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  • IP Address 分类: POJ 2015-06-12 19:34 12人阅读 评论(0) 收藏

    IP Address
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 19125   Accepted: 11053

    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
    
    二进制转化
    #include <cstdio>
    #include <string.h>
    #include <cmath>
    #include <iostream>
    #include <algorithm>
    #define WW freopen("output.txt","w",stdout)
    using namespace std;
    int main()
    {
        char s[100];
        int T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%s",s);
            int len=strlen(s);
            int a[5];
            int sum=0,ans=1;
            for(int j=1,i=len-1;i>=0;i--,j++)
            {
                if(j%8)
                {
                    sum=(s[i]-'0')*ans+sum;
                    ans*=2;
                }
                else
                {
                     sum=(s[i]-'0')*ans+sum;
                     ans=1;
                     a[j/8]=sum;
                     sum=0;
                }
            }
            printf("%d.%d.%d.%d
    ",a[4],a[3],a[2],a[1]);
        }
        return 0;
    }
    


    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/juechen/p/4722035.html
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