poj2105 IP Address（简单题）

题目链接：

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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

题意：非常easy，就是每一个案例给出一个32位的2进制数字串。要求依照每8位转换为8进制输出（中间有‘.’）就可以！

代码例如以下：

#include <iostream>

#include <cmath>

using namespace std;

int main()

{

	int N;

	int i, j;

	char s[117];

	while(cin >> N)

	{

		while(N--)

		{

			cin>>s;

			int ans[4], k = 7, l = 0;

			int sum = 0;

			for(i = 0; i < 32; i++)

			{

				sum +=(s[i]-'0')*pow(2.0,k);

				k--;

				if(i%8==7)

				{

					ans[l++] = sum;

					sum = 0;

					k = 7;

				}

			}

			cout<<ans[0]<<'.'<<ans[1]<<'.'<<ans[2]<<'.'<<ans[3]<<endl;

		}

	}

	return 0;

}

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poj2105 IP Address（简单题）

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