Beginning JavaScript with DOM Scripting and Ajax

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IP v6技术要求――IPV6协议（2001H74）”送审稿-IP v6 technical requirements-- IPv6 protocol (2001H74) "review and

asp方面的参考资料-the reference asp

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Windows程序设计(第5版)中英文电子书+源代码.rar-Windows Programming (Release 5) e-books in both English and Chinese+ Source code. Rar

mapinfo通用符号库安装程序,mapinfo通用符号库安装程序-Common symbol library mapinfo setup, mapinfo setup generic symbol library

JSP数据库开发实例精粹,很好的参考书，入门必备-Pristine examples of JSP database development, good reference books, entry must

J2EE-完全参考手册.pdf,中文版，是学习J2EE的非常好的入门教材。-J2EE-complete reference manual. Pdf, the Chinese version is studying J2EE very good introductory material.

c++ primer 源代码下载(第二章)-c++ primer source code download (chapter II)

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.