POJ 2105 IP Address（我的水题之路——二进制字符转换成十进制）

IP Address

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 15540		Accepted: 8911

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

对于N组数组，每个数据有32位。每8位决定一个数字，将这个32位转换成4个十进制数字，用'.'间隔。

用一个函数，参数为一个字符串，每次取其前八位，对于每一个位的数字进行判断，如果为1，则加上该位表示的数字。

代码（1AC）：

#include <cstdio>
#include <cstdlib>
#include <cstring>

char str[50];
int bin[10] = {128, 64, 32, 16, 8, 4, 2, 1};

int IPAddress(char str[]){
    int sum = 0;
    int i;

    for (i = 0; i < 8; i++){
        if (str[i] == '1'){
            sum += bin[i];
        }
    }
    return sum;
}

int main(void){
    int casenum, ii;

    scanf("%d", &casenum);
    getchar();
    for (ii = 0; ii < casenum ; ii++){
        gets(str);
        printf("%d.%d.%d.%d\n", IPAddress(str), IPAddress(str + 8), IPAddress(str + 16), IPAddress(str + 24));
    }
    return 0;
}