-
地震波有限差分正演
用于地球物理勘探的地震波有限差分正演,可以用于全波形反演的基础程序。
- 2022-01-25 18:26:31下载
- 积分:1
-
基于wcf服务和rest服务的示例代码
应用背景解决如何编写REST和WCF客户端程序,去获取服务端的各种资源的方式,具有普遍意义。关键技术采用C#实现的基于wcf服务和rest服务的示例代码,提供针对get,post,put,delete的资源获取、序列化和反序列化方法,并采用silverlight实现web服务和数据库的连接.
- 2022-03-14 17:49:08下载
- 积分:1
-
胡小玉的PEG算法
PEG构造创建了周长非常大的矩阵。这种构造已被证明可以产生最著名的Gallager码,对于较小的块长度(如500、1000或2000)尤其重要。
- 2022-05-01 01:11:58下载
- 积分:1
-
数值分析第八版的负担和菲尔斯(源程序)
Numerical Anaysis 8th Edition Burden and Faires (Fortran Source)
- 2022-03-25 18:49:00下载
- 积分:1
-
这几天学VC界面编程,在VC在线上狂看教程,觉得有所长进,于是把以前的Java代码用VC改写了一下。问题还是那个老掉牙的问题,八皇后问题,老归老,但我很喜欢...
这几天学VC界面编程,在VC在线上狂看教程,觉得有所长进,于是把以前的Java代码用VC改写了一下。问题还是那个老掉牙的问题,八皇后问题,老归老,但我很喜欢-VC these days learn programming interface, the VC frenzy see online tutorial, some feel that complacency. So, before the Java code with VC rewritten a bit. Problems or those long-standing problems, 8 Queen"s problems and old to the old, but I like
- 2022-01-26 15:53:22下载
- 积分:1
-
这opennl,图书馆很容易构造和求解稀疏线性系统。
This OpenNL, a library to easily construct and solve sparse linear systems.
* OpenNL is supplied with a set of built-in iterative solvers (Conjugate gradient,BICGSTAB, GMRes) and preconditioners (Jacobi, SSOR).
* OpenNL can also use other solvers (SuperLU 3.0 is supported as an OpenNL extension,MUMPS will be supported in a future version)-This is OpenNL. a library to easily construct and solve sparse l inear systems.* OpenNL is supplied with a set of built-in iterative solvers (Conjugate gradie nt, Application of BICGSTAB. GMRes) and preconditioners (Jacobi, SSOR).* OpenNL can also use other solvers (Super rLU 3.0 is supported as an OpenNL extension, MUMPS will be supported in a future version)
- 2023-07-05 19:45:04下载
- 积分:1
-
polynomial matrix
基于多项式的矩阵积分-polynomial matrix-based Integral
- 2022-05-06 20:54:25下载
- 积分:1
-
计算机网络滑动窗口协议C语言源代码(Linux版)
设计一个滑动窗口协议,在仿真环境下编程实现有噪音信道两站点间无差错双工通信
- 2022-08-06 07:18:04下载
- 积分:1
-
全自动3D扫描仪软件
资源描述全自动3D扫描仪软件,直接可以用的,基于Matlab和OpenCV,美国的布朗大学研究成果
- 2022-05-23 05:54:05下载
- 积分:1
-
一个求素数个数的程序 算pi(1000000000)不要1秒每个块如果10 0000, 需要 < 100000次全搜索加上8000 * 8000 <...
一个求素数个数的程序 算pi(1000000000)不要1秒每个块如果10 0000, 需要 < 100000次全搜索加上8000 * 8000 < 6400 0000 总的次数小于7000 0000,就是说,搜索次数大大降低-for a few number-counting procedures pi (1000000000) not a second of each block of 10 0000, the need lt; Search 100,000 plenary with 8,000* 8,000 lt; 6400 0000 total is less than the number 7000 0000, that is to say, greatly reducing the number of search
- 2022-03-22 17:01:12下载
- 积分:1