描述
传送门：我是传送门
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−thi−th chip produced this day has a serial number sisi.
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: max(i,j,k)(si+sj)⊕skmax(i,j,k)(si+sj)⊕sk
which i,j,ki,j,k are three different integers between 1 and n. And ⊕⊕ is symbol of bitwise XORXOR.
Can you help John calculate the checksum number of today?
输入
The first line of input contains an integerTT indicating the total number of test cases.
The first line of each test case is an integer nn, indicating the number of chips produced today. The next line has n integers s1,s2,..,sns1,s2,..,sn separated with single space, indicating serial number of each chip.
1≤T≤10001≤T≤1000
3≤n≤10003≤n≤1000
0≤si≤1090≤si≤109
There are at most 10 testcases with n>100n>100
输出
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
输出
6
400
Note
思路
- 01字典树模板题
- 暴力求解
代码
1 /* 2 * ========================================================================= 3 * 4 * Filename: hdu5536.cpp 5 * 6 * Link: http://acm.hdu.edu.cn/showproblem.php?pid=5536 7 * 8 * Version: 1.0 9 * Created: 2018/08/30 20时46分09秒 10 * Revision: none 11 * Compiler: g++ 12 * 13 * Author: 杜宁元 (https://duny31030.top/), duny31030@126.com 14 * Organization: QLU_浪在ACM 15 * 16 * ========================================================================= 17 */ 18 #include <bits/stdc++.h> 19 using namespace std; 20 #define clr(a, x) memset(a, x, sizeof(a)) 21 #define rep(i,a,n) for(int i=a;i<=n;i++) 22 #define pre(i,a,n) for(int i=a;i>=n;i--) 23 #define ll long long 24 #define max3(a,b,c) fmax(a,fmax(b,c)) 25 #define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0); 26 const double eps = 1e-6; 27 const int INF = 0x3f3f3f3f; 28 const int mod = 1e9 + 7; 29 const int N = 10100; 30 int T,n,a[N]; 31 struct node 32 { 33 int next[2]; 34 int size; 35 }trie[32*N]; 36 37 int tot = 1,root = 1; 38 39 void Init() 40 { 41 rep(i,1,tot) 42 { 43 trie[i].next[0] = 0; 44 trie[i].next[1] = 0; 45 trie[i].size = 0; 46 } 47 tot = 1; 48 } 49 50 void Insert(int x) 51 { 52 int o = root; 53 trie[o].size++; 54 for(int k = 30;k >= 0;k--) 55 { 56 int tmp = 0; 57 if(x&(1<<k)) 58 tmp = 1; 59 if(!trie[o].next[tmp]) 60 trie[o].next[tmp] = ++tot; 61 62 o = trie[o].next[tmp]; 63 trie[o].size++; 64 } 65 } 66 67 void Delete(int x) 68 { 69 int o = root; 70 trie[o].size--; 71 for(int k = 30;k >= 0;k--) 72 { 73 int tmp = 0; 74 if(x&(1<<k)) 75 tmp = 1; 76 o = trie[o].next[tmp]; 77 trie[o].size--; 78 } 79 } 80 81 int Query(int x) 82 { 83 int o = root; 84 for(int k = 30;k >= 0;k--) 85 { 86 int tmp = 0; 87 if(x&(1<<k)) 88 tmp = 1; 89 if(tmp) 90 { 91 if(trie[o].next[0] && trie[trie[o].next[0]].size) 92 o = trie[o].next[0]; 93 else 94 o = trie[o].next[1], x^=(1<<k); 95 } 96 else 97 { 98 if(trie[o].next[1] && trie[trie[o].next[1]].size) 99 o = trie[o].next[1],x^=(1<<k); 100 else 101 o = trie[o].next[0]; 102 } 103 } 104 return x; 105 } 106 int main() 107 { 108 ios 109 #ifdef ONLINE_JUDGE 110 #else 111 freopen("in.txt","r",stdin); 112 // freopen("out.txt","w",stdout); 113 #endif 114 scanf("%d",&T); 115 while(T--) 116 { 117 int maxn = -INF; 118 scanf("%d",&n); 119 rep(i,1,n) 120 { 121 scanf("%d",&a[i]); 122 Insert(a[i]); 123 } 124 rep(i,1,n-1) 125 { 126 Delete(a[i]); 127 rep(j,i+1,n) 128 { 129 Delete(a[j]); 130 maxn = max(maxn,Query(a[i]+a[j])); 131 Insert(a[j]); 132 } 133 Insert(a[i]); 134 } 135 printf("%d ",maxn); 136 Init(); 137 } 138 fclose(stdin); 139 // fclose(stdout); 140 return 0; 141 }