登录
首页 » Visual C++ » lihong-wang-mcml

lihong-wang-mcml

于 2020-10-27 发布 文件大小:3019KB
0 343
下载积分: 1 下载次数: 57

代码说明:

  Texas A&M大学Lihong Wang教授编写的经典,用蒙特卡罗模型模拟光在组织、复杂介质、溶液中的传播。附带详细说明文件。(Classics Professor Lihong Wang Texas A & M University, prepared using Monte Carlo simulation model of light in the organization, the complexity of the medium, the solution spread. With detailed documentation.)

文件列表:

李刚lihong wang mcml
....................\1995LWCMPBMcml.pdf,979325,2010-10-08
....................\1997LWCMPBConv.pdf,582300,2010-10-08
....................\index.htm,12987,2010-10-08
....................\Mcman.pdf,771400,2010-10-08
....................\Mcman.txt,51468,2010-10-08
....................\Mcman。txt.docx,38724,2014-03-18
....................\mcml.exe,254025,2010-10-08
....................\mcmlcitations.htm,315834,2010-10-08
....................\mcR5pc
....................\......\Conv.exe,278601,2000-03-01
....................\......\CONV.H,8044,2000-03-01
....................\......\CONVCONV.C,39600,2000-03-01
....................\......\CONVI.C,14941,2000-03-01
....................\......\CONVISO.C,11671,2000-03-01
....................\......\Convmain.c,4230,2000-03-01
....................\......\CONVNR.C,5230,2000-03-01
....................\......\CONVO.C,32049,2000-03-01
....................\......\Debug
....................\......\.....\MCMLMAIN.obj,10077,2014-03-17
....................\......\.....\MCMLMAIN.pch,220280,2014-03-17
....................\......\.....\MCMLMAIN.pdb,25600,2014-03-17
....................\......\.....\vc60.idb,33792,2014-03-17
....................\......\.....\vc60.pdb,45056,2014-03-17
....................\......\Mcml.exe,254025,2000-03-01
....................\......\MCML.H,7129,2000-02-29
....................\......\MCMLGO.C,21002,2000-02-29
....................\......\MCMLIO.C,29259,2000-02-29
....................\......\MCMLMAIN.C,5699,2000-02-29
....................\......\MCMLNR.C,2355,2000-02-29
....................\......\SAMPLE.MCO,31684,1999-09-23
....................\......\temp1.mco,21817,2010-10-11
....................\......\temp2.mco,155752,2010-10-11
....................\......\TEMPLATE.MCI,965,1992-06-23
....................\......\复件 TEMPLATE.MCI,965,1992-06-23
....................\mcR5pc.zip,222375,2010-10-08
....................\mcR5unix
....................\........\convcode
....................\........\........\conv.h,7803,1994-09-27
....................\........\........\convconv.c,38077,1994-09-27
....................\........\........\convi.c,14341,1994-09-27
....................\........\........\conviso.c,11151,1994-09-27
....................\........\........\convmain.c,4076,1994-09-27
....................\........\........\convnr.c,5008,1994-09-27
....................\........\........\convo.c,29943,1994-09-27
....................\........\........\Makefile,317,1998-05-20
....................\........\........\sample.mco,31055,1994-09-27
....................\........\mcmlcode
....................\........\........\Makefile,272,1998-05-20
....................\........\........\mcml.h,6923,2000-03-01
....................\........\........\mcmlgo.c,20253,2000-03-01
....................\........\........\mcmlio.c,28118,2000-03-01
....................\........\........\mcmlmain.c,5493,2000-03-01
....................\........\........\mcmlnr.c,2264,2000-03-01
....................\........\Sample
....................\........\......\conv,122880,1994-09-27
....................\........\......\conv.bat,1248,1992-10-20
....................\........\......\example.mci,634,1993-05-13
....................\........\......\example.mco,31054,1993-05-13
....................\........\......\mcml,90112,1993-04-03
....................\........\......\mcmlconv.man,51468,1992-10-20
....................\........\......\htm" target=_blank>p1,1634,1992-10-20
....................\........\......\sample.mco,31055,1992-10-20
....................\........\......\template.mci,930,1992-10-20
....................\mcR5unix.tar.gz,162462,2010-10-08
....................\ReadmeR5.txt,1937,2010-10-08
....................\src-mcml
....................\........\src-mcml
....................\........\........\mcml.h,6923,1993-04-05
....................\........\........\mcmlgo.c,19967,1993-04-05
....................\........\........\mcmlio.c,27500,1993-04-05
....................\........\........\mcmlmain.c,5492,1993-04-05
....................\........\........\mcmlnr.c,2264,1993-04-05
....................\src-mcml.tar.gz,16116,2010-10-08
....................\TEMPLATE.MCI,965,1992-06-23
....................\基于C语言的多层组织中光传输的蒙特卡罗模型.doc,22016,2010-10-11
....................\笔记.txt,726,2014-03-18

