FDTD2D_UPML

计算矩阵内空间任意两点之间或者两个矩阵之间的欧式距离的代码。(The space between any two points is calculated matrix code or Euclidean distance between the two matrices.)

bellman-ford算法，典型最短路算法，用于计算一个节点到其他所有节点的最短路径(bellman-ford algorithm, a typical shortest path algorithm for computing a node to all other nodes of the shortest path)

Orthogonal Polynomials Approximation 数值分析，计算正交基多项式的系数 (Given a function f and a set of m >0 distinct points . You are supposed to write a function to approximate f by an orthogonal polynomial using the exact function values at the given m points with a weight assigned to each point . The total error must be no larger than a given tolerance. Format of function int OPA( double (*f)(double t), int m, double x[], double w[], double c[], double*eps ) where the function pointer double (*f)(double t) defines the function f int m is the number of points double x[] contains points double w[] contains the values of a weight function at the given points x[] double c[] contains the coefficients of the approximation polynomial double*eps is passed into the function as the tolerance for the error, and is supposed to be returned as the value of error. The function OPA is supposed to return the degree of the approximation polynomial. Note: a constant Max_n is defined so that if the total error is still not small enough when n = Ma)

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.)

牛顿迭代法，用于求解非线性方程，具体使用方法见程序内部说明。(Newton iteration for solving nonlinear equations, the specific use of the procedure see the internal note.)

矩阵的练习 对于掌握数组的 概念很重要。如何快速掌握数组，灵活运用。(The practice of the matrix is ​ ​ very important to grasp the concept of the array. How to quickly grasp the array of flexible use.)

Fortran算法程序集，徐士良，第二版，包括源码和pdf。(The Fortran algorithm assemblies, XU Shi-liang, second edition, including source code and pdf.)

一维热传导方程的求解，包括显式和隐式，及其对比(One-dimensional heat conduction equations, including explicit and implicit, and the contrast)

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)

PSCAD interface with C