POJ 2105 IP Address

举报
谙忆 发表于 2021/05/28 06:34:01 2021/05/28
【摘要】 Problem 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 d...

Problem 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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main(){ int n; scanf("%d",&n);getchar(); while(n--){ char str[35]; int i; gets(str); // for(i=0;i<32;i++) //   scanf("%c",&str[i]); int sum[4]; int j; //printf("!!!\n"); for(int j=3;j>=0;j--){ sum[j]=0; for(i=1;i<=8;i++){ if(str[(j+1)*8-i]=='1'){ sum[j]+=pow(2,i-1); /**printf("sum[%d]=%d\n",j,sum[j]);putchar(str[(j+1)*8-i]); putchar(10);**/ } } } printf("%d.%d.%d.%d\n",sum[0],sum[1],sum[2],sum[3]); for(i=0;i<4;i++) sum[i]=0; } return 0;
}

  
 
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#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int a[8]={1,2,4,8,16,32,64,128};
int main(){ int n; scanf("%d",&n); while(n--){ char str[35]; scanf("%s",str); int i,sum[5],j=0; for(i=31;i>=0;i--){ if(i%8==7){ sum[++j]=0; } if(str[i]=='1'){ sum[j]+=a[7-i%8]; } } printf("%d.%d.%d.%d\n",sum[4],sum[3],sum[2],sum[1]); }
}


  
 
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文章来源: chenhx.blog.csdn.net,作者:谙忆,版权归原作者所有,如需转载,请联系作者。

原文链接:chenhx.blog.csdn.net/article/details/49428385

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