IP Address

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:

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

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给出一个32位二进制，化成点分十进制的IP地址形式

8位一处理即可

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

#include <set>

#include <stack>

#include <map>

#include <functional>

#include <bitset>

#include <string>

using namespace std;

#define LL long long

#define INF 0x3f3f3f3f

int main()

{

    int T;

    char s[1000];

    int a[10];

    scanf("%d",&T);

    while(T--)

    {

        scanf("%s",s);

         a[8]=1;

        for(int i=7;i>0;i--)

        a[i]=2*a[i+1];

        int a1=0,a2=0,a3=0,a4=0;

        int cnt;

        cnt=1;

        for(int i=0;i<8;i++)

        {

            a1+=(s[i]-'0')*a[cnt++];

        }

        cnt=1;

        for(int i=8;i<16;i++)

        {

            a2+=(s[i]-'0')*a[cnt++];

        }

        cnt=1;

        for(int i=16;i<24;i++)

        {

            a3+=(s[i]-'0')*a[cnt++];

        }

        cnt=1;

        for(int i=24;i<32;i++)

        {

            a4+=(s[i]-'0')*a[cnt++];

        }

        printf("%d.%d.%d.%d\n",a1,a2,a3,a4);

    }

    return 0;

}