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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3388 Accepted Submission(s): 1158 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.
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.
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.
221 241 2 3 431 2 351 2 3 4 5
Case #1:123-1Case #2:0123-1
题意：
给你个n个数，这n个数总共可以构成2^n-1个集合（除掉空集）
每个集合的值为这个集合中所有数字的异或和，算出所有集合的值之后从小到大排序后并去重可以得到一个序列
接下来Q次询问，每次询问给你一个数K，求新序列中第K小的数
特殊的，如果K>序列元素个数，输出-1
线性基模板题
下面就有这道题题解
求出线性基后贪心即可，不过如何求出序列中有多少个数呢？（判断输出-1的情况）
如果线性基有k个元素，那么肯定可以构成2^k个不同值，也就是2^k个不同的数
不过空集不能算，所以应该是2^k-1个数
但是！如果k<n，说明原先n个数存在一个非空集满足异或为0，所以有
当k==n时，共有2^k-1个数
当k<n时，共有2^k个数
特判-1的情况后，就是模板了
#include #include #define LL long longLL a[10005], p[66], jz[66];int main(void){ LL k, siz, ans; int T, n, i, j, cnt, Q, cas; scanf("%d", &T); cas = 1; while(T--) { scanf("%d", &n); for(i=1;i<=n;i++) scanf("%lld", &a[i]); memset(p, 0, sizeof(p)); for(i=1;i<=n;i++) { for(j=62;j>=0;j--) { if(a[i]&(1ll< =1;i--) { for(j=i-1;j>=0;j--) { if(p[i]&(1ll< siz) printf("-1\n"); else { ans = 0; if(siz%2==0) k--; for(i=0;i
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