仿真机器人的经典代码，里面包含以余弦曲线且按给定的速度到定点，PD壁障到顶点，末端处理，点球大战等等，很值得参考...

如做电子书籍一个小软件。-do as a small electronic book software.

《我替姐姐来爱你》原名：《天使与海豚》深爱着你，呼吸停止的瞬间，请让我在梦寐的重逢中，燃尽我的思念，坦然而骄傲的与你告别-"I for my sister to Love You" titled : "Angels and dolphins," I love you, and stopped breathing moments Please let me dream of a reunion, I miss burnout, and frankly proud of you and farewell

c语言游戏开发用 附带 文件相关函数 常规内存操作函数 等函数的简介-c language game development with the accompanying documents related to function of conventional memory operation functions, such as a brief introduction of function

CAN控制器的程序设计,包括两部分，第一部分为中断向量表，第二部分包括系统初始化，CAN模块初始化及主程序等-CAN controller programming, including two parts, the first for the Interrupt Vector Table, Part II, including system initialization, CAN module initialization and the main program, such as

day la cuon sach hoc ve thi truong, ban co the tham khao

此书为电子书，很有名-The book for the

java书籍《thinking in java》，英文版，很好的java书籍之一。-"thinking in java" and the English version, good java books one.

Hello Android 2nd Edition November 2009 Retail Ebook-ATTiCA -"Hello, Android" will get you creating your first working application: Android s version of "Hello, World." From there, you ll build up a more substantial example: an Android Sudoku game.

The book introduces the IP multicast and its applications, such as IPTV

背包问题动态规划算法JAVA 给定n种物品和一背包。物品i的重量是wi，其价值为vi，背包的容量为c。问应如何选择装入背包中的物品，使得装入背包中物品的总价值最大？ 在选择装入背包的物品时，对每种物品i只有两种选择，即装入背包或不装入背包。不能将物品i装入背包多次，也不能只装入部分的物品i。因此，该问题称为0-1背包问题。-Knapsack problem dynamic programming algorithm for a given n species JAVA goods and one backpack. Item i is the weight of wi, the value for vi, knapsack capacity c. Question should be how to choose the items into the backpack, the backpack into the total value of the largest items? In the selection of the items into the backpack, for each of the items i have only two choices, that is loaded or not loaded backpack backpack. Items should not be many times i load backpack can not only part of the goods loaded i. Therefore, the problem known as the 0-1 knapsack problem.