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 = 0;
            for(int i = 63; i>=0; i--){
                x1 = (bool)(a&((ll)1<<i));
                x2 = (bool)(b&((ll)1<<i));
                y1 = (bool)(c&((ll)1<<i));
                y2 = (bool)(d&((ll)1<<i));
                if(x1 == x2&&y1 == y2){
                    if(x1 != y1){
                       ans += ((ll)1<<i);
                    }
                }
                else if(x1 != x2&&y1 == y2){
                    ans += ((ll)1<<i);
                    if(y1 == 1){
                        b = ((ll)1<<i) - 1;
                    }
                    else{
                        a = ((ll)1<<i);
                    }
                }
                else if(x1 == x2&&y1 != y2){
                    ans += ((ll)1<<i);
                    if(x1 == 1){
                        d = ((ll)1<<i) - 1;
                    }
                    else{
                        c = ((ll)1<<i);
                    }
                }
                else if(x1 != x2&&y1 != y2){
                    ans += ((ll)1<<(i+1))-1;
                    break;
                }
            }
            cout<<ans<<endl;
        }
}
int main()
{
    solve();
    return 0;
}