• poj 2105 IP Address


    IP Address
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 18951   Accepted: 10939

    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
    

    Source

     
    分析:
    进制转换
     1 #include <cstdio>
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<string>
     5 #include<cstring>
     6 #include<vector>
     7 using namespace std;
     8 int b[8]={128,64,32,16,8,4,2,1};
     9 int main(){
    10     int n;
    11     string s;
    12     while(cin>>n){
    13         while(n--){
    14             cin>>s;
    15             int i,j,k=0;
    16             for(i=0;i<=31;){
    17                 int sum=0;
    18                 j=0;
    19                 while(j<8){
    20                     if(s[i]=='1')
    21                     sum+=b[i%8];
    22                     i++;
    23                     j++;
    24                 }
    25                 cout<<sum;
    26                 if(k<3){
    27                     cout<<'.';
    28                 }
    29                 else{
    30                     cout<<endl;
    31                 }
    32                 k++;
    33             }
    34         }
    35     }
    36     return 0;
    37 } 
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  • 原文地址:https://www.cnblogs.com/Deribs4/p/4294036.html
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