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GAUSS-JORDAN
gauss jordan method. not my own program but i just want to share it.
- 2012-09-30 09:32:51下载
- 积分：1
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MtlabCFD
基于matlab计算流体力学(CFD)源代码。(Matlab-based computational fluid dynamics (CFD) code.)
- 2007-04-28 11:00:45下载
- 积分：1
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pls2
代码为偏最小二乘法计算过程的代码，主要部分都有中文解释(Code for the calculation process of partial least squares method)
- 2020-06-24 00:40:02下载
- 积分：1
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3DForce
Fortran编写的用来计算集中力在三维半空间弹性体中下表面应力的程序。(Written in Fortran is used to calculate the stress concentration at the surface in three-dimensional space of half an elastomer program.)
- 2016-10-10 09:35:53下载
- 积分：1
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code
用C++编写的大整数计算的模板，可实现加、减、乘、除、取模等运算，可以通过此源码学习高精度的实现。(Written with C++ template large integer can be realized addition, subtraction, multiplication, division, modulus and other operations, you can achieve high-precision study of this source.)
- 2011-06-05 17:19:32下载
- 积分：1
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zernfun
波面的zernike多项式拟合，在光学测试领域应用相当广泛(Wave surface zernike polynomial fitting, in the field of application of a wide range of optical test)
- 2009-10-14 15:05:12下载
- 积分：1
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Newton
C++编写的电力系统潮流计算程序，采用牛顿法，并考虑到节电优化编号，5,14,57,300节点验证正确(Power flow calculation program, C++ written using Newton' s method, taking into account the number to the power-saving optimization, 5,14,57,300 node authentication)
- 2021-04-29 10:28:43下载
- 积分：1
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p2mesh
划分网格一个开源的库，有限元的算法，很好用(Meshing an open source library = for finite element)
- 2013-07-22 15:34:38下载
- 积分：1
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Umat_ElasticPlastic_Explicit
Abaqus UMAT code 适用于平面应变，轴对称弹塑性本构关系，也适用于大应变情况，使用显示积分算法。(Abaqus UMAT code applies to plane strain, axisymmetric elastoplastic constitutive relation, also applicable to large strain, using the integration algorithm.)
- 2013-03-30 03:46:43下载
- 积分：1
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11087 统计逆序对
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.)
- 2019-01-07 23:52:06下载
- 积分：1
