登录
首页 » Visual C++ » guass_call_c

guass_call_c

于 2011-04-13 发布 文件大小:241KB
0 281
下载积分: 1 下载次数: 0

代码说明:

说明:  光斑聚集的C++程序调用,以前MATLAB计算要15分钟,现在只要3分钟就可以算完。运算超快。(Gathering spot called C++ program, MATLAB calculations to 15 minutes, now only 3 minutes. Ultrafast operation.)

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • Win32GUI
    windows gui wrapper for fb
    2013-12-12 15:44:33下载
    积分:1
  • faxie
    用matlab编写得一个发泄工具,通过这个可以看出你得英语水平(Using matlab to prepare for a venting tool, through which we can see that you have the standard of English)
    2008-05-16 09:42:58下载
    积分:1
  • MultiWiiV1_7
    基于 Matlab 的积分分离 PID 控制 算法及仿真 摘 要 MATLAB 是一种集数值计算、符号运算、可视化(flying car)
    2013-10-11 13:55:27下载
    积分:1
  • blobtrack
    opencv自带的跟踪框架程序,方便开发者开发出能够自动检测前景、目标跟踪的算法程序(Opencv own tracking framework application, convenient and developers to develop to be able to automatically detect prospects, the algorithm of target tracking application )
    2014-12-24 15:43:27下载
    积分:1
  • PAntColonySysa
    粒子群算法的matlab程序在有障碍的路径图中寻寻找最短路径的程序。 (Particle swarm algorithm matlab program in obstacles in the road map to find to find the shortest path to the program.)
    2012-08-20 07:30:14下载
    积分:1
  • across
    遗传算法解决作业车间调度问题,这一M文件为遗传算子的交叉操作(Genetic algorithms to solve the job shop scheduling problem, the M-file for the genetic operators crossover operation)
    2015-04-15 15:14:06下载
    积分:1
  • Allan_fenxi
    在陀螺仪测试中,实现ALLAN方差法分析动态数据,得到五个主要参数(Gyroscope test to achieve the the ALLAN variance analysis of dynamic data, five main parameters)
    2020-08-23 16:28:16下载
    积分:1
  • lagrange
    用拉格朗日n次插值方法 对一系列离散点进行插值(N times with Lagrange interpolation method interpolates a series of discrete points)
    2015-01-25 20:02:44下载
    积分:1
  • Software
    软件调试方面的书籍,内容不错,而且只有100多页,便于学习,感觉还不错,与大家共同分享。(Software debugging of books, the content is true, and only 100 pages, easy to learn, feel pretty good, and everyone shared.)
    2010-02-01 12:09:22下载
    积分:1
  • compare
    int compare(const char* a, const char* b, int size) { for (int i = 0 i < size i ++) { if (a[i] != b[i]) { return i } } return -1 } (int compare(const char* a, const char* b, int size) { for (int i = 0 i < size i++) { if (a[i] != b[i]) { return i } } return-1 } )
    2014-09-02 13:53:50下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载