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

最新推荐文章于 2019-11-07 15:50:07 发布

原创最新推荐文章于 2019-11-07 15:50:07 发布 · 1.6k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#numbers #output #math.h #input #system #string

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");
}