艾迪森维斯理

观也u内核和开发法程序/。很好用，希望号郝整理-View also u kernel and Development Act procedures /. Very good with the hope that Hao finishing No.

信息安全3关于密码学-3 on information security cryptography

The_C++_Standard_Library c++标准库文档,开发C++程序的必读电子书籍-The_C++ _Standard_Libraryc++ Standard library documentation, development C++ Procedures must-read e-books

vb。net 2003数据库设计整套源代码-vb. Net 2003 database design package source code

单片机的以太网数据采集与控制电路设计.PDF-Single Chip Ethernet data acquisition and control circuit design. PDF

哈夫曼，程序员至少会想到二叉树和二进制编码。的确，我们一直在哈夫曼编码领域总结哈夫曼在个人计算机领域，特别是在数据压缩领域的突出贡献。我们知道压缩编码=模型，作为一种压缩方法，必须充分考虑两个模型和编码模块的有效性；但同时，模型和编码模块的两个相互独立。

在实时平台上,高斯混合模型(GMM)具有计算有效性和易于实现的优点。最大似然规则中,模型参数不 断更新,但由于爬山特征,任意的原始模型参数估计通常将导致局部最优 遗传算法(GA)适于求解复杂组合优化问 题及非线性函数优化。提出了基于说话人识别的可以解决GMM局部最优问题的GMM/GA新算法,实验结果表明, 提出的GMM/GA新算法比纯粹的GMM算法能获得更优的效果。 - In real-time platform, the Gaussian mixture model (GMM) with the calculation of the effectiveness and easy to realize benefits. Maximum likelihood rule, the model parameters are not Broken updates, but due to climbing features, any of the original model parameter estimation will usually result in local optimum genetic algorithm (GA) is suitable for solving complex combinatorial optimization question Title and non-linear function optimization. Proposed speaker recognition based on GMM can solve the problem of local optimal GMM/GA new algorithm, experimental results show that the Proposed GMM/GA new algorithm than purely GMM algorithm can get better results.

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.

汇编程序设计 觉得可以的你就下-compilation of program design you can think on the next

介绍C++的基础知识，包括类，对象，以及编程的一些基础知识。-Describes the basics of C++, including classes, objects, as well as some basic knowledge of programming.