<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[stone1982的专栏]]></title><description><![CDATA[]]></description><link>https://blog.csdn.net/stone1982</link><language>zh-cn</language><generator>https://blog.csdn.net/</generator><copyright><![CDATA[Copyright &copy; stone1982]]></copyright><item><title><![CDATA[Tomcat 数据源学习笔记]]></title><link>https://blog.csdn.net/stone1982/article/details/867660</link><guid>https://blog.csdn.net/stone1982/article/details/867660</guid><author>stone1982</author><pubDate>Sun, 02 Jul 2006 23:09:00 +0800</pubDate><description><![CDATA[Tomcat 数据源学习笔记 条件1：数据库驱动程序的jar包必须放在Tomcat的common/lib的目录下。条件2： web应用的WEB-INF/lib目录下不存在Tomcat的common/lib目录下的jar包（Tomcat的配置手册说只需不存在Jakarta-Commons DBCP 1.0.jar，Jakarta-Commons Collections 2.0.jar， ]]></description><category></category></item><item><title><![CDATA[活动安排问题]]></title><link>https://blog.csdn.net/stone1982/article/details/448906</link><guid>https://blog.csdn.net/stone1982/article/details/448906</guid><author>stone1982</author><pubDate>Tue, 09 Aug 2005 12:07:00 +0800</pubDate><description><![CDATA[//Arrange action void arrange(int n, Type start[], Type end[], boolean used[]) {int i, j = 0; Sort_ASC(n, start, end);// sort by the array of end[] : ascend used[0] = true; for(i = 1; i = end[j])//if ]]></description><category></category></item><item><title><![CDATA[KMP_Algorithm]]></title><link>https://blog.csdn.net/stone1982/article/details/448903</link><guid>https://blog.csdn.net/stone1982/article/details/448903</guid><author>stone1982</author><pubDate>Tue, 09 Aug 2005 12:04:00 +0800</pubDate><description><![CDATA[//KMP Algorithm referenced to the book of data structure int KMP(int n, const char pri_str[], const char sub_str[]) { int i, j; int next[n];// store the String of sub_strs value of next assert(pri_st]]></description><category></category></item><item><title><![CDATA[背包问题另解]]></title><link>https://blog.csdn.net/stone1982/article/details/21089</link><guid>https://blog.csdn.net/stone1982/article/details/21089</guid><author>stone1982</author><pubDate>Tue, 16 Mar 2004 16:54:00 +0800</pubDate><description><![CDATA[背包问题：有不同价值，不同重量的物品n件，求这 件物品中选取一部分，是选中物品的总重量不超过指定的限制重量，但选取的物品总价值要最大。解题思想：利用递归求出所选取的物品的所有可能，将满足限制重量条件的进行复值，Value始终存放最大的物品总价值，Weight是满足条件的物品的总重量，以下是我自己写的一段代码，与《系统设计师》教程中的思路是不一样的，代码经过编译的，而且运行结果正确：   #in]]></description><category></category></item><item><title><![CDATA[猫吃老鼠问题]]></title><link>https://blog.csdn.net/stone1982/article/details/21088</link><guid>https://blog.csdn.net/stone1982/article/details/21088</guid><author>stone1982</author><pubDate>Mon, 15 Mar 2004 09:17:00 +0800</pubDate><description><![CDATA[#define len sizeof(struct catmouse)struct catmouse{int num;struct catmouse *next;};void main(){struct catmouse *head,*p,*q; int n,m,i,j; int a[7]; printf("/nenter the an integer:"); scanf("%d",&n); p=]]></description><category></category></item><item><title><![CDATA[集合的运算]]></title><link>https://blog.csdn.net/stone1982/article/details/21087</link><guid>https://blog.csdn.net/stone1982/article/details/21087</guid><author>stone1982</author><pubDate>Mon, 15 Mar 2004 09:16:00 +0800</pubDate><description><![CDATA[#define len sizeof(struct order)#includestruct order{char num; struct order *next; };void main(){int i,j;char p1[30],p2[30];struct order *b,*bpre,*bhead,*a,*apre,*ahead,*p; printf("/nenter the first c]]></description><category></category></item></channel></rss>