资源列表
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- sharpzip组件的使用,可以打包和解压文件夹,通过递归-zipsharp components to use, can be packaged and extract the folder, through the recursive
cobol_Algorithm
- 用cobol编写的四则混合运算程序，包括一个JCL和一个Cobol,在IBM S390上调试通过-With cobol written four mixed-op program, including a JCL and a Cobol, debugging through the IBM S390
Otelo3niveles
- CODE OF OTELO THREE LEVELS
DOSmingliing
- DOS学习文章，内容简单易懂，讲解详细丰富，针对注册项做逐个分析，十天可以学会-DOS study articles, content simple and easy to understand, explain in detail rich, one by one against the registered item to do analysis of 10 days can learn to
ball
- 小球在窗口中弹动 可按上下左右键改变运动方向-the running ball
4545456
- 易语言脚本编写模块 完全支持最新易语言软件-Yi language scr ipting module fully supports the latest easy language software
Create_a_ZIP_file_using_java
- 这本是别人的东西，我只是修改了中文问题。在这个基础上改一下就可以压缩多个文件和目录，甚至可以写一个winzip之类的东东哦，有兴趣的可以研究一下。-This is other people s things, I just modified the Chinese problem. On this basis, to change what you can compress multiple files and directories, or even write a winzip-like D
CE
- 效果图多少发生地方都十分都是反对方法第三方的说法-this is picture with software
student
- 一个简易的学生宿舍管理系统，增加开发经验-A simple student hostel management system, increasing development experience
chen
- 有很经典的成功经验 大家一定会有启发的-chen anzhi
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- 2． 用循环链表解Josephus问题。设有n个人围坐在一个圆桌周围，现从第1个人开始报数，数到第m的人出列，然后从出列的下一个人重新开始报数，数到第m的人又出列，…，如此反复直到所有的人全部出列为止。Josephus问题是：对于任意给定的n和m，求出按出列次序得到的n个人员的序列-2. Josephus solution of the problem with the circulation list. With n individuals sitting around a round tab