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
題意：在一個數組中找出 (s[i] + s[j]) ^ s[k] 的最大值，其中 i、j、k 各不相同。
思路：看數據範圍可以想到暴力+字典樹，有一個要注意的就是，s[i]+s[j]不能與自己異或，所以我們要用一個更新操作，將s[i]和s[j]路徑上的值都減一
#include<bits/stdc++.h>
using namespace std;
#define maxn 100010
typedef long long ll;
struct Tire{
ll siz,ch[32*maxn][2],val[32*maxn],vis[32*maxn];
void tire_clear(){siz=1,memset(ch[0],0,sizeof(ch[0]));}
void tire_insert(ll x){
int u=0;
for(int i=32;i>=0;i--){
int c=(x>>i)&1;
if(!ch[u][c]){//節點不存在
memset(ch[siz],0,sizeof(ch[siz]));
val[siz]=0; vis[siz]=0;
ch[u][c]=siz++;//新建節點
}
u=ch[u][c];//往下走
vis[u]++;
}
val[u]=x;//字符串的最後一個字符的附加信息爲v
}
void tire_update(ll x,int d){
int u=0;
for(int i=32;i>=0;i--){
int c=(x>>i)&1;
u=ch[u][c];
vis[u]+=d;
}
}
ll tire_find(ll x){
int u=0;
for(int i=32;i>=0;i--){
int c=(x>>i)&1;
if(ch[u][c^1]&&vis[ch[u][c^1]]) u=ch[u][c^1];
else u=ch[u][c];
}
return val[u]^x;
}
}tire;
ll a[maxn];
int main(){
int t; scanf("%d",&t);
while(t--){
tire.tire_clear();
int n; scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
tire.tire_insert(a[i]);
}
ll mx=0;
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++){
tire.tire_update(a[i],-1);
tire.tire_update(a[j],-1);
ll x=a[i]+a[j];
mx=max(mx,tire.tire_find(x));
tire.tire_update(a[i],1);
tire.tire_update(a[j],1);
}
}
printf("%lld\n",mx);
}
return 0;
}