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

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

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三次提交AC！

//10049.cpp:Definestheentrypointfortheconsoleapplication.
//

//#include"stdafx.h"
#include<iostream.h>
#include<math.h>
//usingnamespacestd;

intOutIP[9][4];

voidChangeToIp(constcharTemp[],intj)
...{
for(intx=0;x<4;x++)
...{
inttemp=0;
for(inty=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;
}
}

intmain()
...{
intN=0;
cin>>N;

if((N>0&&N<10)!=1)
return0;

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

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

//cout<<TempIn<<endl;
return0;
}

附上“马牛不是人”的解法：

/*start17:43*/
/*end18:0825min*/
#include<stdio.h>
#include<string.h>

main()
{
intn,i,j,p,dec=0;
chars[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(" ");
elseprintf(".");
}


}
}

system("PAUSE");
}