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JavaEE学习之路-Sharing Information
Web components, like most objects, usually work with other objects to accomplish their tasks. Web components can do so by doing the following. Using private helper objects (for example, JavaBeans...
分类:编程语言   时间:2014-08-26 09:55:05    阅读次数:340
Non-blocking algorithm(非阻塞算法,非阻塞同步的算法实现)
Non-blocking algorithm In computer science, a non-blocking algorithm ensures that threads competing for a shared resource do not have their execution indefinitely postponed by mutual ex...
分类:其他好文   时间:2014-08-26 09:52:09    阅读次数:270
Fibonacci
#include<iostream> usingnamespacestd; intmain() { intpound,p0,p1,days,result; charans; do { cout<<"Pleaseinputthepoundandthedays:\n"; cin>>pound>>days; intn=days/5; if(0==n||1==n) { result=pound; cout<<"Theresultis"<<..
分类:其他好文   时间:2014-08-26 03:12:26    阅读次数:204
持续运行一个命令,并且将结果输出到文本
#! /bin/shwhile [ 2 > 1 ]do echo `date` >> recode.txt df -h >> recode.txt sleep 5done
分类:其他好文   时间:2014-08-25 22:28:24    阅读次数:207
ZOJ 3805 Machine(树形DP)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3805 Machine Time Limit: 2 Seconds      Memory Limit: 65536 KB In a typical assembly line, machines are connected one b...
分类:其他好文   时间:2014-08-25 21:17:04    阅读次数:296
ZOJ 3798 Abs Problem(规律题)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3798 Abs Problem Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Alice and Bob is playing a game, and ...
分类:其他好文   时间:2014-08-25 21:16:24    阅读次数:293
ZOJ3802 Easy 2048 Again (状压DP)
ZOJ Monthly, August 2014 E题ZOJ月赛 2014年8月 E题http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5334Easy 2048 AgainTime Limit: 2 Seconds Memory ...
分类:其他好文   时间:2014-08-25 21:04:24    阅读次数:233
Reading List
[2014-08-25]things want to do : assembly win32 MFC COM/COM+ ATL ActiveX TCP/IP object-orientied analysis and design with application...
分类:其他好文   时间:2014-08-25 19:02:14    阅读次数:200
Linux脚本(二)
1.for循环以及加法的使用portStr=`lsof -i:56801 | head -2`count=0for str in `lsof -i:56801 | head -2`do((count=count+1))echo "Count: $count"if [ $count == 11 ]; ...
分类:系统相关   时间:2014-08-25 18:59:14    阅读次数:176
近似pi
#include<iostream> #include<cmath> usingnamespacestd; intmain() { intn; charans; doublepi; //for(intn=0;n<=20;n++) //{ do{ cout<<"输入n求pi的近似值:\n"; cin>>n; pi=0.0; for(inti=0;i<=n;i++) { pi+=1.0*pow(-1,i)/(2*i+1); } ..
分类:其他好文   时间:2014-08-25 17:20:14    阅读次数:198
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