一本很好的WINDOWS DRIVER的书

InfoPath开发人员参考,详细描述了开发InfoPath所需要的各个函数和方法，对编程很有帮助。-InfoPath Developer Reference, a detailed description of the development needed InfoPath all function and method of programming helpful.

windows的写字板的vc实现-windows tablet to achieve the vc

包括常用的经典的算法及其实现 非常受用的经典苏澳发达偶在里面

讲讲自己学习RSA中的实践过程，已经对RSA熟悉的看家就不用在此浪费时间了。-the study process of RSA.

Everything是目前最快的本地文件名搜索软件，利用其SDK可以很好地将本地搜索加入到自己的项目中。

The programm calculate Gauss-Markov process 1-st and 2-nd order

通过广度优先算法寻找迷宫(nSize*nSize)的最优路径。-breadth- first through the maze to find algorithm (nSize* nSize) the optimal path.

欢迎使用闰年查询程序，大家好啊！查询一下几时是闰年也是生活中的一种乐趣嘛-welcomed the use of procedures for leap year, we said! Inquiries about the earliest it is a leap year is a life of fun!

部分索尼爱立信手机应用补丁源码，从俄国网站找的。-Part of Sony Ericsson mobile phone application source patch, from Russia to find the site.

0-l背包问题是子集选取问题。一般情况下，0-1背包问题是NP难题。0-1背包 问题的解空间可用子集树表示。解0-1背包问题的回溯法与装载问题的回溯法十分类 似。在搜索解空间树时，只要其左儿子结点是一个可行结点，搜索就进入其左子树。当 右子树有可能包含最优解时才进入右子树搜索。否则将右子树剪去。设r是当前剩余 物品价值总和；cp是当前价值；bestp是当前最优价值。当cp+r≤bestp时，可剪去右 子树。计算右子树中解的上界的更好方法是将剩余物品依其单位重量价值排序，然后 依次装入物品，直至装不下时，再装入该物品的一部分而装满背包。由此得到的价值是 右子树中解的上界。-0-l knapsack problem is selected subset of the problem. Under normal circumstances, 0-1 knapsack problem is NP-hard. 0-1 knapsack problem the solution space available subset of the tree said. Xie 0-1 knapsack problem with the law retroactively loading the retroactive law very similar. The search solution space trees, as long as their son left node is a viable nodes, the search entered its left subtree. When the right subtree is the optimal solution may contain only son into the right tree search. It will cut right subtrees. Suppose that r is the total value of the remaining items; Cp is the current value; Bestp is currently the best value. When cp r bestp, they can cut right subtrees. Calculation right