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

  • sonar
    一种时频域联合捕捉主动声纳信号的方法.介绍了一种可应用于被动声纳的时频域联合捕捉主动声纳信号的方法及其工作流程。仿真相同噪声环境下使用该方法的发现距离与主动声纳探测距离,仿真结果表明使用该方法的发现距离远超过主动声纳探测距离,使用该方法可实现对目标的远程发现。(A time-frequency domain joint capture of active sonar signals. Introduced passive sonar, which can be applied when the frequency domain joint to capture active sonar signal its workflow. The discovery of the method used in the simulation of the same noise environment distance active sonar detection range, simulation results show that using the method found distance than active sonar detection range, the use of this method can be found remote target.)
    2020-11-12 15:49:44下载
    积分:1
  • xinxilun
    排队论模型仿真程序。用MATLAB实现 。欢迎大家一起讨论(Queuing theory model simulation program. Use MATLAB to achieve. Welcome to discuss with)
    2010-01-26 18:06:55下载
    积分:1
  • acou
    交错网格声波方程正演模拟,交错网格有限差分地震波场计算,二维的声波正演,利用空间10阶,时间2阶。(The CalCulation of the SeismiC Wave一 fieldwith Staggered-grid Finite difference Sheme)
    2014-01-09 08:44:16下载
    积分:1
  • hexLiuhaiou
    以二进制显示数据,并将显示内容用文本方式保存以便分析(display data in binary and text content used to analyze ways to preserve)
    2006-12-13 16:43:28下载
    积分:1
  • satelite
    通过卫星的测量方程和状态方程来求解卫星的姿态。包括四元数,角度,角速度等等。(By satellite measurements to solve equations and state equations of satellite attitude. Including the quaternion, angle, angular velocity and so on.)
    2009-11-09 10:13:00下载
    积分:1
  • UART_GUI
    RS232通訊用GUI,可偵測PC上的所有Port供使用者選擇,並提供9600、115200兩個baud rate供使用者選擇(如需其他baud rate可自行於code內添加),提供RX、TX接收傳送功能,下方接收欄可接收第一個0x0A指令之前所有資料,右方接收欄則接收剩餘所有資料。(RS232 communication with the GUI, can be detected on all Port PC for users to choose and provide 9600,115200 two user selectable baud rate (for other baud rate can add their own in the code), to provide RX, TX reception transmission function, the receiver can receive all the information bar below before the first 0x0A command bar on the right to receive all the information you receive surplus.)
    2014-12-22 17:51:39下载
    积分:1
  • CC++bishi
    理论知识是用来指导具体实践的。本文在深刻理解通信系统理论的基础上利用 MATLAB 强大的仿真功能,设计了许多具体的通信系统仿真模型。在仿真模型设计过程中,本文对模型设计的目的、具体的结构组成、仿真流程以及仿真结果都给出了具体详实的分析和说明。(Theoretical knowledge is used to guide the specific practice. In this paper, a deep understanding of the theory of communication systems using MATLAB based on the powerful simulation function, the design of a number of specific communication system simulation model. In the simulation model of the design process, this article is designed on the model, the specific structural components, simulation and flow simulation results are given specific and detailed analysis and description.)
    2008-06-02 15:06:57下载
    积分:1
  • owners
    Script generates list of owners for files in choose directory
    2014-10-27 22:17:41下载
    积分:1
  • samsung_monitor_jig_3.0b_534
    SAMSUNG LCD FIRMWARE
    2012-09-12 07:47:35下载
    积分:1
  • 5
    一道程序编译顺序的考题,涉及到函数调用的先后顺序及运算符号的优先级等问题。下面我展开给你讲。 C的程序编译总是从main函数开始的,这道题的重点在“fun((int)fun(a+c,b),a-c)) ”语句。 系统首先要确定最外层 fun()函数的实参,第一个参数的确定需要递归调用fun()函数(不妨称其为内层函数)。内层函数的两个参数分别为x=a+b=2+8=10、y=b=5,执行函数体x+y=10+5=15,于是得外层函数的参数x=15。其另一个参数y=a-c=2-b=-6,再次执行函数体,得最终返回值x+y=15+(-6)=9。 (Compiling together the sequence of test procedures, involving the sequence of function calls and operator symbols, such as the priority problem. Now I give you to start speaking. Procedures for C compiler always start from the main function and at这道题the focus of " fun ((int) fun (a+ c, b), ac)) " statement. System must first determine the most outer layer of fun () function of real parameters, the first parameters of recursive calls required fun () function (may be called the inner function). Inner function separately for the two parameters x = a+ b = 2+8 = 10, y = b = 5, to execute the function body x+ y = 10+5 = 15, then the outer function parameters were x = 15 . Its another parameter y = ac = 2-b =- 6, once again to execute the function body may eventually return the value of x+ y = 15+ (-6) = 9.)
    2009-03-15 15:36:23下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载