﻿<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:trackback="http://madskills.com/public/xml/rss/module/trackback/" xmlns:wfw="http://wellformedweb.org/CommentAPI/" xmlns:slash="http://purl.org/rss/1.0/modules/slash/"><channel><title>BlogJava-一只牛的异想世界-随笔分类-求职</title><link>http://www.blogjava.net/andyelvis/category/39030.html</link><description>积水成流，聚沙成堆!</description><language>zh-cn</language><lastBuildDate>Thu, 07 May 2009 12:31:44 GMT</lastBuildDate><pubDate>Thu, 07 May 2009 12:31:44 GMT</pubDate><ttl>60</ttl><item><title>一道网易有道2009实习生招聘笔试程序设计题</title><link>http://www.blogjava.net/andyelvis/archive/2009/05/06/269304.html</link><dc:creator>蒋耘</dc:creator><author>蒋耘</author><pubDate>Wed, 06 May 2009 14:29:00 GMT</pubDate><guid>http://www.blogjava.net/andyelvis/archive/2009/05/06/269304.html</guid><wfw:comment>http://www.blogjava.net/andyelvis/comments/269304.html</wfw:comment><comments>http://www.blogjava.net/andyelvis/archive/2009/05/06/269304.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.blogjava.net/andyelvis/comments/commentRss/269304.html</wfw:commentRss><trackback:ping>http://www.blogjava.net/andyelvis/services/trackbacks/269304.html</trackback:ping><description><![CDATA[有一个无向图，写一个函数判断图中是否存在回路。<br />
函数原型如下：<br />
int ring(int[][] array, int n, int k)<br />
n为图中的顶点数，k为图中的边数，array其实是一个k行2列的二维数组，其中的元素比如（0，1）表示顶点0和顶点1之间存在边。<br />
图中如果存在回路，函数返回1，否则返回0。<br />
&nbsp;<br />
不知哪位大牛给高效地实现一下。
<img src ="http://www.blogjava.net/andyelvis/aggbug/269304.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.blogjava.net/andyelvis/" target="_blank">蒋耘</a> 2009-05-06 22:29 <a href="http://www.blogjava.net/andyelvis/archive/2009/05/06/269304.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>一道Google2009夏季实习生招聘笔试算法题</title><link>http://www.blogjava.net/andyelvis/archive/2009/04/14/265592.html</link><dc:creator>蒋耘</dc:creator><author>蒋耘</author><pubDate>Tue, 14 Apr 2009 11:23:00 GMT</pubDate><guid>http://www.blogjava.net/andyelvis/archive/2009/04/14/265592.html</guid><wfw:comment>http://www.blogjava.net/andyelvis/comments/265592.html</wfw:comment><comments>http://www.blogjava.net/andyelvis/archive/2009/04/14/265592.html#Feedback</comments><slash:comments>3</slash:comments><wfw:commentRss>http://www.blogjava.net/andyelvis/comments/commentRss/265592.html</wfw:commentRss><trackback:ping>http://www.blogjava.net/andyelvis/services/trackbacks/265592.html</trackback:ping><description><![CDATA[<span style="widows: 2; text-transform: none; text-indent: 0px; border-collapse: collapse; font: 13px arial; white-space: normal; orphans: 2; letter-spacing: normal; color: rgb(0,0,0); word-spacing: 0px; -webkit-border-horizontal-spacing: 0px; -webkit-border-vertical-spacing: 0px; -webkit-text-decorations-in-effect: none; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0" class="Apple-style-span">有一台机器，上面有m个储存空间。然后有n个请求，第i个请求计算时需要占R[i]个空间，储存计算结果则需要占据O[i]个空间（其中O[i]&lt;R[i]）。问怎么安排这n个请求的顺序，使得所有请求都能完成。你的算法也应该能够判断出无论如何都不能处理完的情况。<br />
比方说，m=14，n=2，R[1]=10，O[1]=5，R[2]=8，O[2]=6。在这个例子中，我们可以先运行第一个任务，剩余9个单位的空间足够执行第二个任务；但如果先走第二个任务，第一个任务执行时空间就不够了，因为10&gt;14-6。<br />
</span><br />
