IP聚合
Accepts: 1898
Submissions: 4972
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem Description
Input
第一行包含一个整数
接下来
第一行两个正整数
再接下来
Output
Sample Input
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
Sample Output
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());
}
}
}
}