链接：http://acm.hdu.edu.cn/showproblem.php?pid=5536
题面;
Chip Factory
Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 6277 Accepted Submission(s): 2847
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:
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?
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
,s
2
,..,s
n
, separated with single space, indicating serial number of each chip.
1≤T≤1000
3≤n≤1000
0≤si
≤10
9
There are at most 10
testcases with n>100
3
1 2 3
3
100 200 300
400
模板题
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e3+;
int tot;
int ch[*M][],vis[*M];
ll val[*M],a[M]; void init(){
memset(vis,,sizeof(vis));
tot = ;
ch[][] = ch[][] = ;
} void ins(ll x){
int u = ;
for(int i = ;i >= ;i --){
int v = (x>>i)&;
if(!ch[u][v]){
ch[tot][] = ch[tot][] = ;
val[tot] = ;
vis[tot] = ;
ch[u][v] = tot++;
}
u = ch[u][v];
vis[u]++;
}
val[u] = x;
} void update(ll x,int c){
int u = ;
for(int i = ;i >= ;i --){
int v = (x>>i)&;
u = ch[u][v];
vis[u] += c;
}
} ll query(ll x){
int u = ;
for(int i = ;i >= ;i --){
int v = (x>>i)&;
if(ch[u][v^]&&vis[ch[u][v^]]) u = ch[u][v^];
else u = ch[u][v];
}
return x^val[u];
} int main()
{
ios::sync_with_stdio();
cin.tie(); cout.tie();
int t,n,m;
cin>>t;
while(t--){
cin>>n;
ll mx = ;
init();
for(int i = ;i <= n;i ++)
cin>>a[i],ins(a[i]);
for(int i = ;i <= n;i ++){
for(int j = i+;j <= n;j ++){
update(a[i],-); update(a[j],-);
mx = max(mx,query(a[i]+a[j]));
update(a[i],); update(a[j],);
}
}
cout<<mx<<endl;
}
}