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） 不同的i,j,k使得 (si,+sj)⊕sk 最大 将所有的数插入字典树中，枚举i,j从字典树中删去，然后找与 si+sj 亦或最大的，贪心就好。
#include<bits/stdc++.h> using namespace std; #define LL long long const int maxnode=1000000; const int sigma_size=2; struct Tire{ int ch[maxnode][sigma_size]; int val[maxnode]; int sz=1; void init(){ sz=1; memset(ch,0,sizeof(ch)); memset(val,0,sizeof(val)); } void T_insert(int x,int p){ int u=0; for(int i=31;i>=0;i--){ int c=(x>>i)&1; if(!ch[u][c]){ memset(ch[sz],0,sizeof(sz)); val[sz]=0; ch[u][c]=sz++; } u=ch[u][c]; val[u]+=p; } } LL query(int x){ int u=0; LL ans=0; for(int i=31;i>=0;i--){ ans*=2; int c=(x>>i)&1; //如果左右两边都存在一定是选择与它相反的分支， if(ch[u][c^1]&&val[ch[u][c^1]]){ u=ch[u][c^1]; ans++; } else u=ch[u][c]; } return ans; } }; Tire a; int num[1005]; int main(){ int t; scanf("%d",&t); while(t--){ int n; scanf("%d",&n); a.init(); for(int i=1;i<=n;i++){ scanf("%d",&num[i]); a.T_insert(num[i],1); } LL ans=0; for(int i=1;i<=n;i++){ a.T_insert(num[i],-1); for(int j=i+1;j<=n;j++){ a.T_insert(num[j],-1); ans=max(ans,a.query(num[i]+num[j])); a.T_insert(num[j],1); } a.T_insert(num[i],1); } printf("%I64d\n",ans); } return 0; }