ECNU-IP Address 1002

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:
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
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 code :

#include<stdio.h> #include<string.h> #define max 32 void set_one( unsigned char *a , int n ) { (*a) |= 1<<(n-1) ; } void set_zero( unsigned char *a , int n ) { (*a)&= ~(1<<(n-1)); } void get_ip( char sbi[max] ) { unsigned char result[4]= {'/0','/0','/0','/0'} ; int i , j ; for ( i = 0 ; i < 32 ; ++ i ) { if ( sbi[i] & 1 ) { set_one( &(result[i/8]) , 8 - i % 8 ) ; } } for ( i = 0 ; i < 3 ; ++ i ) { printf("%d.",(int)result[i]) ; } printf("%d/n", (int) result[i]) ; } int main() { int m ; char s[32] ; scanf("%d",&m) ; while( m -- ) { scanf("%s", s ) ; get_ip( s ) ; } return 0 ; } Problems :

 1、没有用unsigned char 来进行操作，出现错误。

 2、在设置标志位的时候出现了问题。

       set_one( &(result[i/8]) , 8 - i % 8 ) ; //error code below: set_one( &(result[i/8]) , i % 8 ) ;