一种解法：<span style="widows: 2; text-transform: none; text-indent: 0px; border-collapse: collapse; font: 13px arial; white-space: normal; orphans: 2; letter-spacing: normal; color: rgb(0,0,0); word-spacing: 0px; -webkit-border-horizontal-spacing: 0px; -webkit-border-vertical-spacing: 0px; -webkit-text-decorations-in-effect: none; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0" class="Apple-style-span">按照R值和O值之差（即释放空间的大小）从大到小排序，然后依次做。</span><br />
<img src ="http://www.blogjava.net/andyelvis/aggbug/265592.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.blogjava.net/andyelvis/" target="_blank">蒋耘</a> 2009-04-14 19:23 <a href="http://www.blogjava.net/andyelvis/archive/2009/04/14/265592.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>一道Google2009夏季实习生招聘笔试程序设计题</title><link>http://www.blogjava.net/andyelvis/archive/2009/04/14/265496.html</link><dc:creator>蒋耘</dc:creator><author>蒋耘</author><pubDate>Tue, 14 Apr 2009 06:07:00 GMT</pubDate><guid>http://www.blogjava.net/andyelvis/archive/2009/04/14/265496.html</guid><wfw:comment>http://www.blogjava.net/andyelvis/comments/265496.html</wfw:comment><comments>http://www.blogjava.net/andyelvis/archive/2009/04/14/265496.html#Feedback</comments><slash:comments>15</slash:comments><wfw:commentRss>http://www.blogjava.net/andyelvis/comments/commentRss/265496.html</wfw:commentRss><trackback:ping>http://www.blogjava.net/andyelvis/services/trackbacks/265496.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 要求：写一个函数void count(char* input,int len)，此函数的功能是计算出一个字符串中每个字符的个数，不区分大小写，输出结果时按字符在字符串中出现的先后顺序。使用程序语言不限。&nbsp;&nbsp;<a href='http://www.blogjava.net/andyelvis/archive/2009/04/14/265496.html'>阅读全文</a><img src ="http://www.blogjava.net/andyelvis/aggbug/265496.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.blogjava.net/andyelvis/" target="_blank">蒋耘</a> 2009-04-14 14:07 <a href="http://www.blogjava.net/andyelvis/archive/2009/04/14/265496.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>三道Google2009夏季实习生招聘笔试选择题</title><link>http://www.blogjava.net/andyelvis/archive/2009/04/14/265482.html</link><dc:creator>蒋耘</dc:creator><author>蒋耘</author><pubDate>Tue, 14 Apr 2009 05:19:00 GMT</pubDate><guid>http://www.blogjava.net/andyelvis/archive/2009/04/14/265482.html</guid><wfw:comment>http://www.blogjava.net/andyelvis/comments/265482.html</wfw:comment><comments>http://www.blogjava.net/andyelvis/archive/2009/04/14/265482.html#Feedback</comments><slash:comments>10</slash:comments><wfw:commentRss>http://www.blogjava.net/andyelvis/comments/commentRss/265482.html</wfw:commentRss><trackback:ping>http://www.blogjava.net/andyelvis/services/trackbacks/265482.html</trackback:ping><description><![CDATA[1.从以下四项中选出两项Google工程师&#8220;20%时间项目&#8221;<br />
A、移动搜索<br />
B、音乐搜索<br />
C、整合搜索<br />
D、地图搜索<br />
<br />
2.有一个网络带宽为"下行速率为2Mbps和上行速率为512Kbps"的ADSL，请问下载一个120Mb的东西，至少需要<br />
A、4分钟<br />
B、8分钟<br />
C、16分钟<br />
D、32分钟<br />
<br />
3.只有两个结点的二叉树有两种形状，请问五个结点的二叉树有多少种形状<br />
A、22<br />
B、30<br />
C、42<br />
D、46<br />
<br />
知道答案的讨论下！<br />
<br />
<img src ="http://www.blogjava.net/andyelvis/aggbug/265482.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.blogjava.net/andyelvis/" target="_blank">蒋耘</a> 2009-04-14 13:19 <a href="http://www.blogjava.net/andyelvis/archive/2009/04/14/265482.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>