8078349

说明：  特征值分解，HouseHolder分解等(matlab Linear Algebra Package)

由美国圣母大学Jena教授及其学生编写的关于计算半导体迁移率的matla脚本文件。(Jena professor from the University of Notre Dame and his students prepare for the calculation of a series of semiconductor mobility matlab script file.)

这是另一个改进的关于数字预失真功能仿真的matlab程序，效果不错，希望对大家有所帮助。(This is a improved matlab program about function estimation of digital estimation,and it has a good performance,really hope it can help you.)

全局牛顿-拉普森方法求解非线性方程组，自动搜索步长，进一步保证了程序的收敛(Global Newton-Raphson method for solving nonlinear equations. The steps can be found automatically. The convergence is better than the common Newton-raphson method.)

说明：  金属断裂的数值计算提供理论依据及实际详例解读(The numerical calculation of metal fracture provides theoretical basis and interpretation of practical examples)

随机加减法计算，可以用来检测加减法口算能力Random add and subtract calculation(Random add and subtract calculation)

求二元一次方程组的解,简便算法的过程及计算(Solving linear equations group，Process and simple calculation algorithm)

说明：  本书详细介绍了格子波尔兹曼法的原理，边界条件，具体模型及数值方法，非常适合学习与应用。(This book introduce the theory of lattice boltzmann methods alone with it's boundary conditions,models and numerical methods. It's really a book for us to learn and use.)

用matlab实现的改进的布谷鸟算法——SDCS，它是将最速下降法与CS相结合的(Matlab implementation of the improved algorithm of the cuckoo- SDCS , it will steepest descent method with CS )

说明：  Description 设a[0…n-1]是一个包含n个数的数组，若在ia[j]，则称(i, j)为a数组的一个逆序对（inversion）。 比如 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, 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.)