Claris and XOR

Problem Description

Claris loves bitwise operations very much, especially XOR, because it has many beautiful features. He gets four positive integers a,b,c,da,b,c,d that satisfies a\leq bab and c\leq dcd. He wants to choose two integers x,yx,y that satisfies a\leq x\leq baxb and c\leq y\leq dcyd, and maximize the value of x~XOR~yx XOR y. But he doesn't know how to do it, so please tell him the maximum value of x~XOR~yx XOR y.

Input

The first line contains an integer T\left(1\leq T\leq10,000\right)T(1T10,000)——The number of the test cases. For each test case, the only line contains four integers a,b,c,d\left(1\leq a,b,c,d\leq10^{18}\right)a,b,c,d(1a,b,c,d1018​​). Between each two adjacent integers there is a white space separated.

Output

For each test case, the only line contains a integer that is the maximum value of x~XOR~yx XOR y.

Sample Input
2
1 2 3 4
5 7 13 15
Sample Output
6
11
Hint
In the first test case, when and only when x=2,y=4x=2,y=4, the value of x~XOR~yx XOR y is the maximum. In the second test case, when and only when x=5,y=14x=5,y=14 or x=6,y=13x=6,y=13, the value of x~XOR~yx XOR y is the maximum.
 
 
 
这游戏真难,之前写过一道类似的,就直接敲了...o(n)...超时....
 
 
TLE代码:
复制代码
 1 #include <vector>
 2 #include <map>
 3 #include <set>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <cstdio>
 7 #include <cmath>
 8 #include <cstdlib>
 9 #include <string>
10 #include <cstring>
11 #include <queue>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 #define ll long long
15 
16 int const MAX = 100000005;
17 int n;
18 
19 struct Trie
20 {
21     int root, tot, next[MAX][2], end[MAX];
22     inline int node()
23     {
24         memset(next[tot], -1, sizeof(next[tot]));
25         end[tot] = 0;
26         return tot ++;
27     }
28 
29     inline void Init()
30     {
31         tot = 0;
32         root = node();
33     }
34 
35     inline void insert(ll x)
36     {
37         int p = root;
38         for(int i = 31; i >= 0; i--)
39         {
40             int ID = ((1 << i) & x) ? 1 : 0;
41             if(next[p][ID] == -1)
42                 next[p][ID] = node();
43             p = next[p][ID];
44         }
45         end[p] = x;
46     }
47 
48     inline int search(int x)
49     {
50         int p = root;
51         for(int i = 31; i >= 0; i--)
52         {
53             int ID = ((1 << i) & x) ? 1 : 0;
54             if(ID == 0)
55                 p = next[p][1] != -1 ? next[p][1] : next[p][0];
56             else
57                 p = next[p][0] != -1 ? next[p][0] : next[p][1];
58         }
59         return x ^ end[p];
60     }
61 
62 }trie;
63 
64 int a[2],b[2];
65 int main()
66 {
67     int t;
68     scanf("%d",&t);
69     while(t--)
70     {
71         int n=4;
72         int WTF = 0,ans=0, x;
73         trie.Init();
74         for(int i = 0; i < 2; i++)
75         {
76             scanf("%d", &a[i]);
77         }
78         for(int i = 0; i < 2; i++)
79         {
80             scanf("%d", &b[i]);
81         }
82         for(int i=a[0]; i<=a[1]; i++){
83             //WTF=0;
84             //trie.insert(1);
85             //WTF = WTFx(WTF, trie.search(1));
86             for(int j=b[0]; j<=b[1]; j++){
87                 trie.Init();
88                 trie.insert(i);
89                 WTF = max(WTF, trie.search(i));
90                 trie.insert(j);
91                 WTF = max(WTF, trie.search(j));
92                 ans=max(WTF,ans);
93             }
94         }    
95         printf("%d\n", WTF);
96     }
97 }
复制代码

 

 

posted @   Vmetrio  阅读(250)  评论(0编辑  收藏  举报
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