19暑假线性基A

A - XOR
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 2³4=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.
类比与线性代数，线性基就是求出基向量，通过线性基，可得到这个空间所有数字
异或第k小板题

#include<bits/stdc++.h>
#define ll long long
#define MAXN 10005
using namespace std;
int t,n,q;
ll k,tmp;
struct L_B
{
	ll b[65],p[65];
	int cnt,flag;
	L_B()
	{
		memset(p,0,sizeof(p));
		memset(b,0,sizeof(b));
		cnt = flag = 0;
	}
	inline bool insert(ll x)
	{
		for(int i = 62;i >= 0;--i)
			if(x & (1ll <<i))
			{
				if(b[i])
					x ^= b[i];
				else
				{
					b[i] = x;
					return true;
				}
			}
		flag = 1;
		return false;
	}
	ll get_max()
	{
		ll ret = 0;
		for(int i = 62;i >= 0;--i)
			if((ret^b[i]) > ret)
				ret ^= b[i];
		return ret;
	}
	ll get_min()
	{
		if(flag)
			return 0;
		for(int i = 0;i <= 62;++i)
			if(b[i])
				return b[i];
		return 0;
	}
	inline void rebuild()
	{
		for(int i = 62;i >= 1;--i)
			if(b[i])
				for(int j = i-1;j >= 0;--j)
					if(b[i] & (1ll << j))
						b[i] ^= b[j];
		for(int i = 0;i <= 62;++i)
			if(b[i])
				p[cnt++] = b[i];
	}
	ll kth(ll k)
	{
		if(flag)
			--k;
		if(k == 0)
			return 0;
		ll ret = 0;
		if(k >= (1ll << cnt))
			return -1;
		for(int i = 0;i <= cnt-1;++i)
			if(k & (1ll << i))
				ret ^= p[i];
		return ret;
	}
};
L_B merge(const L_B &n1,const L_B &n2)
{
	L_B ret = n1;
	for(int i = 0;i <= 62;++i)
		if(n2.b[i])
			ret.insert(n2.b[i]);
	ret.flag = n1.flag | n1.flag;
	return ret;
}
int main()
{
	scanf("%d",&t);
	for(int cas = 1;cas <= t;++cas)
	{
		L_B lis;
		scanf("%d",&n);
		for(int i = 1;i <= n;++i)
		{
			scanf("%lld",&tmp);
			lis.insert(tmp);
		}
		
		scanf("%d",&q);
		lis.rebuild();
		printf("Case #%d:\n",cas);		
		while(q--)
		{
			scanf("%lld",&k);	
			printf("%lld\n",lis.kth(k));
		}
	}
}