HUNAN UNIVERSITY ACM/ICPC Judge Online —— Problem 10049 IP Address

本文介绍了一种将32位二进制数转换为点分十进制IP地址的方法。通过实例演示了如何逐段读取二进制序列,并将其转换为相应的十进制数值。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

这次注意到了两个问题:首先是连续输入二进制数的时候使用整型的数组是不方便的,所以后来改用了char。第二就是char型数组需要预留一个元素用来存储'/0'的结束字符,我之前的几次提交失败的原因就是在这。

问题描述:

IP Address 
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB 
Total submit users: 405, Accepted users: 388 
Problem 10049 : No special judgement 
Problem description
Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form by grouping 8 bits at a time and converting the binary representation to decimal representation. Any 8 bits is a valid part of an IP address. To convert binary numbers to decimal numbers remember that both are positional numerical systems, where the first 8 positions of the binary systems are:

27 26 25 24 23 22 21 20 128 64 32 16 8 4 2 1

 
Input
The input will have a number N (1<=N<=9) in its first line representing the number of streams to convert. N lines will follow.

 
Output
The output must have N lines with a doted decimal IP address. A dotted decimal IP address is formed by grouping 8 bit at the time and converting the binary representation to decimal representation.

 
Sample Input
4
00000000000000000000000000000000
00000011100000001111111111111111
11001011100001001110010110000000
01010000000100000000000000000001
 
Sample Output
0.0.0.0
3.128.255.255
203.132.229.128
80.16.0.1
 
Problem Source
MCA 2004

 
Submit   Discuss   Judge Status  Problems  Ranklist  

三次提交AC!

// 10049.cpp : Defines the entry point for the console application.
//

//#include "stdafx.h"
#include <iostream.h>
#include 
<math.h>
//using namespace std;

int OutIP[9][4];

void ChangeToIp(const char Temp[],int j)
{
    
for(int x=0;x<4;x++)
    
{
        
int temp=0;
        
for(int y=0;y<8;y++)
        
{
            
if(Temp[8*x+y]=='1')
                temp
+=1*pow(2,7-y);
            
else //if(Temp[8*x+y]=='1')
                temp+=0*pow(2,7-y);
        }

        OutIP[j][x]
=temp;
    }

}

 
int main()
{
    
int N=0;
    cin
>>N;
    
    
if((N>0&&N<10)!=1)
        
return 0;

    
char TempIn[33];
    
for(int i=0;i<N;i++)
    
{
        cin
>>TempIn;
        ChangeToIp(TempIn,i);
    }


    
for(int a=0;a<N;a++)
    
{
        cout
<<OutIP[a][0]<<"."<<OutIP[a][1]<<"."<<OutIP[a][2]<<"."<<OutIP[a][3]<<endl;
    }


//    cout<<TempIn<<endl;
    return 0;
}

 

附上“马牛不是人”的解法:

/*start 17:43*/
/*end 18:08  25min*/
#include 
<stdio.h>
#include 
<string.h>

main()
{
      
int n,i,j,p,dec=0;
      
char s[32];
      scanf(
"%d",&n);     
      
      
for( i = 0 ; i < n ; i++ ){
           scanf(
"%s",s);
           
for( j = 0,p = 7 ; j < 32 ; j++,p--){

  
                    
if(s[j]=='1')dec+=pow(2,p);    
                    
if( p == 0 ){
                        p
=8;
                        printf(
"%d",dec);
                        dec
=0;
                        
if(j == 31)printf(" ");
                        
else printf(".");
                    }  
 
                
           }
      }
      
      system(
"PAUSE");
}
 
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值