ZOJ2482 IP Address 2017-04-18 23:11 44人阅读 评论(0) 收藏

IP Address

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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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;
}