POJ2105:IP Address
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
我擦，用int数组的话居然一直没弄出答案，改用char很快就正确了，什么原因？
[cpp]
#include <stdio.h>
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int n;
char a[35];
cin >> n;
getchar();
while(n--)
{
int i;
gets(a);
for(i = 0; i<32; i++)
{
if((i+1)%8 == 0)
{
int j,sum = 0;
for(j = i-7; i-j>=0; j++)
{
if(a[j]!='0')
sum = sum+pow(2.0,(i-j));
}
cout << sum;
if(i!=31)
cout << ".";
}
}
cout << endl;
}
return 0;
}
补充：软件开发 , C++ ,
