HDU 3949 XOR (linear base, template)

Description

XOR is a kind of bit operator, we define that as follow:for the binary base number A and B, let C=a XOR b Bit of C, we can get it value by check for 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 is given N numbers, and you can choose some of them (even a single number) to doing XOR on them, and you can get the man Y different numbers. Now I want your tell me which number is the k-th smallest number among them.

Input

First line of the input was a single integer T (t<=30), indicates there is t test cases.

For each test case, the first line was 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

The For each test Case,first output case #C: A line,c means the number of the "the" the "Test Case" which is from 1 to T. Then for each query, you should output a contains the ki-th smallest number in them, if there is 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

Test Instructions

Given an array of length n n N, there is a q Q-q query, and each time a query is made of a subset of the array or the result of the K K K decimal value is what.

Ideas

Inserts all the elements of the array into a linear base and then rebuild them: starting from the high to the low, enumeration J<i J < I j for each I I I, and if d[i] d [i] d[i] is 1 1 1, make it different Or once d[j] d [j] D[j] (eliminates the current bit).

After the operation, for D[i] d [i] d[i] only I I i this bit is 1 1 1, the remaining bits are 0 0 0, we put the non 0 elements in the order from small to large in P p p.

At the time of querying the K-K k large difference or value, we construct according to the binary system K K K, for 1 1 1 bit, will be different or on the corresponding p[i] p [i] p[i], think about such a construction method is correct.

It is important to note that the result of the difference or out of the linear basis is non-zero, and the value of K K K in the query may be zero, so we need to construct a linear basis to determine whether the current number can be different or 0 0 0, and then the final query to the results of a special sentence.

AC Code

 #include <bits/stdc++.h> #define IO \ Ios::sync_with_stdio (FALSE) ;                  \ cin.tie (0);
\ cout.tie (0);
using namespace Std;
typedef long Long LL;
const int MAXN = 1e5 + 10;

int n, q;
    Class Linearbase {public:const static int maxn = 64-1;
    LL D[MAXN], P[MAXN], tot;
        void Clear () {tot = 0;
        memset (d, 0, sizeof (d));
    memset (p, 0, sizeof (p));
                } bool Insert (LL val) {for (int i = maxn-1; I >= 0; i--) {if (Val >> i) & 1) { if</