620 - Cellular Structure

All about problems in Volume 6. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

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Post by mamun » Tue Dec 27, 2005 11:37 am

Thanks. I understand it now. I'll give it a try.

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Post by Psyco » Wed Feb 15, 2006 10:58 am

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.

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Post by jan_holmes » Sat Jul 08, 2006 7:26 pm

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"


Raiyan Kamal
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Post by Raiyan Kamal » Fri Jan 05, 2007 8:36 am

"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.

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