<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[lkbsbird的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/lkbsbird</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; lkbsbird]]></copyright><item><title><![CDATA[个人学习的链接]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52974136</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52974136</guid><author>lkbsbird</author><pubDate>Sun, 30 Oct 2016 20:56:22 +0800</pubDate><description><![CDATA[maven配置springbatch]]></description><category></category></item><item><title><![CDATA[java　线程间通信同步代码块]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52373038</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52373038</guid><author>lkbsbird</author><pubDate>Tue, 30 Aug 2016 21:13:33 +0800</pubDate><description><![CDATA[import java.lang.*;
class Res
{
	String name;
	String sex;
	boolean flag=false;
	public synchronized void set(String name,String sex)
    {
		if(flag)
		 try
		 {
			this.wait();
		 }
		 catch (Except]]></description><category></category></item><item><title><![CDATA[指针复习之单链表实现]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52291647</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52291647</guid><author>lkbsbird</author><pubDate>Tue, 23 Aug 2016 15:53:22 +0800</pubDate><description><![CDATA[#include
#include
#include
typedef struct node
{
    int data;
    struct node *next;
} node;
node *init()
{
    node *head;
    head=(node*)malloc(sizeof(node));
    head->next=NULL;
    return head;]]></description><category></category></item><item><title><![CDATA[程序员必学的linux命令（1）]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52234567</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52234567</guid><author>lkbsbird</author><pubDate>Wed, 17 Aug 2016 20:55:36 +0800</pubDate><description><![CDATA[1.cp -airusdf
2.rm -fir(recursion)(\rm -r 不需要询问) (rm -aaa- -> rm ./-aaa-or rm -- -aaa-)
3.rmdir 空目录
4.mv -fiu (也有重命名)=rename
5.basename(取得文件名),dirname(取得目录名)
6.cat -n
7.tac
8.nl
9.more
10.od ]]></description><category></category></item><item><title><![CDATA[Linux init命令]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52049712</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52049712</guid><author>lkbsbird</author><pubDate>Wed, 27 Jul 2016 21:36:44 +0800</pubDate><description><![CDATA[切换执行等级： init
本章上头有谈到过关亍 run level 的问题。乀前谈到的是系统运作的模式，分为纯文本(run level 3)及图
形接口模式(run level 5)。除了这两种模式外，有没有其他模式呢？其实 Linux 共有七种执行等级， 七
种等级的意丿我们在后面会再谈到。本章你只要知道底下四种执行等级就好了：
 run level 0：关机
 run level]]></description><category></category></item><item><title><![CDATA[linux 正常关机]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52049691</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52049691</guid><author>lkbsbird</author><pubDate>Wed, 27 Jul 2016 21:34:55 +0800</pubDate><description><![CDATA[正确的关机方法
OK！大概知道开机的方法，也知道基本的挃令操作，而且还已经知道在线查询了，好累呦！ 想去休息
呢！那举如何关机呢？我想，径多朊友在 DOS 的年代已经有在玩计算机了！ 在当时我们关掉 DOS 的
系统时，常常是直接关掉电源开关，而 Windows 在你丌爽的时候，挄着电源开关四秒也可以关机！但
是在 Linux 则相当的丌建议这举做！
Why？在 Windows (非 N]]></description><category></category></item><item><title><![CDATA[混合结构]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52046259</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52046259</guid><author>lkbsbird</author><pubDate>Wed, 27 Jul 2016 15:52:07 +0800</pubDate><description><![CDATA[]]></description><category></category></item><item><title><![CDATA[p2p结构]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52046194</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52046194</guid><author>lkbsbird</author><pubDate>Wed, 27 Jul 2016 15:48:45 +0800</pubDate><description><![CDATA[没有永远在线的服务器
任意端系统/节点之间可以直接通讯
节点间歇性接入网络
节点可能改变IP地址
优点：高度可伸缩
缺点：难于管理]]></description><category></category></item><item><title><![CDATA[客户端/服务器结构]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52046144</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52046144</guid><author>lkbsbird</author><pubDate>Wed, 27 Jul 2016 15:46:03 +0800</pubDate><description><![CDATA[服务器
 7*24小时提供服务
 永久性访问地址/域名
 利用大量服务器实现可扩展性
客户机
 与服务器通信，使用服务器提供的服务
 间歇性接入网络
 可能使用动态IP地址
 不会与其他客户机直接通信]]></description><category></category></item><item><title><![CDATA[计算机网络的体系结构]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52046109</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52046109</guid><author>lkbsbird</author><pubDate>Wed, 27 Jul 2016 15:41:29 +0800</pubDate><description><![CDATA[]]></description><category></category></item><item><title><![CDATA[网络应用的体系结构]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52046080</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52046080</guid><author>lkbsbird</author><pubDate>Wed, 27 Jul 2016 15:39:40 +0800</pubDate><description><![CDATA[]]></description><category></category></item><item><title><![CDATA[hdu 5752]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/52038010</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/52038010</guid><author>lkbsbird</author><pubDate>Tue, 26 Jul 2016 17:40:29 +0800</pubDate><description><![CDATA[题意：求能否在5次开根号，回到1（水题）
#include
#include
#include
#include
using namespace std;
const int maxn=10009;
char s[maxn];
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        int len=strlen(s);
        i]]></description><category></category></item><item><title><![CDATA[hdu 5666]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51993036</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51993036</guid><author>lkbsbird</author><pubDate>Fri, 22 Jul 2016 11:20:08 +0800</pubDate><description><![CDATA[题意：求x+y=q与坐标轴围成的三角形中，有多少个整数点坐标（不包括边界）
解析：注意数据很大，要用到快速乘
#include
#include
#include
#include
using namespace std;
typedef long long ll;
long long q,p;
int t;
ll qmul(ll a,ll b,ll m) {
    ll ans=0;

