﻿<?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>C++博客-飘泊永恒</title><link>http://www.cppblog.com/sa389708585/</link><description>Forever Ever</description><language>zh-cn</language><lastBuildDate>Tue, 09 Jun 2026 22:42:16 GMT</lastBuildDate><pubDate>Tue, 09 Jun 2026 22:42:16 GMT</pubDate><ttl>60</ttl><item><title>一个找路径的问题</title><link>http://www.cppblog.com/sa389708585/archive/2010/05/20/115980.html</link><dc:creator>飘泊</dc:creator><author>飘泊</author><pubDate>Thu, 20 May 2010 14:33:00 GMT</pubDate><guid>http://www.cppblog.com/sa389708585/archive/2010/05/20/115980.html</guid><wfw:comment>http://www.cppblog.com/sa389708585/comments/115980.html</wfw:comment><comments>http://www.cppblog.com/sa389708585/archive/2010/05/20/115980.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/sa389708585/comments/commentRss/115980.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/sa389708585/services/trackbacks/115980.html</trackback:ping><description><![CDATA[<p>编程实现下图的查找路径</p>
<img border=0 alt="" src="http://www.cppblog.com/images/cppblog_com/sa389708585/1.jpg" width=441 height=476><br>我的想法是构造一个数组a[5][5],然后赋值：a[i][j]=((i+1)*10+(j+1))<br>然后构造另一个数组b[24]，将a中的值随机不重复的覆给b，<br>当b满足：1、b中没有a[3][1]值<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2、b中相邻的两个值都相差1或10<br>的时候，将b的值输出。<br>问题是给b随机赋值和b的可能性能否全涉及到。<br>希望高手给看看。
<img src ="http://www.cppblog.com/sa389708585/aggbug/115980.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/sa389708585/" target="_blank">飘泊</a> 2010-05-20 22:33 <a href="http://www.cppblog.com/sa389708585/archive/2010/05/20/115980.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>