simple_beam

fitness,适应度函数，计算适应度函数，(fitness, fitness function, calculate the fitness function,)

几个使用的解线性方程组的程序，包括高斯消去法，高斯列主元法，高斯——赛德尔消去法。简明实用。(Gauss elimaination, Gauss_seidel ellimination)

分别采用Roe格式和AUSM格式求解一维黎曼问题，并将计算结果与解析解进行比较，发现这两种计算格式都能获得高的计算精度。(Roe scheme were used in this procedure and AUSM format to solve a 1-D RIEMANN problem, and numerical results were compared with the analytical solution, the results can be seen from these two formats have a solution with high accuracy.)

对于空间上的任意几个点所在的直线或曲线，该程序可以拟合出的方程。(For any of several points on the space where the line or curve, the program can fit out of the equation.)

2015年全国大学生数学建模比赛创意折叠桌，利用Mathematica实现(2015 National Undergraduate Mathematical Modeling Competition creative folding table, using Mathematica to achieve)

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：  一维波动方程的间断有限元计算，包括网格的划分，数值积分的计算等。(Discontinuous finite element method for one-dimensional wave equation, including mesh division, numerical integration calculation and so on.)

说明：  c语言编写的高效用LU分解法求任意维数的矩阵的逆，它是经典书籍《算法导论》（第二版）中矩阵求逆的c语言实现。 (The algorithm of matrix inverse is based on LU decomposition which is proposed by the book of introduction to algorithm (second edition).)

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.)

在visual c++6.0的环境下，使用C++语言编写的最小二乘法程序。(In visual c++6.0 environment, using C++ language least squares procedure.)