Huawei' s program training materials, a number of programming issues that nee...

SPY++的使用帮助-SPY++ Use help

java的交流文档。-exchange files.

MIT "perceptual and artistic principles for effective computer depiction" 计算机视觉相关课程讲稿-第一课-This course presents connections between human visual perception and the art of picture production. Findings from perceptual and cognitive sciences are used to explore pictorial techniques used by artists as they address the challenges raised by the depiction of three-dimensional scenes on a flat media. The course introduces perceptual explanations for a variety of artistic styles. Finally, we present mechanisms that can make an image more effective, and demonstrate the adaptation of these mechanisms to computer graphics. The course is intended for both artists and scientists. Although it offers some practical insights, it is intended more as an in-breadth overview.

本次上传的是一个关于e-learning学习的流程图，在此图上有基本的建立e-learning的一些内容-This upload is a study on the flow chart of e-learning in this diagram are the basic building e-learning some of the content of

HTML文档，英文！ 10多篇文章，主要讲的是电脑游戏中的人工智能制作。-HTML documents, in English! More than 10 articles, mainly talking about artificial intelligence in computer games production.

21天学会sql教程-21 days Institute sql Guide

Abstract―In this paper, we propose transform-domain algorithms to effectively classify the characteristics of blocks and estimate the strength of the blocky effect. The transform-domain algorithms require much lower computational complexity and much less memory than the spatial ones. Along with the estimated blocky strength,

delphi学习-delphi study

主要介绍C＃的主要特点，需要有C基础的人才适于阅读，也便于学习.net的人参考。-C# introduces the main features of the need for C based talent suitable for reading, but also to facilitate learning. Net of reference.

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.