TOJ4413: IP address

传送门:http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=4413

时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte

描述

An IP address can be described as dotted decimal string format or 32-bit binary number format.

Now, given an IP address, if the IP address is in dotted decimal, output the 32-bit binary format ,On the contrary, output dotted decimal format.for example,the dotted decimal IP address 192.168.0.1，192->11000000，168->10101000，0->00000000，1->00000001，and the result is11000000101010000000000000000001 。

输入

Multiple inputs, each input has a string，the length of the string is less than 33;

输出

Output requirements of the subject, if you enter an IP address in dotted decimal, then the output 32-bit binary IP addresses, vice versa, output dotted decimal IP address.

样例输入

192.168.0.1

样例输出

11000000101010000000000000000001

题目意思是，如果给的是32位的01串，那么让你把它转化为十进制IP地址的形式，如果是十进制IP地址的形式，则转化为01串形式输出。

本题用java的BigInteger以及String可以快速实现十进制和二进制相互转换。于是就偷懒用java啦。注意java写十进制的时候读取的是一个字符串，分割到字符串数组里面可以调用split()函数，具体的看代码！

代码：

import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
public class Main {
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        while(input.hasNext()){
            String str = input.nextLine();
            String []s= str.split("\.");
            if(s[0].length() > 8){
                StringBuilder sb=new StringBuilder();
                sb.append(str);
                sb.insert(8,".");
                sb.insert(17,".");
                sb.insert(26,".");
                str = sb.toString();
                String []s1 = str.split("\.");
                for(int i = 0 ; i < s1.length; i++){
                    String num = new BigInteger(s1[i],2).toString(10);
                    System.out.print(num);
                    if(i != s1.length-1) System.out.print(".");
                }
                System.out.println();
            }
            else{
                for(int i = 0 ; i < s.length; i++){
                    String num = new BigInteger(s[i]).toString(2);
                    for(int i1 = 0 ; i1 < 8 - num.length() ; i1++){
                        System.out.print("0");
                    }
                    System.out.print(num);
                }
                System.out.println();
            }    
        }    
    }
}