Claris and XOR(模拟）

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。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 void solve(){
         int t;
         scanf("%d",&t);
         while(t--){
             ll a,b,c,d;
             cin>>a>>b>>c>>d;
             int x1,x2,y1,y2;
             ll ans = ;
             for(int i = ; i>=; i--){
                 x1 = (bool)(a&((ll)<<i));
                 x2 = (bool)(b&((ll)<<i));
                 y1 = (bool)(c&((ll)<<i));
                 y2 = (bool)(d&((ll)<<i));
                 if(x1 == x2&&y1 == y2){
                     if(x1 != y1){
                        ans += ((ll)<<i);
                     }
                 }
                 else if(x1 != x2&&y1 == y2){
                     ans += ((ll)<<i);
                     if(y1 == ){
                         b = ((ll)<<i) - ;
                     }
                     else{
                         a = ((ll)<<i);
                     }
                 }
                 else if(x1 == x2&&y1 != y2){
                     ans += ((ll)<<i);
                     if(x1 == ){
                         d = ((ll)<<i) - ;
                     }
                     else{
                         c = ((ll)<<i);
                     }
                 }
                 else if(x1 != x2&&y1 != y2){
                     ans += ((ll)<<(i+))-;
                     break;
                 }
             }
             cout<<ans<<endl;
         }
 }
 int main()
 {
     solve();
     return ;
 }

卷珠帘