poj 2105 IP Address

原题链接：http://poj.org/problem?id=2105

IP Address

Time Limit: 1000MS Memory Limit: 30000KTotal 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

400000000000000000000000000000000 00000011100000001111111111111111 11001011100001001110010110000000 01010000000100000000000000000001 

Sample Output

0.0.0.03.128.255.255203.132.229.12880.16.0.1

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;}