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POJ 2105 IP Address C语言版

POJ题解 Slyar 2073浏览 0评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

水题,不过蛮实用的。

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.

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

Slyar:恩,就是把二进制的IP转换成十进制,注意数据可能不等长,有的后面有空格,有的没有。

#include 
#include 

int value[8] = {128, 64, 32, 16, 8, 4, 2, 1};

int main()
{
    int i, j, k, n;
    int ip;
    char bits[50];
    scanf("%d", &n);
    getchar();
    while (n--)
    {
        gets(bits);
        i = 0;
        
        /* 4段IP */
        for (k = 1; k <= 4; k++)
        {
            j = ip = 0;
            
            /* 每段8位 */
            while (j < 8)
            {
                if (bits[i] == '1') ip += value[j];
                i++;
                j++;
            }
            
            printf("%d", ip);
            if (k != 4) printf(".");
        }
        printf("\n");
    }
    //system("pause");
    return 0;
}

转载请注明:Slyar Home » POJ 2105 IP Address C语言版

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