python 解决 啤酒2块钱一瓶，2个酒瓶可以换一瓶酒，4个瓶盖可以换一瓶酒 问10块钱可以喝几瓶酒

<pre name="code" class="python">
#!/usr/bin/env python
#coding:utf-8
#啤酒2块钱一瓶，2个酒瓶可以换一瓶酒，4个瓶盖可以换一瓶酒
#问10块钱可以喝几瓶酒
def jipingjiu(money):
	beer = money / 2  # 不考虑找零  money%2
	btt, cap = money/2, money/2   #买过酒得到的酒瓶及瓶盖
	flag = 1
	brr = 0  #如果可以赊酒的话
	while flag:
		a = btt/2   #酒瓶换酒
		b = cap/4   #瓶盖换酒
		tmp = a + b - brr
		if tmp > 0 or brr == 2:
			beer += a + b
			btt = a + b - brr + btt%2 
			cap = a + b - brr + cap%4  
			brr = 0
			flag = btt + cap
		#	flag = btt/2 + cap/4   #如果不能赊酒的话
		else:
			btt += 1 #赊一瓶酒
			cap += 1
			brr += 1  
		print "喝了{}瓶酒，赊了{}瓶酒，剩余{}酒瓶及{}瓶盖".format(beer, brr, btt,cap)
		
jipingjiu(10)

如果可以赊酒的话： 不可以赊酒的话：

SAS代码：

   

data work.test;
	beer = 5;
	btt = 5;
	cap = 5;
	flag = 1;
	brr = 0;
	do until (flag = 0);
		a = int(btt/2);
		b = int(cap/4);
		tmp = a + b - brr;
		if tmp > 0 or brr = 2 then do;
			beer = beer +  a + b;
			btt = ( a + b - brr + mod(btt, 2));
			cap = ( a + b - brr + mod(cap,4));
			brr = 0;
			flag = btt + cap;
		end;
		else do;
			brr + 1;
			btt + 1;
			cap + 1;
       end;
	   output;
	end;
run;