[x][son];
}else{
clear(++tot);
x=a[x][son]=tot;

}
sum[x]++;
}
}
void update(unsigned int v,unsigned int f1,unsigned int f2){
int x=0;unsigned int tmp=0;
bool k1=1,k2=1;
// printf("%u %u %u\n",v,f1,f2);
for(int i=31;i>=0;i--){
bool son=((v>>i)&1);
// printf("%d %d %d %d %d %d %d %u\n",i,son,(((f1>>i)&1)^son),(((f2>>i)&1)^son),sum[a[x][son^1]],k1,k2,1u<<i);
if(a[x][son^1]&&sum[a[x][son^1]]-k1*(((f1>>i)&1)^son)-k2*(((f2>>i)&1)^son)>0){
x=a[x][son^1];
if((((f1>>i)&1)^son)==0)k1=0;

if((((f2>>i)&1)^son)==0)k2=0;
tmp|=(1u<<i);
}else{
x=a[x][son];
if(((f1>>i)&1)!=son)k1=0;
if(((f2>>i)&1)!=son)k2=0;
}
}
ans=max(ans,tmp);
//  printf("tmp = %d\n",tmp);
}
}T;
void work(){
T.init();
scanf("%d",&n);
ans=0;
for(int i=1;i<=n;i++){
scanf("%u",&a[i]);
T.insert(a[i]);
}
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++){
unsigned int t=a[i]+a[j];
T.update(t,a[i],a[j]);
}
printf("%u\n",ans);
}
int main(){
int t;scanf("%d",&t);
while(t--)work();
return 0;
}



                                            

                                        

                        
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