The Problem With #

PureHeart

Mar 19 2008, 02:29 AM

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?

pyost

Mar 19 2008, 02:21 PM

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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The Problem With # - Algorithm problem

The Problem With # - Algorithm problem