Яка команда в Скретч відповідає алгоритмічній структурі повторення?

Можно по русски?

Объяснение:

((3 + 1) * 3) * 3 = 36
((3) * 3 + 1 + 1 + 1) * 3 = 36
((3) * 3 + 1 + 1) * 3 + 1 + 1 + 1 = 36
((3) * 3 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) * 3 = 36
((3) * 3) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
ответ 16

def f0(number, log) #
  v = 1
  n = number + v
  log += " + #{v}"
  # log += "[" + n.to_s + "] "
  return [n, log]
end

def f1(number, log) #
  v = 3
  n = number * v
  log = "(#{log}) * #{v}"
  # log += "[" + n.to_s + "] "
  return [n, log]
end

def countWays(start_num, end_num, op_numbers, max_steps = 0)
  ways = {}
  ways.store(start_num.to_s, start_num)

  max_steps = max_steps == 0 ? (start_num - end_num).abs : max_steps
  count = 0

  for steps in 1..max_steps
      # puts "steps = #{steps}"
      new_ways = {}
      ways.each_pair{|log, num|

          for k in 0..op_numbers-1
              num1, log1 = f0(num, log) if k == 0
              num1, log1 = f1(num, log) if k == 1
              num1, log1 = f2(num, log) if k == 2

              if num1 == end_num
                  # and log.include?('[8]')
              then
                  log1 += " = " + end_num.to_s
                  count += 1
                  puts log1
              elsif num1.between?(start_num, end_num)
                  new_ways.store(log1, num1)
              else
                  # log1 = log1 + " = " + num1.to_s + " BAD "
                  # puts log1
              end
          end
      }
      # p [steps, ways.size, new_ways.size]
      ways = new_ways
  end
  return count
end

((3 + 1) * 3) * 3 = 36
((3) * 3 + 1 + 1 + 1) * 3 = 36
((3) * 3 + 1 + 1) * 3 + 1 + 1 + 1 = 36
((3) * 3 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) * 3 = 36
((3) * 3) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3 + 1) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
(3) * 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 36
ответ 16

def f0(number, log) #
  v = 1
  n = number + v
  log += " + #{v}"
  # log += "[" + n.to_s + "] "
  return [n, log]
end

def f1(number, log) #
  v = 3
  n = number * v
  log = "(#{log}) * #{v}"
  # log += "[" + n.to_s + "] "
  return [n, log]
end

def countWays(start_num, end_num, op_numbers, max_steps = 0)
  ways = {}
  ways.store(start_num.to_s, start_num)

  max_steps = max_steps == 0 ? (start_num - end_num).abs : max_steps
  count = 0

  for steps in 1..max_steps
      # puts "steps = #{steps}"
      new_ways = {}
      ways.each_pair{|log, num|

          for k in 0..op_numbers-1
              num1, log1 = f0(num, log) if k == 0
              num1, log1 = f1(num, log) if k == 1
              num1, log1 = f2(num, log) if k == 2

              if num1 == end_num
                  # and log.include?('[8]')
              then
                  log1 += " = " + end_num.to_s
                  count += 1
                  puts log1
              elsif num1.between?(start_num, end_num)
                  new_ways.store(log1, num1)
              else
                  # log1 = log1 + " = " + num1.to_s + " BAD "
                  # puts log1
              end
          end
      }
      # p [steps, ways.size, new_ways.size]
      ways = new_ways
  end
  return count
end