linux,shell编程。讲解详细，包括awk,sed,tr等工具的使用。对于初学者而言，非常有帮助。
上传时间： 2013-12-17
上传用户：www240697738
linux,shell编程。讲解详细，包括awk,sed,tr等工具的使用。对于初学者而言，非常有帮助。
上传时间： 2014-01-01
上传用户：小码农lz
e3tree开发文档，e3tree的使用手册，协助开发高效率的tr
上传时间： 2016-09-22
上传用户：731140412
《An Introduction to Ant Colony Optimization》 作者：Marco Dorigo and Krzysztof Socha 时间：Technical Report No.TR/IRIDIA/2006-010April 2006 一份介绍蚁群系统的英文原版论文报告
标签： Introduction Optimization Krzysztof Technical
上传时间： 2013-12-12
上传用户：heart520beat
Linux下的tree命令，同window，功能更强大！ 1、下载tree-1.5.1.1源程序 2、解压源码到工作目录tar zxvf tree-1.5.1.1.tgz 3、在tree-1.5.1.1目录交叉编译：make CC=arm-linux-gcc 4、安装：make prefix=/home/xxxx/rootfs install Ubuntu下 直接 apt-get install tree 即可安装最新的tr
上传时间： 2016-09-27
上传用户：wangzhen1990
一被控对象 ，给定为阶跃给定，幅值为500，设计一个两维模糊PI型控制器，输入语言变量和输出语言变量均取7个值{NB，NM，NS，ZE，PS，PM，PB}，模糊论域为{-6，-5，-4，-3，-2，-1，0，1，2，3，4，5，6}，用matlab编程仿真研究。
标签： 对象
上传时间： 2013-12-16
上传用户：大融融rr
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：zhyiroy
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2016-10-31
上传用户：sunjet
private double PointToAngle(Point AOrigin, Point APoint) { if (APoint.X == AOrigin.X) if (APoint.Y > AOrigin.Y) return Math.PI * 0.5f else return Math.PI * 1.5f else if (APoint.Y == AOrigin.Y) if (APoint.X > AOrigin.X) return 0 else return Math.PI else {
标签： AOrigin APoint Point PointToAngle
上传时间： 2013-12-18
上传用户：rocketrevenge
For solving the following problem: "There is No Free Lunch" Time Limit: 1 Second Memory Limit: 32768 KB One day, CYJJ found an interesting piece of commercial from newspaper: the Cyber-restaurant was offering a kind of "Lunch Special" which was said that one could "buy one get two for free". That is, if you buy one of the dishes on their menu, denoted by di with price pi , you may get the two neighboring dishes di-1 and di+1 for free! If you pick up d1, then you may get d2 and the last one dn for free, and if you choose the last one dn, you may get dn-1 and d1 for free. However, after investigation CYJJ realized that there was no free lunch at all. The price pi of the i-th dish was actually calculated by adding up twice the cost ci of the dish and half of the costs of the two "free" dishes. Now given all the prices on the menu, you are asked to help CYJJ find the cost of each of the dishes.
标签： Limit following solving problem
上传时间： 2014-01-12
上传用户：362279997
