【hdu3949】XOR

2014年12月11日3,7310
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.

题解
话说我不会高斯消元T T
高斯消元求线性基,将k二进制拆分。。。
线性基好像2014集训队论文有
大概也可以这样理解
高斯消元会消出这样一个01矩阵,右边是最低位
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
(这样答案很显然)
或者可能是这样的
1 0 0 1 0 0 — 36
0 1 0 0 0 0 — 16
0 0 1 1 0 0 — 12
0 0 0 0 0 0 — 0
0 0 0 0 1 0 — 2
0 0 0 0 0 1 — 1
(当然这不是很准确。。只是为了美观)
第四列出现了若干1,在这些1前面必然有1。。。
如果我们只看前三行的前三列,那么可以得到大小确定的 7 个数
8 16 24 32 40 48 56
考虑下第四列,则这7个数可能有一些这一位异或完得到1,一些是0,差别是2^2
考虑第四位之后变为
8+4 16 22+4 32+4 40 48+4 56
不影响这7个数的相对大小关系
而第四行以后的,五六两行的数取否不影响前面确定的二进制位,也不影响大小关系
即每个数x都可以有x,x+1,x+2,x+3
反正无视第四位就好了
我是语文渣。。。
0要特判一下能否取到