zoj2482 IP Address (water problem, but there is a grammar can't get through)

IP Address

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld &%llu

Submit Status

Description

Suppose you is reading byte streams from any device, representing IP addresses. Your task is to convert a by characters long sequence of ' 1s ' and ' 0s ' (bits) to a dotted decimal format. A dotted decimal format for an IP address was form by grouping 8 bits at a time and converting the binary representation to Decimal representation. Any 8 bits are a valid part of an IP address. To convert binary numbers to decimal numbers remember that both is positional numerical systems, where the first 8 Positi ONS of the binary systems are:

27	26	25	24	23	22	21st	20
128	64	32	16	8	4	2	1

Input

The input'll has a number N (1 <= N <= 9) in its first line representing the number of streams to convert. N lines would follow.

Output

The output must has 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 re Presentation.

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 Mexico and Central America 2004 Analysis: Water problem. But there is a grammatical problem that can't be figured out. Cin>>c can be used in conjunction with the c-= ' 0 '. But I don't know why ....
AC Code:

#include <iostream>
#include <cstdio>
using namespace Std;

int main ()
{
int pos[8]={128,64,32,16,8,4,2,1};//order in turn
char c;
int n,i,j;
int sum;
scanf ("%d", &n);
while (n--)
{
for (i=0;i<4;i++)
{
sum=0;
for (j=0;j<8;j++)
{
scanf ("%c", &c),//cin>>c to match c-= ' 0 ' usage. But I don't know why ....
cin>>c;
c-= ' 0 ';//remember this usage when C is converted to int type. or c-=48;
c-=48;
if (c)//This time C is converted to int type
SUM+=POS[J];
}
if (i)
printf (".");
printf ("%d", sum);
}
printf ("\ n");
}
return 0;
}