因为输入的参数都是要从excel表格里导入的向量，而且其中两部分的参数维数还不同，再加上又需要蒙特卡洛方法（即生成10000个路径），所以我只能编两个for循环：
下面是我写的code:
function [price] = mcmc(S,K,B,T,sigma,v,theta,r,d,M,N,seed)
%M is the number of time steps,
%N is the number of paths,
M = 100;
N = 10000;


MCpayoff = zeros(N,M+1);
S=zeros(N,M+1);
rand('seed',0);


%S is the current price of the underlying asset,
%K is the exercise price of the option,
%B is the barrier of the option,
%r is the risk-free rate,
%d is the dividend,
%T is the time to maturity,
%sigma is the volatility of the asset price,
NUM = xlsread('EuropeanUpAndOutCall.xls');
MCprice = zeros(26,5);


for i = 1:26
B = NUM(i,1);
K = NUM(i,2);
%这两个参数的数据是26行的向量，
for ii = 1:5
T = NUM(ii,3);
sigma = NUM(ii,4);
v = NUM(ii,5);
theta = NUM(ii,6);
r = NUM(ii,7);
d = NUM(ii,8);
S0 = NUM(ii,9);
%这几个参数却是5行的向量，
%choose a time discretization;
Deltat=T/M;


%simulate a random number of gamma processes,independent of all other yet
%simulated random number,
G = gamrnd(Deltat/v,v,N,M);
%simulate a standard normally distributed random number Z_i,
Z = normrnd(0,1,N,M);
%the return is
deltaX = theta*G+sigma*sqrt(G).*Z;
X0 = zeros(N,M);
X = X0+cumsum(deltaX,2);
%the stock prices
S(:,1) = S0*ones(N,1);
t = (Deltat:Deltat:T);
t = repmat(t,N,1);
w = (1/v)*log(1-theta*v-sigma^2*v/2);
S(:,2:M+1) = S0.*exp(r*t+X+w*t);
%the payoff for a Down and Out Put Option
ST = S(:,M+1);
Max = max(S,[],2);
for j=1:length(Max)
ifMax(j,1) < B && K < S(j,M+1)%因为此处需要10000个元素的列向量的每个元素与B，K（26个元素的列向量）的每个元素比较，又设了个for循环和if循环
MCpayoff(j,1) = S(j,M+1)-K;
else
MCpayoff(j,1) = 0;
end;
end;
%the option price is
MCprice = exp(-r*T)*(1/N)*sum(MCpayoff(:,1));%此处是得出10000个路径的平均值，之后要针对有5个元素的列向量循环一次，然后是26个元素的列向量再循环一次。。
end
end
我的问题是：类似这种代入参数维数不同的情况，有没有什么简化的方法？如果没有，循环的顺序是否先内再外呢？好复杂啊，我这个新手撑不住了。。光end都end了好几个，已经分不清了。。