C: 1, 3, 6
C++: 1, 3, 4, 6
Cobol:
Fortran:
Java: 1, 3, 4, 6
Lisp:
ML:
Perl:
Python: 1, 3, 4, 5, 6, 8
VB:
1)
1.因为a是S,由xxx可知aa+是SS+
2.因为aa+是SS+,由xxx可知aa+是S
3.因为aa+是S,且a是S,由xxx可知aa+a*是SS*
4.因为aa+a*是SS*,由xxx可知aa+a*是S
3)
用归纳法可得出,L1 = { a(a+)n(a*)m | n,m=>1 }
S=>0S1=>00S11=>0...01...1(n个0,n个1,n=>1)
S=>+SS=>+S(+SS)
S=>
S=>aSbS=>abSaSbS