Solution for Problem A (Osmos), of Round1B, CodeJam 2013: https://code.google.com/codejam/contest/2434486/dashboard#s=p0

osmos.rb

# Solution for Problem A (Osmos), of Round1B, CodeJam 2013: 
# https://code.google.com/codejam/contest/2434486/dashboard#s=p0

def solve(a, mots)

  # If the list of mots is empty, we are done.
  return 0 if mots == []

  # If a is 1, the only way to make it solvable
  # is to delete all the mots, because it cannot
  # absorb any mot and cannot inreasi its length.
  return mots.length if a == 1

  # If a > mots[0], we can absorb it
  # without any additions or deletions.
  if a > mots[0]
    return solve(a + mots[0], mots[1..-1])
  end

  ### At this point we have a <= mots[0] .
  ### We can either add mots (in front of mots[0])
  ### until we can absorb it, or delete all the
  ### remaining mots. Deletting just the first mot
  ### will not improve the situation, since the next
  ### one is not smaller than it. The same goes for
  ### deletting any other mots.
  
  # Let's find the number of mots we have to insert
  # in order to absorb mots[0].
  nr = 0   # nr of mots we have to insert  
  while a <= mots[0]
    nr += 1
    a += a - 1
  end

  # If number of motes we have to insert is greater
  # than the remaining mots, we better just delete
  # the remaining mots, and then we are done.
  return mots.length if nr >= mots.length
    
  # Find the number of changes if we absorb the first mot and continue
  nr1 = solve(a + mots[0], mots[1..-1])

  # Return the smallest number of deleting all mots
  # or absorbing the first one and continuing.
  return [nr + nr1, mots.length].min
end

T = gets.to_i
for t in 1..T
  a, n = gets.chomp.split.map &:to_i
  mots = gets.chomp.split.map &:to_i
  mots.sort!
  puts "Case ##{t}: #{solve a, mots}"
end