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７＊７方格去掉四角各四个各，剩下３３个格除中间以外其他都放满牌，一长牌跳过另一张牌，被跳过的牌会被移走，求最后只剩中间一张牌的算法？-7* 7 1.40 removed the box four of the remaining 33 in addition to the middle lattice outside are filled with other licensing, a long licensing skip another card, skip the license will be removed and left for the middle of a licensing algorithm?

游戏编程入门的书籍，主要说明遗传算法和神经网络 -Introduction to the book Game Programming

一个初学者谁学习mplab c编译器手册。

软件测试与junit实践一书的源代码，需要的请下载-The practice of software testing with junit book source code, need to download

用C实现图象编程-C Programming Image

Understanding linux kernel，2nd edition 学习linux内核最好的教材 -Understanding, 2nd edition linux kernel best learning materials

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.

java 与 xml 技术的确融合，众多详实的代码-java and xml technology integration Indeed, many detailed code

C++大学教程(中文版）-A C++ tutorial for college (Chinese).

C#程序开发范例宝典 图书光盘附带源码 第五章-C# Program development paradigm Baodian Book CD-ROM attached to Chapter V source