USDFLD HASHIN 2D & 3D

矩阵,复数自定义数据类型的C++实现,矩阵求逆采用牛顿法,(The matrix, complex number custom data type C realization, the matrix inverse uses the Newton method,)

改程序可用于绘制热加工图，用于热加工领域(It is a code using for making hot processing maps used to characterize the hot deformation behavior)

calculate maximum volume of cuboid given conditions.. a+b+c<s ab+bc+ca<d where s and d are constants..

fft 利用DSP Builder 在MATLAB下产生的模型(DSP Builder fft using the model generated in MATLAB)

对于应用数学专业学习非线性混沌控制的学生非常有用，可以下载并尝试自己动手改变后半部分，就能应用于新系统中(For students applied mathematics learning nonlinear chaotic control is very useful, you can download and try yourself to change the second half, will be able to be applied to the new system)

说明：  9跨每跨35米桥梁响应的车桥耦合程序，使用Fortran编写(9 of 35 meters span bridge across every bridge coupled response procedures, the use of Fortran written)

多点高斯平稳随机过程模拟，运用了快速傅里叶变换进行模拟。(Multi-Gaussian stationary random process simulation, using a fast Fourier transform simulation.)

处理任意两点间多条公交、地铁线路的到达问题此，改程序将一二问综合在一起，方便精简，运行速度快(Treatment between any two many bus, subway lines to reach this issue, change the program integrated with the twelve asked to facilitate streamlined, fast)

Description 设a[0…n-1]是一个包含n个数的数组，若在i<j的情况下，有a[i]>a[j]，则称(i, j)为a数组的一个逆序对（inversion）。 比如 <2,3,8,6,1> 有5个逆序对。请采用类似“合并排序算法”的分治思路以O(nlogn)的效率来实现逆序对的统计。 一个n个元素序列的逆序对个数由三部分构成： （1）它的左半部分逆序对的个数，（2）加上右半部分逆序对的个数，（3）再加上左半部分元素大于右半部分元素的数量。 其中前两部分（1）和（2）由递归来实现。要保证算法最后效率O(nlogn)，第三部分（3）应该如何实现？ 此题请勿采用O(n^2)的简单枚举算法来实现。 并思考如下问题： （1）怎样的数组含有最多的逆序对？最多的又是多少个呢？ （2）插入排序的运行时间和数组中逆序对的个数有关系吗？什么关系？ 输入格式 第一行：n，表示接下来要输入n个元素，n不超过10000。 第二行：n个元素序列。 输出格式 逆序对的个数。 输入样例 5 2 3 8 6 1 输出样例 5(Set a[0... N-1] is a n array containing n numbers. If there is a [i] > a [j] i n the case of I < j, then (i, j) is a n inversion pair of a array. For example, <2,3,8,6,1> has five reverse pairs. Please use the idea of "merge sorting algorithm" to achieve the statistics of inverse pairs with O (nlogn) efficiency. The number of inverse pairs of a sequence of n elements consists of three parts: (1) The number of reverse pairs in the left half, (2) the number of reverse pairs in the right half, (3) the number of elements in the left half is greater than that in the right half. The first two parts (1) and (2) are implemented by recursion. To ensure the final efficiency of the algorithm O (nlogn), how should the third part (3) be implemented? Do not use O (n ^ 2) simple enumeration algorithm to solve this problem.)

全面介绍有限元网格生成算法的英文教材，可以作为网格生成算法的编程资料(finite element mesh generation has not been taken as a formal subject of teaching in universities,as it encompasses several disciplines including classical geometry, computational geometry and topology, finite element method, data structures and algorithms, computer programming and, to a certain extent, even computer graphics.)