题目：Chip Factory
题解：这题其实用暴力是可以解决的。但我是为了练习01字典树。先将n个数组建成一个01字典树，我是用num数组来统计有多少个数经过某个结点。然后来一个双重循环，计算(a[i]+a[j]),但题目要求i,j,k不能相等，所以需要在上面建的树，将第i个数和第j个数删点，这时候就用num数组了，将经过第i个数和第j个数的结点都进行-1。然后在树中找最大值的时候num[x]为0的话，说明已经走不下去了。
#include <cstdio>
#include <stack>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 1e3+10;
typedef long long ll;ll a[N];
int cnt;
int num[N*35];
int tree[N*35][2];//2000000000 31位
void insert(ll y){int p = 1;for(int i = 32;i >= 0;i--){int x = (y>>i)&1;if(!tree[p][x]) {tree[p][x] = ++cnt;}num[tree[p][x]]++;p = tree[p][x];}
}
ll query(ll y){int p = 1;ll ans = 0;for(int i = 32;i >= 0;i--){int x = (y>>i)&1;if(tree[p][x^1] && num[tree[p][x^1]]) {ans = ans*2+1;p = tree[p][x^1];}else if(num[tree[p][x^0]]){ans = ans*2;p = tree[p][x^0];}}return ans;
}void update(ll y,int c){int p = 1;for(int i = 32;i >= 0;i--){int x = (y>>i)&1;num[tree[p][x]] += c;p = tree[p][x];}
}
int main(){int t;scanf("%d",&t);while(t--){int n;cnt = 1;memset(num,0,sizeof num);memset(tree,0,sizeof tree);scanf("%d",&n);for(int i = 1;i <= n;i++){scanf("%lld",&a[i]);insert(a[i]);}/*for(int i = 0;i < 40;i++){cout<<i<<" "<<num[i][0]<<" "<<num[i][1]<<endl;}*/ll ans = 0;for(int i = 1;i <= n;i++){update(a[i],-1);for(int j = i+1;j <= n;j++){update(a[j],-1);ans = max(ans,query(a[i]+a[j]));update(a[j],1);}update(a[i],1);}cout<<ans<<endl;}return 0;
}