关于python之字典的嵌套,递归调用方法

>s1={'no surface'：{}}
>s1['no surfacce'][0]='no'
>s1
{'no surface':{0:'no'}}
>s2={'flipper':{}}
>s2['flipper'][0]='no'
>s2['flipper'][1]='yes'
>s1['no surface'][1]==s2
>s1
{'no surface':{0:'no',1:{'flipper':{0:'no,1:'yes''}}}}

def fact():
if n==1:
  return 1
return n*fact(n-1)

===> fact(5) 
===> 5 * fact(4) 
===> 5 * (4 * fact(3)) 
===> 5 * (4 * (3 * fact(2))) 
===> 5 * (4 * (3 * (2 * fact(1)))) 
===> 5 * (4 * (3 * (2 * 1))) 
===> 5 * (4 * (3 * 2)) 
===> 5 * (4 * 6) 
===> 5 * 24 
===> 120