HDU3949 XOR (线性基)

HDU3949 XOR

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 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.

Solution

简述一下题意:n个权值,m个询问,每次一个k,表示询问n个元素可以异或出来的所有值中的第k小值

线性基求k小值...

我们需要改进一下线性基,将每一位都独立,如果j<i，且p[i]的第j位是1，就把p[i]^p[j].这样，对于二进制的每一位i.只有p[i]这一位是1，其他的都是0。

我们查询的时候，将k进行二进制拆分，如果第i位是1，就异或上线性基中第i个元素(注意不是第i位,俄日是从小到大第i个)，最终得出的答案就是k小值,怎么证?我也不会....

il void rebuild() {//重构函数
    for(rg lol i=63;i>=0;i--)
        for(rg lol j=i-1;j>=0;j--)
            if(arr[i]>>j&1) arr[i]^=arr[j];
    for(int i=0;i<=63;i++) if(arr[i]) arr[cnt++]=arr[i];
}

然后,要注意一个点,因为线性基的性质:线性基没有异或和为0的子集

所以默认的最小值是没有0的,但是如果有一个值插入失败了,说明这个值已经可以通过线性基里面的值异或出来,那么最小值肯定是0,所以们要把查询的k-1

Code

#include<bits/stdc++.h>
#define lol long long
#define il inline
#define rg register
using namespace std;

void in(lol &ans) {
    ans=0; lol f=1; char i=getchar();
    while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
    while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+i-'0',i=getchar();
    ans*=f;
}

lol n,q,T,cnt,flag;
lol arr[65];

il void init(lol x) {
    for(lol i=63;i>=0;i--) {
        if((x>>i&1)==0) continue;
        if(!arr[i]) {arr[i]=x;break;}
        else x^=arr[i];
    }if(!x) flag=1;//插入失败
}

il void query(lol k,lol ans=0) {
    if(k>>cnt) {puts("-1");return;}
    for(rg lol i=0;i<cnt;i++) {
        if(k>>i&1) ans^=arr[i];
    }printf("%lld\n",ans);
}

il void rebuild() {
    for(rg lol i=63;i>=0;i--)
        for(rg lol j=i-1;j>=0;j--)
            if(arr[i]>>j&1) arr[i]^=arr[j];
    for(int i=0;i<=63;i++) if(arr[i]) arr[cnt++]=arr[i];
}

int main()
{
    in(T);
    for(int t=1;t<=T;t++) {
        flag=0; printf("Case #%d:\n",t);
        in(n);
        for(lol i=1,x;i<=n;i++) in(x),init(x);
        rebuild(); in(q);
        for(lol i=1,x;i<=q;i++) {
            in(x); if(flag) x--; query(x);//查询的k的范围-1
        }
        memset(arr,0,sizeof(arr)); cnt=0;
    }return 0;
}