2
5 2
192.168.1.0
192.168.1.101
192.168.2.5
192.168.2.7
202.14.27.235
255.255.255.0
255.255.0.0
4 2
127.127.0.1
10.134.52.0
127.0.10.1
10.134.0.2
235.235.0.0
1.57.16.0

Case #1:
3
2
Case #2:
3
4

解题思路： 可以直接模拟，关键用java注意超时、

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;


public class Main2
{
	public static void main(String[] args)
	{
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
      for (int p = 0; p < n; p++)
      {
        int i1 = sc.nextInt();
        int i2 = sc.nextInt();
        sc.nextLine();
        String a1[] = new String[i1];
        String a2[] = new String[i2];
        for (int i = 0; i < a1.length; i++)
        	a1[i]=sc.nextLine();
        for (int i = 0; i < a2.length; i++)
        	a2[i]=sc.nextLine();
        System.out.println("Case #"+(p+1)+":");
        	for (int i = 0; i < i2; i++)
        	{
        		String s[] = a2[i].split("[.]");
        		int A = Integer.parseInt(s[0]);
    			int B = Integer.parseInt(s[1]);
    			int C = Integer.parseInt(s[2]);
    			int D = Integer.parseInt(s[3]);
        		Set<String>list=new HashSet<String>();
        		for (int j = 0; j < i1; j++)
        		{
        			s = a1[j].split("[.]");
        			int a = Integer.parseInt(s[0]);
        			int b = Integer.parseInt(s[1]);
        			int c = Integer.parseInt(s[2]);
        			int d = Integer.parseInt(s[3]);
        			StringBuffer sb = new StringBuffer();
        			sb.append(a&A).append(".").append(b&B).append(".").append(c&C).append(".").append(d&D);
        			list.add(sb.toString());
        		}
        		System.out.println(list.size());
        	}
      }
	}
}