hdu3949:XOR

Problem Description

XOR is a kind of bit operator, we define that as follow: for two binary base number A and B, let C=A XOR B, then for each bit of C, we can get its value by check the digit of corresponding position in A and B. And for each digit, 1 XOR 1 = 0, 1 XOR 0 = 1, 0 XOR 1 = 1, 0 XOR 0 = 0. And we simply write this operator as ^, like 3 ^ 1 = 2,4 ^ 3 = 7. XOR is an amazing operator and this is a question about XOR. We can choose several numbers and do XOR operatorion to them one by one, then we get another number. For example, if we choose 2,3 and 4, we can get 2^3^4=5. Now, you are given N numbers, and you can choose some of them(even a single number) to do XOR on them, and you can get many different numbers. Now I want you tell me which number is the K-th smallest number among them.

Input

First line of the input is a single integer T(T<=30), indicates there are T test cases.
For each test case, the first line is an integer N(1<=N<=10000), the number of numbers below. The second line contains N integers (each number is between 1 and 10^18). The third line is a number Q(1<=Q<=10000), the number of queries. The fourth line contains Q numbers(each number is between 1 and 10^18) K1,K2,......KQ.

Output

For each test case,first output Case #C: in a single line,C means the number of the test case which is from 1 to T. Then for each query, you should output a single line contains the Ki-th smallest number in them, if there are less than Ki different numbers, output -1.

Sample Input

2 2 1 2 4 1 2 3 4 3 1 2 3 5 1 2 3 4 5

Sample Output

Case #1: 1 2 3 -1 Case #2: 0 1 2 3 -1

Hint

If you choose a single number, the result you get is the number you choose. Using long long instead of int because of the result may exceed 2^31-1.

好玄学啊这玩意儿

首先判一下边界：如果能得到0，就特殊处理答案，因为题目不支持自己异或自己这种操作

高斯消元，边消元边排序，如果得到了0，就说明可以通过异或得到这个数，那么这个数就是无意义的

那么剩下来的肯定对于二进制中的某一位，只有他为1其他都为0，要不然会被消掉

所以他们的特殊位也是从大到小排序的，那么感性的yy一下：
当前的a1，异或了之后的a2一定更大，然后数学归纳一下

就可以按照排名来取a异或，得出答案

代码如下：

//MT_LI
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a[110000];
int n,m;
int main()
{
    int T,Case=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
        bool bk=0;int t=n;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
                if(a[j]>a[i])
                    swap(a[j],a[i]);
            if(a[i]==0){bk=1;t=i-1;break;}
            for(int k=63;k>=0;k--)
                if(a[i]>>k&1)
                {
                    for(int j=1;j<=n;j++)    
                        if(i!=j&&a[j]>>k&1)
                            a[j]^=a[i];
                    break;
                }
        }
        scanf("%d",&m);
        printf("Case #%d:\n",++Case);
        while(m--)
        {
            ll x,ans=0ll;
            scanf("%lld",&x);
            if(bk==true)x--;
            if(x>1ll<<t)printf("-1\n");
            else
            {
                for(int i=t-1;i>=0;i--)
                    if(x>>i&1)
                        ans^=a[t-i];
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}

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