24 hours learning how to use PHP MySQL and Apache

方格识别算法：输入具有外部轮廓和外部标记的九宫格图像，标记设定在右上角，个小格的位置从左到右，从下到上一次表示成0-8位的二进制数，其中黑―1，白―0.输出对应的十进制数，对任意角度放置的图像能准确识别。-Grid recognition algorithm: Enter an external contour and external marking squares image, mark set at the upper right corner, a small grid position from left to right, from bottom to top into a 0-8 first express-bit binary number, in which black-1,-0 white. output corresponding to a decimal number, for any angle to place the image can accurately identify.

可以自动设置表格，表格框架线的灰度，表格具有的行数和列数！-Can automatically set up tables, forms the framework of the gray line, the form has a number of rows and columns!

Mantis 中文用户手册，详细介绍Mantis使用方法-Mantis user manual

一个基于pci2000数据采集卡的数据采集系统的界面-a data acquisition card data acquisition system interface

文件上传的一个程序-file uploads a procedure

这是一个用控制台做的螺旋线的动画，帮你改变背景，实现简单动画-This is a console to do with the spiral of animation, to help you change the background, the realization of a simple animation

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

VC与halcon混合编程，用处很大的哦-helpful for you Programming。it is a example of halcon and VC6.0。

wince5.0 下，用于获取sdcard 的 ID. 用c++ 开发，vs2005 -wince5.0 used for sdcard access to the ID. with c++ development, vs2005

香农费诺编码输入为+7个信源符号,其概率分布为：(其中信源符号用A+B、、G表示)A_0.2++B_0.19++C_0.18++D_0.17++E_0.15++F_0.10++G_0.01然后输入程序中得到结果（详细文档，值得一下）香农-费诺编码-Shannon Fenaux+7 encoded input source symbols, the probability distribution as follows: (where source symbols with the A+ B,, G indicated) A_0.2++ B_0.19++ C_0.18++ D_0.17++ E_0.15++ F_0.10++ G_0.01 and then enter the program to get results (detailed documentation, it is worth it) Shannon- Fano coding