network address是前(32-n)随意 后n位全零 network mask是前(32-n)全一 后n位全零 本题主要利用位移操作,1ULL表示无符号长整型的常数1,这样写可防止不必要的溢出,取反后可以作为mask的枚举然后拿mask和mins或者maxs并一下就得到address了. 代码: #include <cstdio> #include <iostream> using namespace std; const int maxn = 70; int main(…

书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 代码:(Accepted,0 ms) //UVa1590 - IP Networks #include<iostream> unsigned i, m, num, ip[4], ipmax[4], ipmin[4], mask[4]; int cmp(unsigned *a, unsigned *b) {//compare for (int i = 0;i < 4;++i) { if (a[i] <…

习题4-5 IP网络(IP Networks, ACM/ICPC NEERC 2005, UVa1590) 可以用一个网络地址和一个子网掩码描述一个子网(即连续的IP地址范围).其中子网 掩码包含32个二进制位,前32-n位为1,后n位为0,网络地址的前32-n位任意,后n位为0. 所有前32-n位和网络地址相同的IP都属于此网络. 例如,网络地址为194.85.160.176(二进制为11000010|01010101|10100000|10110000), 子网掩码为255.255.255.…

 Alex is administrator of IP networks. His clients have a bunch of individual IP addresses and he decided to group all those IP addresses into the smallest possible IP network.  Each IP address is a 4-byte number that is written byte-by-byte in a dec…

1.链接地址: http://poj.org/problem?id=2105 http://bailian.openjudge.cn/practice/2105 2.题目: IP Address Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 18146   Accepted: 10456 Description Suppose you are reading byte streams from any device, r…

How endless looping of packets in a TCP/IP network might occur? Router is a device used to interconnect two or more computer networks and routing is the process of properly, forwarding traffic between related computer networks. Endless looping of pac…

https://en.wikipedia.org/wiki/Network_address_translation In the face of the foreseeable global IP address space exhaustion, NAT was increasingly used since the late 1990s in conjunction with IP masquerading, which is a technique that hides an entire…

题意:输入m代表接下来的数据个数,计算接下来输入数据的网络掩码,和最小网络地址. 思路:①子网掩码:先将数据转为二进制,判断从哪一位开始有数据不一样,记下下标index,则子网掩码是index的前面是1,后面是0(2#),就是子网掩码,再转为10#输出: ②最小网络地址:用任意一个IP与子网掩码进行按位与运算得出最小IP,与运算的符号为&. import java.util.Scanner; public class UVA1590 { public static void main(Strin…

IP Address Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 18951   Accepted: 10939 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'…

这道题目是一道关于IP地址的题目,要深入理解这道题需要有一定的网络基础. 这道题目我第一次做的时候虽然也AC了,但代码写的比较复杂,不够精炼.近期刚刚参加了网络方面的培训,在有一定知识的基础上,又重写了这道题目.将很多步骤通过位运算(如移位,异或)进行了简化,在此贴一下前后两次的代码. 第二次代码: #include <cstdio> + ; ]; int find_firstdiff(int m){ ;i<;i++) ;j<m;j++) ][i])return i; ; } vo…