poj 2015 IP Address

IP Address

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 15331		Accepted: 8772

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:

2

⁷

   2

⁶

  2

⁵

  2

⁴

  2

³

   2

²

  2

¹

  2

⁰

 

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

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    char input[9][33];
    int output[9][4];
    int n;
    int sum;
    int i,j,k;
    int count;
    cin>>n;
    for(i=0;i<n;i++)
        cin>>input[i];
    for(i=0;i<n;i++)
    {
        sum=0;
        k=0;
        count=128;
        for(j=0;j<8;j++)
        {
            sum+=count*(input[i][j]-'0');
            output[i][k]=sum;
            count/=2;    
        }
        k++;
        sum=0;
        count=128;
        for(j=8;j<16;j++)
        {
            sum+=count*(input[i][j]-'0');
            output[i][k]=sum;
            count/=2;    
        }
        k++;
        sum=0;
        count=128;
        for(j=16;j<24;j++)
        {
            sum+=count*(input[i][j]-'0');
            output[i][k]=sum;
            count/=2;    
        }
        k++;
        sum=0;
        count=128;
        for(j=24;j<32;j++)
        {
            sum+=count*(input[i][j]-'0');
            output[i][k]=sum;
            count/=2;
        }
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<4;j++)
        {
            if(j<3)
                cout<<output[i][j]<<".";
            else
                cout<<output[i][j];
        }
        cout<<endl;
    }
    return 0;
}