poj 2105 IP Address

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#poj #2105 IP Address

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该博客介绍了如何解决POJ上的2105题，即从32位二进制数转换为IPv4地址。题目要求将32位的二进制数分成四段，每八位转换为十进制，形成标准的IP地址格式。文章提供了通过编程实现这一转换的AC代码。

原题链接：http://poj.org/problem?id=2105

IP Address

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 19317		Accepted: 11156

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

Source

México and Central America 2004

题意：根据32位二进制01代码，转化为4段十进制IP地址（每八位二进制对应一个十进制）

附上AC代码：

#include <stdio.h>
#include <math.h>
int main()
{
    int i,n,sum;
    char a[35];
    scanf("%d",&n);
    getchar();//消除回车，否则对下面的字符串有影响
    while(n--)
    {
        gets(a);
        sum=0;
        for (i=0;i<32;i++)
        {
            if (a[i]=='1')
            {
                sum+=pow(2,(7-i%8));//循环，标准二进制转十进制
            }

            if ((i+1)%8==0&&i!=0)
            {
                if (i==31)
                {
                   printf("%d\n",sum);
                }
                else
                {
                    printf("%d.",sum);
                }
                sum=0;
            }
        }
    }
    return 0;
}