Integrald1.gz

C-means in 2 types. C=2

TGM.m \传统的Galerkin方法 MD.m \时滞惯性流形方法 brownian.m \演示布朗运动 randomwalk.m \ 演示随机游走 tumor.m \ 演示tumor演化(TGM.m traditional Galerkin method MD.m Delays inertial manifold method brownian.m demo Brownian motion randomwalk.m demo random walk tumor.m demo tumor evolution)

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

求解非线性方程组的NewRaphson方法(Solving nonlinear equations NewRaphson method)

王兆华关于apFFT原理的一篇文章，写的很简单明了，方便大家更好理解全相位FFT(WangZhaoHua about apFFT principle of an article, written in very simple and clear, facilitate everybody better understand all phase FFT )

说明：  Dsimplex.m 对偶单纯形法求解线性规划问题(Dual simplex method for solving linear programming problems)

实现卡尔曼滤波算法，可以进行参数估计、插值、滤波和预报(The Kalman filter algorithm, parameter estimation, interpolation, filtering and prediction)

预估校正法，关于解分数阶微分方程的算法程序，适合非线性方程(Solves an initial value problem for a non-linear differential equation of fractional order (FDE). The code implements the predictor-corrector PECE method of Adams-Bashforth-Moulton )

适用于ABAQUS的自定义材料子程序，适用于平面单元的随动强化J2模型。(Applies the custom material subroutine in ABAQUS, kinematic hardening J2 model applies to plane unit.)

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.)