http://hi.baidu.com/tianfan/blog/item/1c637f81c96990dcbd3e1e20.html
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我也想知道......
简单的来说,只要按拿悔照彩票的玩法规则写一个运算过程就OK了.都是随机数,没什么太大意义
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我刚写旅滑了个在PHP环境下获取双色球号码的程序。
不知道你会PHP不。可以看看
福利彩票选号的Matlab模拟程序%
axes('position',[0.1,0.8,0.8,0.1])
text(0,0,'模拟3D彩票','fontsize',18)
axis off
axes('position',[0.6,0.8,0.3,0.2],'Visible','off')
bb=text(0.1,0.5,'0','fontsize',24)q=0
axes('position',[0.3,0.23,0.6,0.1])
tt=text(0,0,'please press "space" to stop!',...
'fontsize',18)
axis off
axes('position',[0.1,0.4,0.2,0.3])
t1=text(0.3,0.5,'3','fontsize',60)
box on
set(gca,'xtick',[],'ytick',[])
set(gca,'xticklabel',[],'yticklabel',[])
axes('position',[0.4,0.4,0.2,0.3])
t2=text(0.3,0.5,'3','fontsize',60)
box on
set(gca,'xtick',[],'ytick',[])
set(gca,'xticklabel',[],'yticklabel',[])
axes('position',[0.7,0.4,0.2,0.3])
t3=text(0.3,0.5,'3','fontsize',60)
box on
set(gca,'xtick',[],'ytick',[])
set(gca,'xticklabel',[],'yticklabel',[])
set(gcf,'doublebuffer','on')
k=1DD=[]
fid = fopen('save_data.txt','wt')
while k
s=get(gcf,'currentkey')
if strcmp(s,'space')
clck=0
end
d=fix(rand(1,3)*10*(1-eps))
a=num2str(d(1))
b=num2str(d(2))
c=num2str(d(3))
set(t1,'string',a)
set(t2,'string',b)
set(t3,'string',c)
set(tt,'color',rand(1,3))
q=q+1set(bb,'string',num2str(q))
p=num2str(d)
fprintf(fid,'%c',p)
fprintf(fid,'%c\慎腊n',' ')
DD=[DDd]
pause(0.2)
end
fclose(fid)
figure(gcf)
figure
plot(1:size(DD,1),DD(:,1),'rs','MarkerFaceColor','r')
legend('first')
figure
plot(1:size(DD,1),DD(:,2),'gs','MarkerFaceColor','g')
legend('second')
figure
plot(1:size(DD,1),DD(:,3),'bs'禅圆,'MarkerFaceColor','b')
legend('贺孝塌third')
dos('save_data.txt')
delete save_data.txt
一等奖的概率为(1/10)×(1/10)×(1/10)×(1/10)×(1/10)×(1/春察10)×(1/5)=1/5000000
二等奖的概率为(1/10)×(1/10)×(1/10)×(1/10)×(1/10)×(1/10)
=1/1000000
三等奖的概率为(1/10)×(1/10)×(1/10)×(1/扒敬茄10)×(1/10)×(9/10)×6
=54/1000000=27/500000
四等奖的概率为(1/10)×(1/10)×(1/10)×(1/10)×(9/10)×(9/10)×15
=1215/100000=243/200000
五等奖的概率为(1/10)×(1/10)×(1/10)×(9/10)×(9/10)×(9/10)×20
=14580/1000000=729/50000
六等奖的概率为稿御(1/10)×(1/10)×(9/10)×(9/10)×(9/10)×(9/10)×15
=98415/1000000=19683/200000
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