The 2010 ACM-ICPC Asia Chengdu Regional Contest - C Binary Number 水题，枚举~

The 2010 ACM-ICPC Asia Chengdu Regional Contest - C Binary Number 水题，枚举~

Binary Number Time Limit: 2 Seconds      Memory Limit: 65536 KB

For 2 non-negative integers x and y, f(x, y) is defined as the number of different bits in the binary format of x and y. For example, f(2, 3)=1, f(0, 3)=2, f(5, 10)=4.

Now given 2 sets of non-negative integers A and B, for each integer b in B, you should find an integer a in A such that f(a, b) is minimized. If there are more than one such integers in set A, choose the smallest one.

Input

The first line of the input is an integer T (0 < T ≤ 100), indicating the number of test cases. The first line of each test case contains 2 positive integers m and n (0 < m, n ≤ 100), indicating the numbers of integers of the 2 sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than 1000000. The first m lines are the integers in set A and the other n lines are the integers in set B.

Output

For each test case you should output n lines, each of which contains the result for each query in a single line.

Sample Input

2
2 5
1
2
1
2
3
4
5
5 2
1000000
9999
1423
3421
0
13245
353

 

Sample Output

1
2
1
1
1
9999
0

不解释了，N2

  
  
  
  

   
   
   
    1
   
   
   
   
   
   
   
   # include 
   
   
   
   <
   
   
   
   iostream
   
   
   
   >
   
   
   
   

   
   
   
    2
   
   
   
   # include 
   
   
   
   <
   
   
   
   algorithm
   
   
   
   >
   
   
   
   

   
   
   
    3
   
   
   
   
   
   
   
   using
   
   
   
    
   
   
   
   namespace
   
   
   
    std;

   
   
   
    4
   
   
   
   
   
   
   
   int
   
   
   
    a[
   
   
   
   101
   
   
   
   ],n,m;

   
   
   
    5
   
   
   
   
   
   
   
   int
   
   
   
    main()

   
   
   
    6
   
   
   
   
   
   
   
   
   
   
   
   {
 7    int t;
 8    cin>>t;
 9    while(t--)
10    {
11        cin>>n>>m;
12        for(int i=0;i<n;i++)
13           cin>>a[i];
14        sort(a,a+n);
15        for(int i=0;i<m;i++)
16        {
17           int t,res=0xfffffff,num;
18           cin>>t;
19           for(int j=0;j<n;j++)
20           {
21              int tmp1=a[j],tmp2=t,diff=0;
22              while(tmp1||tmp2)
23              {
24                 diff+=((tmp1&1)!=(tmp2&1));
25                 tmp1>>=1;
26                 tmp2>>=1;
27              }
28              if(diff<res) res=diff,num=a[j];
29           }
30           cout<<num<<endl;
31        }
32    }
33  //  system("pause");
34    return 0;
35}
   
   
   
   

   
   
   
   36