hdu 5661 Claris and XOR

Claris and XOR

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 744    Accepted Submission(s): 330

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.

 

数据范围是10^18,转化为二进制是2^62左右,检查每一位的取值
1.如果只能取1那就取
2.如果可以取1或者取零那么就让x,y中的一方取0，另一方取1

#include <cstdio>
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long a,b,c,d;
        scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
        long long ans,ansx=0,ansy=0;
        for(int i=62;i>=0;i--)
        {
            ans=1LL<<i;
            if(ansx+ans<=b&&ansy+ans<=d)
            {
                if((ansx+ans<a)||(ansy+ans<c))   //必选，不然无法出现在范围内
                {
                  if(ansx+ans<a) ansx+=ans;
                  if(ansy+ans<c) ansy+=ans;
                }
                else
                {
                   ansx+=ans;   //可选其中之一
                }
            }
            else
            {
                if(ansx+ans<=b) ansx+=ans;
                else if(ansy+ans<=d) ansy+=ans;
            }
        }
        printf("%lld
",ansx^ansy);
    }
    return 0;
}