Claris and XOR

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 332    Accepted Submission(s): 135

Problem Description

Claris
loves bitwise operations very much, especially XOR, because it has many
beautiful features. He gets four positive integers a,b,c,d that satisfies a≤b and c≤d. He wants to choose two integers x,y that satisfies a≤x≤b and c≤y≤d, and maximize the value of x XOR y. But he doesn‘t know how to do it, so please tell him the maximum value of x XOR y.

Input

The first line contains an integer T(1≤T≤10,000)——The number of the test cases.
For each test case, the only line contains four integers a,b,c,d(1≤a,b,c,d≤1018). 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 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 $x=2,y=4$, the value of $x~XOR~y$ is the maximum.
In the second test case, when and only when $x=5,y=14$ or $x=6,y=13$, the value of $x~XOR~y$ is the maximum.

昨天的B题，我的理解力，我都惭愧了，自己写了好几个样例才完全搞懂。

从最高位到最低位贪心。

贪心时，如果x走到此地方

x                   y

b 1 1 0 0 0    d  1 0 0 1 1

a 1 0 0 0 0    c   1 0 0 0 1

此时x1 = x2, y1 =y2  而且x1 = y1,所以有唯一解，取 0. 如果y系列等于0 有唯一解 1

x                   y

b 1 0 1 0 1 1    d  1 0 0 0 0 0

a    1 0 0 0 0    c   1 0 0 0 0 0

x1 != x2, y1 == y2，贪心使其为1.  如果y = 1，所以x = 0，此时二进制有五位数，设其为C， 所以10000 <= c <= 11111, 所以让b = ((ll)1<<i) - 1. 同理。

x                   y

b 1 0 1 0 1 1    d  1 0 0 0 0 0

a    1 0 0 0 0    c   0 1 0 0 0 0

此时x1 != x2, y1 != y2, 0 1可任意取， 前面取 11111 后面取100000 ，跳出。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<queue>
 6 #include<algorithm>
 7 using namespace std;
 8 typedef long long ll;
 9 void solve(){
10         int t;
11         scanf("%d",&t);
12         while(t--){
13             ll a,b,c,d;
14             cin>>a>>b>>c>>d;
15             int x1,x2,y1,y2;
16             ll ans = 0;
17             for(int i = 63; i>=0; i--){
18                 x1 = (bool)(a&((ll)1<<i));
19                 x2 = (bool)(b&((ll)1<<i));
20                 y1 = (bool)(c&((ll)1<<i));
21                 y2 = (bool)(d&((ll)1<<i));
22                 if(x1 == x2&&y1 == y2){
23                     if(x1 != y1){
24                        ans += ((ll)1<<i);
25                     }
26                 }
27                 else if(x1 != x2&&y1 == y2){
28                     ans += ((ll)1<<i);
29                     if(y1 == 1){
30                         b = ((ll)1<<i) - 1;
31                     }
32                     else{
33                         a = ((ll)1<<i);
34                     }
35                 }
36                 else if(x1 == x2&&y1 != y2){
37                     ans += ((ll)1<<i);
38                     if(x1 == 1){
39                         d = ((ll)1<<i) - 1;
40                     }
41                     else{
42                         c = ((ll)1<<i);
43                     }
44                 }
45                 else if(x1 != x2&&y1 != y2){
46                     ans += ((ll)1<<(i+1))-1;
47                     break;
48                 }
49             }
50             cout<<ans<<endl;
51         }
52 }
53 int main()
54 {
55     solve();
56     return 0;
57 }

卷珠帘