国外以为牛人编的，我也不怎么懂。。可以用来做课程设计，折实作者后来司升级后的软件...

资源描述写一个简单的Hough特征检测器，可以检测输入图像中的任意大小的圆。

通过这个项目你可以识别字符，数字等你想要的语言。首先你要画出每个字母的地方，

char *MD5_file (char *path, int md5_len)   {       FILE *fp = fopen (path, "rb");       MD5_CTX mdContext;       int bytes;       unsigned char data[1024];       char *file_md5;       int i;          if (fp == NULL) {           fprintf (stderr, "fopen %s failed ", path);           return NULL;       }          MD5Init (&mdContext);       while ((bytes = fread (data, 1, 1024, fp)) != 0)       {           MD5Update (&mdContext, data, bytes);       }       MD5Final (&mdContext);              file_md5 = (char

常用算法的C程序。主要内容包括多项式的计算、复数运算、随机数的产生、矩阵运算、矩阵特征值与特征向量的计算、线性代数方程组的求解、非线性方程与方程组的求解、插值与逼近、数值积分、常微分方程组的求解、数据处理、极值问题的求解、数学变换与滤波、特殊函数的计算、排序和查找。-Commonly used algorithm of C procedures. Include polynomial terms, complex computation, random number generation, matrix operations, matrix eigenvalue and eigenvector calculations, linear algebra equations, nonlinear equations with the equations, interpolation and approximation, numerical integration, solving ordinary differential equations, data processing, for solving extremum problems, mathematical transformation and filtering, the calculation of special functions, sorting and search.

倒酒问题描述： 设有两个能装8两的酒杯（称为１号，２号）装满了酒, 和1个能装3两的空酒杯（称为３号）, 问怎样用这3个酒杯向4个人 敬酒, 使得每个人都喝4两酒. 要求： 用程序计算出可行方案。 输入： 无 输出： 每一步决策．-down quite Problem description : two can be loaded with two of the eight glasses (known as 1, 2) filled with the wine, and one can hold three two empty glasses (known as the 3rd) and asked how to use it three to four glasses and toast the individual, so that each person maximum servings of both types 4 2 liquor. Request : procedures calculated options. Input : None output : each step of the decision-making.

用opencv实现的图像直方图均衡化处理，用HistogramStat函数统计图像中三个通道中的像素总数，用Histogram函数对其进行直方图均衡化处理

ApMl provides users with the ability to crawl the web and download pages to their computer in a directory structure suitable for a Machine Learning system to both train itself and classify new documents. Classification Algorithms include Naive Bayes, KNN -ApMl provides users with the ability to crawl the web and download pages to their computer in a directory structure suitable for a Machine Learning system to both train itself and classify new documents. Classification Algorithms include Naive Bayes, KNN

这种算法是与所谓的印章加密....密封件也使用四个32位寄存器A，B，C和D，其初始值是由n和第k派生表R和T.这些寄存器被修改过几次迭代，每一个都涉及到8发来确定。在每一轮的9位的第一寄存器（无论是A，B，C或D）用于索引表T在T中检索到的值然后被添加到或异或同的第二寄存器的内容：再次之一的A，B，C或D的第一寄存器，然后由圆9的位置错开。在一些回合第二寄存器通过加入或与异或它进一步修饰的（现在的移位）第一个寄存器。后8轮的这一点，A，B，C和D添加到密钥流，每通过添加或具有一定字从S的迭代是通过向A和C依赖于n个附加的值完成异或它掩蔽第一，N1，N2，N3，N4;到底是哪一个取决于的迭代次数的奇偶性。

Visual C++常微分方程解法（转载）-Visual C Ordinary Differential Equations Solution (reproduced)

计算方法《牛顿差值》-calculation method, "Newton margin"