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   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"); }