HDU 4081-Parsing URL(水)

Parsing URL

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1575    Accepted Submission(s): 796

Problem Description

In computing, a Uniform Resource Locator or Universal Resource Locator (URL) is a character string that specifies where a known resource is available on the Internet and the mechanism for retrieving it.
The syntax of a typical URL is:
scheme://domain:port/path?query_string#fragment_id
In this problem, the scheme, domain is required by all URL and other components are optional. That is, for example, the following are all correct urls:
http://dict.bing.com.cn/#%E5%B0%8F%E6%95%B0%E7%82%B9
http://www.mariowiki.com/Mushroom
https://mail.google.com/mail/?shva=1#inbox
http://en.wikipedia.org/wiki/Bowser_(character)
ftp://fs.fudan.edu.cn/
telnet://bbs.fudan.edu.cn/
http://mail.bashu.cn:8080/BsOnline/
Your task is to find the domain for all given URLs.

 

Input

There are multiple test cases in this problem. The first line of input contains a single integer denoting the number of test cases.
For each of test case, there is only one line contains a valid URL.

 

Output

For each test case, you should output the domain of the given URL.

 

Sample Input

3
http://dict.bing.com.cn/#%E5%B0%8F%E6%95%B0%E7%82%B9
http://www.mariowiki.com/Mushroom
https://mail.google.com/mail/?shva=1#inbox

 

Sample Output

Case #1: dict.bing.com.cn
Case #2: www.mariowiki.com
Case #3: mail.google.com
训练一下高速找到水题并切掉的能力。。毕竟现场赛仅仅能做水题了。
。7分钟1A
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <cmath>
#include <map>
using namespace std;
#define LL long long
char s[10010],ans[10010];
int main()
{
	//ios::sync_with_stdio(false);
	int t,p,cas=1;
	scanf("%d",&t);
	getchar();
	while(t--)
	{
		gets(s);
		int len=strlen(s);
		for(int i=0;i<len;i++)
		{
			if(s[i]=='/'&&s[i-1]=='/')
			{
				int j=i+1;p=0;
				while(s[j]!='/'&&s[j]!=':')
					ans[p++]=s[j++];
				break;
			}
		}
		printf("Case #%d: ",cas++);
		for(int i=0;i<p;i++)
			printf("%c",ans[i]);
		puts("");
	}
	return 0;
}

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