poj 2105 IP Address

最新推荐文章于 2021-07-25 00:37:32 发布

转载最新推荐文章于 2021-07-25 00:37:32 发布 · 35 阅读

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原文链接：http://www.cnblogs.com/Deribs4/p/4294036.html

本文介绍了一种将32位二进制IP地址转换为十进制点分表示法的方法，并提供了一个C++实现示例。该算法通过逐个读取8位二进制数并将其转换为十进制数来完成这一过程。

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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:

⁷

⁶

⁵

⁴

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

México and Central America 2004

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 }