hdu 3949 XOR （高斯消元求线性基）
XOR
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2469 Accepted Submission(s): 838
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.
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.
Author
elfness
Source
2011 Multi-University Training Contest 11 - Host by UESTC
Recommend
xubiao | We have carefully selected several similar problems for you: 3946 3948 3947 3945 3944
题目大意：给出ｎ个数，可以选择任意个异或，求排名第Ｋ的数
题解：高斯消元求线性基
线性基
线性基异或得到的子集与原数异或得到的子集是相同的。
而且线性基从高到低排列的，所以在求解第K大之类的问题时可以直接对k进行二进制分解，将1对应位置的基底异或得到答案。
#include
#include
#include
#include
#include
#define LL long long
#define N 11003
using namespace std;
int n,m,T,zero,tot;
LL b[N],a[N];
void gauss()
{
tot=zero=0;
for (LL i=b[60];i;i>>=1){
int j=tot+1;
while (!(i&a[j])&&j<=n) j++;
if (j==n+1) continue;
tot++;
swap(a[tot],a[j]);
for (int k=1;k<=n;k++)
if (k!=tot&&(a[k]&i)) a[k]^=a[tot];
}
if (tot!=n) zero=1;
}
LL solve(LL x)
{
x-=(LL)zero; LL ans=0;
if (!x) return 0;
if (x>=b[tot]) return -1;
for (int i=1;i<=tot;i++)
if (x&(b[tot-i])) ans^=a[i];
return ans;
}
int main()
{
freopen("a.in","r",stdin);
freopen("my.out","w",stdout);
scanf("%d",&T);
b[0]=1;
for (int i=1;i<=60;i++) b[i]=b[i-1]*2;
for (int t=1;t<=T;t++){
scanf("%d",&n);
printf("Case #%d:\n",t);
memset(a,0,sizeof(a));
for (int i=1;i<=n;i++) scanf("%I64d",&a[i]);
gauss();
//for (int i=1;i<=tot;i++) cout<