HDU3949 XOR

题意

XOR

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5489    Accepted Submission(s): 1928

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.

 

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

 

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分析

先利用高斯消元，找出异或基，然后由于基底内二进制最高位位互不相干，所以二进制才分询问即可。注意题目中不允许自己异或自己以达到0，所以矩阵中有0行的时候找第k-1大(从0开始编号)，没有的时候找第k大，相当于跳过了这个0。

时间复杂度\(O(64 (N+Q))\)

以前学过一个线性基，现在看来完全没用。

代码

#include<iostream>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;
    rg char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-') w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
        data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x){
    return x=read<T>();
}
typedef unsigned long long ull;
ull a[10010];
int n,m,t,T;
int main()
{
//  freopen(".in","r",stdin),freopen(".out","w",stdout);
    read(T);
    for(int C=1;C<=T;++C){
        read(n);
        for(int i=1;i<=n;++i) read(a[i]);
        bool zero=0;
        t=n;
        for(int i=1;i<=n;++i){
            for(int j=i+1;j<=n;++j)
                if(a[j]>a[i]) std::swap(a[j],a[i]);
            if(a[i]==0) {zero=1,t=i-1;break;}
            for(int k=63;k>=0;--k) if(a[i]>>k&1){
                for(int j=1;j<=n;++j)
                    if(i!=j&&a[j]>>k&1) a[j]^=a[i];
                break;
            }
        }
        read(m);
        printf("Case #%d:\n",C);
        while(m--){
            ull k=read<ull>()-zero,ans=0;
            if(k>=1LLU<<t) {puts("-1");continue;}
            for(int i=t-1;i>=0;--i)
                if(k>>i&1) ans^=a[t-i];
            printf("%llu\n",ans);
        }
    }
    return 0;
}