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Posted: Tue Dec 27, 2005 11:37 am
by mamun
Thanks. I understand it now. I'll give it a try.


Posted: Wed Feb 15, 2006 10:58 am
by Psyco
I think this problem's tip is..

len=string length

Mutant : len is 1, and str[0]='B'

Simple : if len is 1, and str[0]='A'

Fully-Grown : str[len-1]='B', and str[len-2]='A'

Mutagenic : str[0]='B', and str[len-1]='A'

Else : Mutant

In case of Mutant, it has 2 cases to print.

I got AC thie problem in C, line 18.

Posted: Sat Jul 08, 2006 7:26 pm
by jan_holmes
How could "BAAB" become "MUTANT" ??? Shouldn't it be "FULLY-GROWN" ??? because in the problem statement :

fully-grown stage O = OAB

and in this case, "BAAB", BA = O, and AB,so BAAB should be "FULLY-GROWN"


Posted: Fri Jan 05, 2007 8:36 am
by Raiyan Kamal
"BAAB", BA = O, and AB,so BAAB should be "FULLY-GROWN"
How do you know BA = o ?

BA does not match with any of the given patterns.

Code: Select all

A <- simple
OAB <- fully-grown
BOA <- mutagenic
So it must be treated as 'mutant'. And since the input string has a substring who is mutant, the entire string is actually mutant.