poj2105 IP Address（简单题）

题目链接：http://poj.org/problem?id=2105

----------------------------------------------------------------------------------------------------------------------------------------------------------

欢迎光临天资小屋







：http://user.qzone.qq.com/593830943/main

----------------------------------------------------------------------------------------------------------------------------------------------------------

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;
}