八数码难题解决实现方法

fft的文件，fortran环境下运行！欢迎指正！-fft documents, fortran environment running! Please correct me!

对于给定电路系统的 ：C X+G X=B U(t)；Y=LT X; 用GCR迭代法求解GX0=BU(0)中的X0，然后用后向Euler以及GCR迭代法求解上述的方程得到矩阵X。其中矩阵C、G、B、LT以及输入函数U(t)为已知。

一个简单的多线程的例子-a simple example Multithreading

SVD的应用，奇异值超平面正交投影证明 A中每个向量xi在随机阵B上的正交投影之和的最小值-Application of SVD, singular value hyperplane orthogonal projection proves A vector xi for each B in a random array of the orthogonal projection and the minimum

用vc++实现马尔科夫过程，并利用其原理实现简单的语音识别功能。感觉最重要的是要有一定的数学和编程功底，值得搞开发的人员使用。涉及的函数较多，需要耐心分析才能有一定的收获。

这是SIR微分方程模型对初值问题的求解和作图-This is the SIR model differential equations to solve the problem of initial value and mapping

Vondrak算法:主要用于未知函数测量数据滤波-Vondrak algorithm: being used in the filtering of measuring data whose function is unknown.

一个acm算法题目 问题描述： 有一个黑盒子，可视为一个15×10的二维矩阵，里面有一些空隙，在二维矩阵中用0表示空隙，用1表示被占有。现有一个仪器，可以从4个方向对盒子进行分行透视：从左到右，从左下到右上，从下到上，从右下到左上。透视可得到该二维矩阵的该行1的个数。要求，还原此矩阵。 输入：4行，每一行分别代表从左到右，从左下到右上，从下到上，从右下到左上透视所得的数据 输出：该矩阵 使用的方法是回溯法，具体见程序-a acm algorithm topic Problem description : a black box, as a 15 x 10 two-dimensional matrix, there are some gaps in two-dimensional matrix using 0 indicates the gap, with a show that occupies. An existing equipment, from the direction of four branches box for perspective : from left to right, from the lower left to upper right, from bottom to top, from the lower right to upper left. Perspective availability of the two-dimensional matrix of the line a number. Demand reduction of this matrix. Input : 4 firms on behalf of every line from left to right, from the lower left to upper right, from bottom to top, from the upper left to lower right perspective on data obtained from the output : the matrix method is the use of backtracking, see the specific procedures

Cantilever simulations in COMSOL Multiphysics

java 计算器,功能较强大,可进行大多数科学计算-java calculator, has very powerful functions, able to perform most of scientific calculation