HDU

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.

InputFirst 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.OutputFor 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

题意：给定N个数，所有集合的不同异或和中，求从小到大第K个，不存在则输出-1。

思路：我们知道线性基可以表示用不超过64个数，表示出所有集合的异或和，那么为0的部位不考虑，我们求第K个，就是等效表示成二进制。。。ok了。

先求线性基，得到p数组。然后把为0的忽略，并且前面的p对后面的效果求出来。 有个注意的问题就是0，因为线性基我们没有考虑0，所以0单独考虑，如果线性基的大小和原数组大小一样，则可以表示出来，那么K--；

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep2(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int maxn=100010;
ll p[66],x;
int main()
{
    int T,N,Q,Cas=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        rep(i,0,63) p[i]=0;
        rep(i,1,N) {
            scanf("%lld",&x);
            rep2(j,63,0){
                if(x&(1LL<<j)){
                    if(p[j]) x^=p[j];
                    else { p[j]=x;break;}
                }
            }
        }
        ll num=0,ans,K;
        rep(i,0,63) if(p[i]){
             p[num++]=p[i];
             rep(j,i+1,63) if((p[j]>>i)&1) p[j]^=p[i];
        }
        scanf("%d",&Q);
        printf("Case #%d:
",++Cas);
        while(Q--){
            scanf("%lld",&K); if(N!=num) K--; //here，notice!考虑0的存在性
            if(K>=(1LL<<num)) puts("-1");
            else {
                ans=0;
                rep(j,0,63) {
                    if(K&(1LL<<j)) ans^=p[j]; //不能加，还是用异或，可能有尾巴，相互抵消
                }
                printf("%I64d
",ans);
            }
        }
    }
    return 0;
}