问题补充：

能提供一个例子看看吗？我就不懂MATLAB，想知道具体代码。因为现在写论文急，也没时间仔细看书了

蒙特卡罗模拟

就是随机数相关的东西，你只要知道随机数是怎么得到。其它的事就要好办了。

rand(m,n)产生m*n均匀随机数。

ex:

用概率方法求pi

N=100000;

x=rand(N,1);

y=rand(N,1);

count=0;

for i=1:N

if (x(i)^2+y(i)^2<=1)

count=count+1;

end

end

PI=4*count/N

-------------------------------

有关使用matlab进行蒙特卡罗模拟的程序问题

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2010-10-1 15:37

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浏览次数：1425次

已经分布是均匀分布(连续)，区间为(12,62)，请问各位大侠，如何用matlab编程实现此蒙特卡罗模拟，我想模拟2000次，得到概率密度图与累积概率密度图，程序应该如何编，麻烦大家指教，期待帮助，谢谢。小妹现在没有分数，还是希望大家能帮我，谢谢1

问题补充：

n=2000; %随机点数(可增加点数)

x=12+(62-12)*rand(1,n); %产生2000个12到62的随机数

xx=12:2:62; %画概率密度图的区间

nx=histc(x,xx); %计算x在xx每个小区间内的点数。

px=nx/n;

sumpx=cumsum(px);

subplot(1,2,1)

bar(xx(1:end-1),px(1:end-1));

title('概率密度')

subplot(1,2,2)

plot(xx(1:end-1),sumpx(1:end-1));

title('累积概率密度')

-----------------------------------

我这儿正好有份程序，希望有所帮助。

代码：

一份蒙特卡洛程序

count=input('input the count:');%输入模拟粒子数

sigmaedata=[28370,13845,6908,2555,1223,2602,1925,905.3,479.7,164.5,74.24,23.86,66.60,36.62,22.29,9.978,5.298,1.668,0.7378,0.2361,0.1099,0.06211,0.03939,0.02030];

sigmacdata=[0.0220,0.0393,0.0568,0.0904,0.1209,0.1479,0.1722,0.2136,0.2480,0.3092,0.3486,0.3932,0.4153,0.4268,0.4319,0.4291,0.4215,0.3969,0.3691,0.3269,0.2944,0.2709,0.2512,0.2209];

Edata=[1,1.5,2,3,4,5,6,8,10,15,20,30,40,50,60,80,100,150,200,300,400,500,600,800];%截面数据

channel=zeros(1,ceil(662/5)+10);%多道数组

nget=0;%探测到的总计数

ntotal=0;%进入探测器的总计数

for ii=1:count%count个粒子循环

collidetime=0;%当前粒子碰撞次数

%粒子状态初始化

E0=622;

E=E0;

z=-2;

r=0;

theta=2*pi*rand(1);% 源抽样，z,r,theta坐标

miu=2*rand(1)-1;

fai=2*pi*rand(1);%方向角抽样

if miu

continue;

else

z=0;

r=2*sqrt(1-miu^2)/miu;

theta=fai;

end

while E>1%一个粒子在闪烁体中的输运过程

sigmae=interp1(Edata,sigmaedata,E,'linear');

sigmac=interp1(Edata,sigmacdata,E,'linear');

sigmat=sigmae+sigmac;%线性插值得到截面数据

L=-log(rand(1))/sigmat;%下次碰撞的距离

%计算下次碰撞位置坐标

rnew=sqrt(r^2+L^2*(1-miu^2)+2*r*L*sqrt(1-miu^2)*cos(fai-theta));

z=z+L*miu;

cdth=(rnew^2+r^2-L^2*(1-miu^2))/2/r/rnew;

sdth=L*sqrt(1-miu^2)*sin(fai-theta)/rnew;

dtheta=asin(sdth);

if cdth<0

dtheta=pi-dtheta;

end

theta=theta+dtheta;

r=rnew;

if(r>2)|(z>=4)|(z<0)%判断是否在闪烁体内

break;

else

collidetime=collidetime+1;

end

if rand(1)

E=0;

else%康普顿散射

alpha=E/511;

flag=0;

while flag==0

if rand(1)<=27/(4*alpha+29)

x=(1+2*alpha)/(1+2*alpha*rand(1));

if rand(1)<=0.5*((alpha+1-x/alpha)^2+1)

flag=1;

end

else

x=1+2*alpha*rand(1);

if rand(1)<=27/4*((x-1)^2)/x^3

flag=1;

end end

end E=E/x;

alphat=alpha/x;

miuL=1-1/alphat+1/alpha;%散射后的方向

a=miuL;

b=sqrt(1-a^2);

randangle=2*pi*rand(1);

miunew=a*miu+b*sqrt(1-miu^2)*cos(randangle);

sdf=b*sin(randangle)/sqrt(1-miunew^2);

cdf=(a-miu*miunew)/sqrt(1-miu^2)/sqrt(1-miunew^2);

sfn=sdf*cos(fai)+cdf*sin(fai);

cfn=cdf*cos(fai)-sdf*sin(fai);

fainew=asin(sfn);

if(cfn<0)