 ]]></description><category></category></item><item><title><![CDATA[hdu 题目分类]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51991310</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51991310</guid><author>lkbsbird</author><pubDate>Fri, 22 Jul 2016 08:43:55 +0800</pubDate><description><![CDATA[基础题：
1000、1001、1004、1005、1008、1012、1013、1014、1017、1019、1021、1028、1029、1032、1037、1040、1048、1056、1058、1061、1070、1076、1089、1090、1091、1092、1093、1094、1095、1096、1097、1098、1106、1108、1157、1163、1164、1170、1194]]></description><category></category></item><item><title><![CDATA[欧几里德和扩展欧几里德]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51991292</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51991292</guid><author>lkbsbird</author><pubDate>Fri, 22 Jul 2016 08:40:17 +0800</pubDate><description><![CDATA[#include
#include
#include
#include
using namespace std;
const int maxn=1009;

//欧几里德算法又称辗转相除法，用于计算两个整数a,b的最大公约数。
//基本算法：设a=qb+r，其中a，b，q，r都是整数，
//则gcd(a,b)=gcd(b,r)，即gcd(a,b)=gcd(b,a%b)。
//递归
int gcd(]]></description><category></category></item><item><title><![CDATA[hdu 5744]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51991278</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51991278</guid><author>lkbsbird</author><pubDate>Fri, 22 Jul 2016 08:35:45 +0800</pubDate><description><![CDATA[题意：给出字符的个数，组成回文串，找回文串中最小值的最大值。
解析：将奇数分成1和2，分别统计1和2的个数，判断下就出来了。
#include
#include
#include
#include
using namespace std;
const int maxn=100009;
int b[maxn];
int a[maxn];
int t,n;
bool cmp(int a,int b]]></description><category></category></item><item><title><![CDATA[hdu 5742]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51991250</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51991250</guid><author>lkbsbird</author><pubDate>Fri, 22 Jul 2016 08:27:49 +0800</pubDate><description><![CDATA[题意：求a1+a2/a1+a2+..+an的最大值，其中只给出部分值，ax=y;
#include
#include
#include
#include
using namespace std;
const int maxn=109;
int t;
int n,m;
int a[maxn];
int gcd(int a,int b)
{
    return b ? gcd(b,a%b):a;
]]></description><category></category></item><item><title><![CDATA[hdu 1019]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51979627</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51979627</guid><author>lkbsbird</author><pubDate>Thu, 21 Jul 2016 09:41:42 +0800</pubDate><description><![CDATA[题意：求M个数 的最小公倍数
#include
#include
#include
using namespace std;
const int maxn=1000009;
long long g[maxn];
int n;
int m;
long long gcd(long long a,long long b)
{
    return b ? gcd(b,a%b):a;
}
int mai]]></description><category></category></item><item><title><![CDATA[Poj 1840]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51952524</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51952524</guid><author>lkbsbird</author><pubDate>Tue, 19 Jul 2016 09:01:33 +0800</pubDate><description><![CDATA[Description

Consider equations having the following form: 
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 
The coefficients are given integers from the interval [-50,50]. 
It is consider a solution a syste]]></description><category></category></item><item><title><![CDATA[java EE基础知识]]></title><link>https://blog.youkuaiyun.com/lkbsbird/article/details/51945030</link><guid>https://blog.youkuaiyun.com/lkbsbird/article/details/51945030</guid><author>lkbsbird</author><pubDate>Mon, 18 Jul 2016 20:07:48 +0800</pubDate><description><![CDATA[开发一个Servlet程序的步骤：
                                                        1）创建一个java类，继承HttpServlet类
                                                        2）重写HttpServlet类的doGet方法
               ]]></description><category></category></item></channel></rss>