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> The Problem With #, Algorithm problem
PureHeart
post Mar 19 2008, 02:29 AM
Post #1


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Given A,K, the definition
A#B = ( Sum of all digits of A * Maximum digit of B )+Minimum digit of B

and following formulas
A
A#A
(A#A)#A
(A#(A#A))#A

Determine and return the minimal number of necessary # to generate K from A or return -1 if there is no possible solution at all.

I am still looking for a solution to this. Any idea?
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pyost
post Mar 19 2008, 02:21 PM
Post #2


Nenad Bozidarevic
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QUOTE
A
A#A
(A#A)#A
(A#(A#A))#A


Each line must have a greater value than the previous one, so we just need to use the formula until the value becomes K, or exceeds it.

CODE
value = a
counter = 0

while (value < k)
   value = calculate(value, a)
   counter = counter + 1

if (value == k)
   return counter
else
   return -1


"calculate" whould be the function you use to calculate result of the # operator. This is quite easy to achieve, as you can take out the digits from a number by using "% 10 (mod 10)", which gives you the remainder when dividing by 10 (the last digit).
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