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 nn 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:
display maxi,j,k(si+sj)⊕sk
whichi,j,k are three different integers between 1 and n. And oplus⊕ is symbol of bitwise XOR.
Can you help John calculate the checksum number of today?
Input Format
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 nn 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 Format
For each test case, please output an integer indicating the checksum number in a line.
样例输入
2 3 1 2 3 3 100 200 300
样例输出
6 400
题目来源
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 #include <cstdio> 5 #include <vector> 6 #include <queue> 7 #include <stack> 8 #include <cstdlib> 9 #include <iomanip> 10 #include <cmath> 11 #include <cassert> 12 #include <ctime> 13 #include <map> 14 #include <set> 15 using namespace std; 16 #define ull unsigned long long 17 #define ll long long 18 #define N 30009 19 int num[N],tree[N][2]; 20 int t,n,a[1100]; 21 int pos; 22 //01字典树 23 void insert(int x,int rt) 24 { 25 for(int i=30;i>=0;i--) 26 { 27 int y=0; 28 if(x>>i&1) y=1; 29 if(!tree[rt][y]) tree[rt][y]=pos++; 30 rt=tree[rt][y]; 31 num[rt]++; 32 } 33 } 34 void dele(int x,int rt) 35 { 36 for(int i=30;i>=0;i--) 37 { 38 int y=0; 39 if(x>>i&1) y=1; 40 rt=tree[rt][y]; 41 num[rt]--;//tree[rt][y] 还在，并不是真正的删除 42 } 43 } 44 int solve(int x,int rt) 45 { 46 int ret=0; 47 /* 48 00001010 49 10100100是倒着来的 50 那么只要字典树里首位有1，就不可能找到10100100 51 */ 52 for(int i=30;i>=0;i--)//一定要倒过来，因为贪心，高位大，最终的结果才大 53 { 54 int y=0; 55 if(x>>i&1) y=1; 56 if(tree[rt][y^1]&&num[tree[rt][y^1]])//num[tree[rt][y^1]]才有意义 { 57 rt=tree[rt][y^1]; 58 ret+=(1<<i);//该位的异或为1 59 } 60 else{ 61 rt=tree[rt][y]; 62 } 63 } 64 return ret; 65 } 66 int main() 67 { 68 scanf("%d",&t); 69 while(t--) 70 { 71 for(int i=0;i<N;i++) 72 { 73 num[i]=0; 74 for(int j=0;j<=1;j++) 75 { 76 tree[i][j]=0; 77 } 78 } 79 pos=1; 80 scanf("%d",&n); 81 for(int i=0;i<n;i++){ 82 scanf("%d",&a[i]); 83 insert(a[i],0); 84 } 85 int ans=0; 86 //每次要删除a[i],a[j] 因为i!=j!=k 87 //当然每次查询后，还要再次插入字典树 88 for(int i=0;i<n;i++) 89 { dele(a[i],0); 90 for(int j=i+1;j<n;j++) 91 { 92 dele(a[j],0); 93 ans=max(ans,solve(a[i]+a[j],0) ); 94 insert(a[j],0); 95 } 96 insert(a[i],0); 97 } 98 printf("%d ",ans); 99 } 100 return 0; 101 }
1 //9s 的暴力解法 2 #define N 1009 3 int t,n,a[N]; 4 int ans; 5 int solve(int x,int y,int z) 6 { 7 return (x+y)^z; 8 } 9 int main() 10 { 11 scanf("%d",&t); 12 while(t--) 13 { 14 scanf("%d",&n); 15 for(int i=0;i<n;i++) 16 { 17 scanf("%d",&a[i]); 18 } 19 int ans=0; 20 for(int i=0;i<n;i++) 21 { 22 for(int j=i+1;j<n;j++) 23 { 24 for(int k=j+1;k<n;k++) 25 { 26 ans=max(ans,solve(a[i],a[j],a[k]) ); 27 ans=max(ans,solve(a[i],a[k],a[j]) ); 28 ans=max(ans,solve(a[k],a[j],a[i]) ); 29 } 30 } 31 } 32 printf("%d ",ans); 33 } 34 return 0; 35 }