UVALive3093 POJ2105 ZOJ2482 IP Address
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 20070 | Accepted: 11589 |
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
Source
Regionals 2004 >> Latin
America - Mexico and Central America
问题链接:UVALive3093 POJ2105 ZOJ2482 IP Address
问题描述:参见上文。
问题分析:一个简单的数字串切割和进制转换问题。
程序说明:(略)
参考链接:(略)
AC的C++语言程序:
/* UVALive3093 POJ2105 IP Address */ #include <iostream> #include <cstdio> using namespace std; int main() { int n; char c; scanf("%d", &n); c = getchar(); while(n--) { for(int i=1; i<=4; i++) { if(i != 1) putchar('.'); int v = 0; for(int j=1; j<=8; j++) { c = getchar(); v = v * 2 + c - '0'; } printf("%d", v); } printf("\n"); while((c = getchar()) != '\n' && c != EOF); } return 0; }