poj2105 IP Address（简单题）

题目链接：http://poj.org/problem?id=2105

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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

400000000000000000000000000000000 00000011100000001111111111111111 11001011100001001110010110000000 01010000000100000000000000000001 

Sample Output

0.0.0.03.128.255.255203.132.229.12880.16.0.1

题意：很简单， 就是每个案例给出一个32位的2进制数字串， 要求按照每8位转换为8进制输出（中间有‘.’）即可！

代码如下：

#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;}