Chip Factory
http://acm.hdu.edu.cn/showproblem.php?pid=5536
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 280 Accepted Submission(s): 158
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:
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?
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#define maxn 1005
using namespace std;
int tol;
long long val[maxn*];
int tree[maxn*][];
int num[maxn*][];
long long a[maxn]; void init(){
tol=;
tree[][]=tree[][]=;
memset(num,,sizeof(num));
} void add(long long x,int k){
int u=;
for(int i=;i>=;i--){
int v=(x>>i)&;
num[u][v]+=k;
if(!tree[u][v]){
tree[tol][]=tree[tol][]=;
val[tol]=-;
tree[u][v]=tol++;
}
u=tree[u][v];
}
val[u]=x;
} long long query(long long n){
int u=;
for(int i=;i>=;i--){
int v=(n>>i)&;
if(num[u][v^]) u=tree[u][v^];
else u=tree[u][v];
}
return val[u];
} int main(){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
init();
for(int i=;i<=n;i++){
scanf("%lld",&a[i]);
add(a[i],);
}
long long tmp;
long long ans=;
for(int i=;i<=n;i++){
add(a[i],-);
for(int j=i+;j<=n;j++){
add(a[j],-);
tmp=a[i]+a[j];
ans=max(ans,tmp^query(tmp));
add(a[j],);
}
add(a[i],);
}
printf("%lld\n",ans);
} }