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蒲丰氏问题
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%蒙特卡洛模拟计算π的值
%2018-11-29 20:28
%Win10+Matlab2018a
%@moli99
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A = 6; %平行线线距一半
L = 6; %针长
N = 100000000; %模拟次数
count = 0; %记录与平行线相交的针的数量
for i = 1:N
k1 = rand(); %产生均匀分布的随机数
k2 = rand();
x = A * k1;
s = pi * k2;
temp = L * sin(s);
if x <= temp
count = count + 1;
end
end
val = double(count)/N; %针与平行线相交的概率P的估计值P = 2L/(pi*A)
result = double(2*L)/(A*val)
