/*
In the early nineties, the World Wide Web (WWW) was invented. Nowadays, most people think that the WWW simply consists of all the pretty (or not so pretty) HTML-pages that you can read with your WWW browser. But back then, one of the main intentions behind the design of the WWW was to unify several existing communication protocols.

Then (and even now), information on the Internet was available via a multitude of channels: FTP, HTTP, E-Mail, News, Gopher, and many more. Thanks to the WWW, all these services can now be uniformly addressed via URLs (Uniform Resource Locators). The syntax of URLs is defined in the Internet standard RFC 1738. For our problem, we consider a simplified version of the syntax, which is as follows:

“://” [ ":" ] [ "/" ]

The square brackets [] mean that the enclosed string is optional and may or may not appear. Examples of URLs are the following:

http://www.informatik.uni-ulm.de/acm

ftp://acm.baylor.edu:1234/pub/staff/mr-p
gopher://veryold.edu

More specifically,

is always one of http, ftp or gopher.

is a string consisting of alphabetic (a-z, A-Z) or numeric (0-9) characters and points (.).

is a positive integer, smaller than 65536.

is a string that contains no spaces.

You are to write a program that parses an URL into its components.

Input

The input starts with a line containing a single integer n, the number of URLs in the input. The following n lines contain one URL each, in the format described above. The URLs will consist of at most 60 characters each.

Output

For each URL in the input first print the number of the URL, as shown in the sample output. Then print four lines, stating the protocol, host, port and path specified by the URL. If the port and/or path are not given in the URL, print the stringinstead. Adhere to the format shown in the sample output.

Print a blank line after each test case.

Sample Input

3
ftp://acm.baylor.edu:1234/pub/staff/mr-p

http://www.informatik.uni-ulm.de/acm

gopher://veryold.edu

Sample Output

URL #1
Protocol = ftp
Host = acm.baylor.edu
Port = 1234
Path = pub/staff/mr-p

URL #2
Protocol = http
Host = www.informatik.uni-ulm.de
Port = 
Path = acm

URL #3
Protocol = gopher
Host = veryold.edu
Port = 
Path = 
*/

#include <string>
#include <iostream> using namespace std; int main()
{
int n; string s; cin >> n;
for(int i = 1; i <= n; i++)
{
int p;
cin >> s; p = s.find("://"); string protocol = s.substr(0, p);
string host = s.substr(p + 3, s.size());
string port;
string path; p = host.find('/');
if(p != string::npos)
{
path = host.substr(p + 1, host.size());
host = host.substr(0, p);
} p = host.find(':');
if(p != string::npos)
{
int k;
for(k = p + 1; isdigit(host[k]); k++)
port += host[k];
host = host.substr(0, p);
} cout << "URL #" << i << endl
<< "Protocol = " << protocol << endl
<< "Host = " << host << endl
<< "Port = " << (port == "" ? "<default>" : port) << endl
<< "Path = " << (path == "" ? "<default>" : path) << endl << endl;
} return 0;
}

ZOJ 1243 URLs的更多相关文章

  1. POJ题目细究

    acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP:  1011   NTA                 简单题  1013   Great Equipment     简单题  102 ...

  2. 【转】POJ百道水题列表

    以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight ...

  3. ZOJ题目分类

    ZOJ题目分类初学者题: 1001 1037 1048 1049 1051 1067 1115 1151 1201 1205 1216 1240 1241 1242 1251 1292 1331 13 ...

  4. ZOJ People Counting

    第十三届浙江省大学生程序设计竞赛 I 题, 一道模拟题. ZOJ  3944http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=394 ...

  5. Django基础,Day2 - 编写urls,views,models

    编写views views:作为MVC中的C,接收用户的输入,调用数据库Model层和业务逻辑Model层,处理后将处理结果渲染到V层中去. polls/views.py: from django.h ...

  6. ZOJ 3686 A Simple Tree Problem

    A Simple Tree Problem Time Limit: 3 Seconds      Memory Limit: 65536 KB Given a rooted tree, each no ...

  7. ZOJ Problem Set - 1394 Polar Explorer

    这道题目还是简单的,但是自己WA了好几次,总结下: 1.对输入的总结,加上上次ZOJ Problem Set - 1334 Basically Speaking ac代码及总结这道题目的总结 题目要求 ...

  8. ZOJ Problem Set - 1392 The Hardest Problem Ever

    放了一个长长的暑假,可能是这辈子最后一个这么长的暑假了吧,呵呵...今天来实验室了,先找了zoj上面简单的题目练练手直接贴代码了,不解释,就是一道简单的密文转换问题: #include <std ...

  9. ZOJ Problem Set - 1049 I Think I Need a Houseboat

    这道题目说白了是一道平面几何的数学问题,重在理解题目的意思: 题目说,弗雷德想买地盖房养老,但是土地每年会被密西西比河淹掉一部分,而且经调查是以半圆形的方式淹没的,每年淹没50平方英里,以初始水岸线为 ...

随机推荐

  1. Android 4.3正式发布:四大新功能一览

    在旧金山举行的新品发布会上,Google正式发布了Android 4.3,代号仍为“Jelly Bean”.此次更新并没有太大改变,只是紧跟4.1.4.2步伐, 新增了低功耗蓝牙.多用户登录等一系列功 ...

  2. mysql.default_socket 和 mysqli.default_socket

    php.ini 里面mysql.default_socket 和 mysqli.default_socket 设置必须要和mysql里面的值一样,否则连接mysql会提示错误

  3. dedecms二级菜单实现

    修改channelartlist.lib.php if($typeid==0 || $typeid=='top') { $tpsql = " reid=0 AND ispart<> ...

  4. 【STL】-迭代器的用法

    初始化: list<char>::iterator pos; 算法: 1. 遍历 for(pos = col1.begin(); pos != col1.end(); ++pos){... ...

  5. WP8.1 Study4:WP8.1中控件集合应用

    1.AutoSuggestBox的应用 在xaml里代码可如下: <AutoSuggestBox Name="autobox" Header="suggestion ...

  6. new 动态分配数组空间

    (一)定义一个整数              int *p =new int;        int *p =new int(4); //赋初值4 (二)定义一个一维数组                ...

  7. typedef定义函数类型或函数指针

    转载请标明出处: 最近在看redis的代码,发现了有关函数指针的部分,想把它记下来. 在redis中有类似下面的定义,利用typedef 定义了一个新的类型,这种类型是一个函数: typedef vo ...

  8. 《JAVA学习笔记 (final关键字)》

    [14-9]面向对象-final关键字 /* 继承的弊端,打破封装性. 不让其他类继承该类,就不会有重写. 怎么能实现呢?通过Java中的一个关键子来实现,final(最终化). [final关键字] ...

  9. BZOJ 2200 道路与航线

    好厉害呀这道题,有种豁然开朗的感觉.... 按拓扑顺序跑最短路. 然后注意细节,像WA的代码犯下的错是一笔带过没有丝毫考虑的...然而就是错了. 考试的时候一定要拍啊. #include<ios ...

  10. 联合与枚举 、 高级指针 、 C语言标准库(一)

    1 输入一个整数,求春夏秋冬 1.1 问题 在实际应用中,有的变量只有几种可能取值.如人的性别只有两种可能取值,星期只有七种可能取值.在 C 语言中对这样取值比较特殊的变量可以定义为枚举类型.所谓枚举 ...