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以一种"廉价"的方式获取页面的标题(downmoon)

有朋友问到：“我需要得到一个webpage的title，因为这个title一般都比较靠前，只要取得html的前面少量内容就可以了。因为要取得很多个页面的title,如果下载整个htmlcode，肯定比较浪费时间。.net好像没有现成的类可以干这件事情(取得部分html)，我应该如何去实现？”一种比较"廉价"(即较小成本条件下)的解决思路：第一步：取出含有页面title部分的最小集合。这是“廉价”的关键！第二步：用正则表达式取出和</title</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-29 22:21:16</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2939</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/11343" title="虚拟机手机连接上的图形标识不见了解决" target="_blank" onclick="return recordId('',)">虚拟机手机连接上的图形标识不见了解决 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/11343" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/3.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/11343';}else{window.open('/article/11343');}" style="cursor: pointer;">奇怪的事情天天有，今天尤其多。使用虚拟机开发，当插上手机后，虚拟机会抢先获得手机的开发访问权限。在虚拟机里断开连接，删除掉手机之后，在宿主机xp下才能正常用如360手机助手来访问手机。正常情况下，手机插入后再虚拟机的右下脚会有一个方形的标示，如下图：奇怪的是今天竟然莫名奇妙的不见了，要在虚拟机和xp下切换手机的访问权很不方便。解决：点击虚拟机的最上面的“虚拟机”----------“移除设备”，然后找到手机设备后，在“状态栏显示”前面打勾就可以了。虚拟机手</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 09:29:41</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2939</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/11008" title="四大美女元宵齐斗艳秦岚王丽坤佟丽娅马苏舞值" target="_blank" onclick="return recordId('',)">四大美女元宵齐斗艳秦岚王丽坤佟丽娅马苏舞值 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/11008" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/12.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/11008';}else{window.open('/article/11008');}" style="cursor: pointer;">http://www.tu12138.com/comment/comment.do?method=queryComment&id=15030610280689400001&type=佟丽娅马苏四大美女元宵齐斗艳秦岚王丽坤佟丽娅马苏舞值颜值pk</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 09:29:28</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2939</span></span>  </p> </article><article class="excerpt"> <header> <a title="各行各业" class="label label-important zj-del-target" href="/category/38">各行各业<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/10255" title="（原创）OpenStack服务如何使用Keystone(三)---" target="_blank" onclick="return recordId('',)">（原创）OpenStack服务如何使用Keystone(三)--- </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/10255" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/11.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/10255';}else{window.open('/article/10255');}" style="cursor: pointer;">（一）Keystone端的操作（二）如何在OpenStack服务上部署Keystone中间件（三）详细配置keystonemiddleware前文我们介绍了如何部署Keystone中间件以及中间件的配置点.这里给出所有Keystone中间件的配置项.概括起来,Keystone中间件的配置项具有以下特点,1.以"auth_"开头的配置项需要指向Keystone的Admin服务,Keystone中间件使用这些配置信息从Keystone服务处获取令牌的有效信息.</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 09:27:20</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2939</span></span>  </p> </article><article class="excerpt"> <header> <a title="数据库相关" class="label label-important zj-del-target" href="/category/32">数据库相关<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/8293" title="uva 10051 Tower of Cubes(DAG最长路）" target="_blank" onclick="return recordId('',)">uva 10051 Tower of Cubes(DAG最长路） </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/8293" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/13.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/8293';}else{window.open('/article/8293');}" style="cursor: pointer;">题目连接：10051-TowerofCubes题目大意：有n个正方体，从序号1~n，对应的每个立方体的6个面分别有它的颜色（用数字给出），现在想要将立方体堆成塔，并且上面的立方体的序号要小于下面立方体的序号，相邻的面颜色必须相同。输出最高值和路径。解题思路：因为立方体可以旋转，所以一个序号的立方体对应这6种不同的摆放方式，可以将问题理解成DAG最长路问题，只是搜索范围是从i+1开始到n。然后记录路径要开两个2维数组。路径不唯一，随便输出一条。#includ</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 01:55:00</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2939</span></span>  </p> </article><article class="excerpt"> <header> <a title="数据库相关" class="label label-important zj-del-target" href="/category/32">数据库相关<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/6682" title="SZU : A11 Sequence" target="_blank" onclick="return recordId('',)">SZU : A11 Sequence </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/6682" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/15.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/6682';}else{window.open('/article/6682');}" style="cursor: pointer;">DescriptionWearegivenaintegersequence,yourjobisfindthelengthofthelongestcontiguoussubsequencethatisstrictlyincreasingorstrictlydecreasing.InputFirstnumber,representhowmanytestcases.Foreachtestcasethefirstnumberis.Thenpositiveinteg</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 01:53:22</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2939</span></span>  </p> </article><div class="pagination" style="display:;"> <ul><li><a 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