2

我正在研究 Ruby Koans，我在弄清楚我编写的方法出了什么问题时遇到了一些麻烦。我在 about_scoring_project.rb 中，我已经为骰子游戏编写了 score 方法：

def score(dice)
  return 0 if dice == []
  sum = 0
  rolls = dice.inject(Hash.new(0)) { |result, element| result[element] += 1; result; }
  rolls.each { |key, value| 
    # special condition for rolls of 1
    if key == 1  
      sum += 1000 | value -= 3 if value >= 3
      sum += 100*value
      next
    end
    sum += 100*key | value -= 3 if value >= 3
    sum += 50*value if key == 5 && value > 0
  }
  return sum
end

对于不熟悉这项运动的人：

贪婪是一个骰子游戏，您最多可以滚动五个骰子来累积积分。下面的“分数”函数将用于计算单次掷骰子的分数。

贪婪滚动评分如下：

  • 一组三个是1000分

  • 一组三个数字（除了一个）的价值是该数字的 100 倍。（例如三个五是500分）。

  • 一个（不是一组三个的一部分）值 100 分。

  • 一个五（不是一组三的一部分）值 50 分。

  • 其他的都值0分。

例子：

score([1,1,1,5,1]) => 1150 分 score([2,3,4,6,2]) => 0 分 score([3,4,5,3,3]) => 350 分得分([1,5,1,2,4]) => 250 分

下面的测试中给出了更多评分示例：

你的目标是编写 score 方法。

当我尝试运行文件中的最后一个测试时遇到了麻烦：assert_equal 550, score([5,5,5,5])

出于某种原因，我返回 551 而不是 550。感谢您的帮助！

4

14 回答 14

4

这是我的方法：

def score(dice)
  # Count how many what
  clusters = dice.reduce(Hash.new(0)) {|hash, num| hash[num] += 1; hash }

  # Since 1's are special, handle them first
  ones = clusters.delete(1) || 0
  score = ones % 3 * 100 + ones / 3 * 1000

  # Then singular 5's
  score += clusters[5] % 3 * 50

  # Then the triples other than triple-one
  clusters.reduce(score) {|s, (num, count)| s + count / 3 * num * 100 }
end
于 2015-08-19T11:18:55.607 回答
3

我的方法使用两个查找表 - 一个包含三倍的分数，另一个包含单打的分数。我使用表格计算出每个数字的分数，并使用以下方法累积总分inject：

def score(dice)
  triple_scores = [1000, 200, 300, 400, 500, 600]
  single_scores = [100, 0, 0, 0, 50, 0]
  (1..6).inject(0) do |score, number|
    count = dice.count(number)
    score += triple_scores[number - 1] * (count / 3)
    score += single_scores[number - 1] * (count % 3)
  end
end
于 2017-08-24T16:36:45.573 回答
2

这是因为您实际上是将|运算符（按位或）的结果添加到总分中：

sum += 100*key | value -= 3 if value >= 3 # This is 501 in your case

证明：

irb(main):004:0> value = 4
=> 4
irb(main):005:0> 100 * 5 | value -= 3 # This should be read as (500) | 1 which is 501
=> 501

所以像这样重写它：

if value >= 3
  sum += 100 * key
  value -= 3
end
于 2012-07-22T20:21:47.583 回答
2

我和

def score(dice)
  dice.uniq.map do |die|
    count = dice.count die
    if count > 2
      count -= 3
      die == 1 ? 1000 : 100 * die
    else 0
    end + case die
          when 1 then count * 100
          when 5 then count * 50
          else 0
          end
  end.inject(:+) || 0
end
于 2015-12-23T19:16:08.643 回答
1

我的方法是：

def score(dice)
  calculator = ->(no, group_multipler, individual_multipler) { (no / 3 * group_multipler) + (no % 3 * individual_multipler) }
  dice.group_by {|i| i % 7 }.inject(0) do |total, (value, scores)|
    group_multipler, individual_multipler = case value
    when 1
      [1000, 100]
    when 5
      [500, 50]
    else 
      [value * 100, 0]
    end
    total += calculator.call(scores.size, group_multipler, individual_multipler)
  end
end
于 2014-01-30T14:39:14.253 回答
1

我的做法：

     def score(dice)
        score = 0
        score += dice.count(1) >= 3? (1000+ (dice.count(1) -3)*100): dice.count(1) * 100
        score += dice.count(5) >= 3 ? (500 + (dice.count(5) -3)*50): dice.count(5) * 50
       [2,3,4,6].each {|x| dice.count(x) >=3? score+= x*100:0}
       return score
    end
于 2015-11-17T05:46:00.457 回答
1

这是我的答案：

def score(dice)
  frequency = dice.inject(Hash.new(0)) do |h, el|
    h[el] += 1
    h
  end

  score_triples = { 1 => 1000 }
  score_singles = { 1 => 100, 5 => 50 }

  score = 0
  frequency.each do |k, v|
    score += v / 3 * score_triples.fetch(k, 100 * k)
    score += v % 3 * score_singles.fetch(k, 0)
  end
  score
end
于 2016-12-12T13:58:08.017 回答
1

我的方法使用整数除法和模除法：

def score(dice)
    points = 1000 * (dice.count(1) / 3)
    points += 100 * (dice.count(1) % 3)
    points += 50 * (dice.count(5) % 3)
    (2..6).each do |i|
        points += (100 * i) * (dice.count(i) / 3)
    end
    points
end
于 2017-04-04T19:00:12.347 回答
0

这是我的解决方案。

def score(dice)
    score = 0
    # grab all the numbers and their amounts
    number_amounts = dice.reduce(Hash.new(0)) { |hash, numb| hash[numb] += 1; hash }

    # iterate through each pair
    number_amounts.each do |key, value|
      # case with number 1
      score += (value % 3) * 100 + value / 3 * 1000 if (key == 1)
      # case with number 5
      score += (value % 3) * 50 + value / 3 * key * 100 if (key == 5)
      # all numbers except 1 and 5
      score += (value / 3) * key * 100 if (key != 1 && key != 5)
    end
    score
end
于 2016-05-11T18:27:22.810 回答
0 
    def score(dice)
  # Set up rules Hash
  rules = { 1 => {:triples => 1000, :singles => 100}, 5 => {:triples => 100, :singles => 50} }
  [2,3,4,6].each {|i| rules[i] = {:triples => 100, :singles => 0} }
  # Count all ocourencies
  counts = dice.each_with_object(Hash.new(0)) {|e, h| h[e] += 1}

  #calculate total
  total = 0
  counts.each_pair{ | key, value | 
    total += value >= 3? (rules[key][:triples]*key + (value -3)*rules[key][:singles]): value * rules[key][:singles]
  }
  return total
end
于 2016-07-09T10:07:23.253 回答
0

这是我自己编写的第一段代码（当然，在 stackoverflow 的大量帮助下。）在看了所有其他答案之后，我意识到这特别过分，因为它适用于 9 数字骰子（存在吗？ )

  def score(dice)
    
  if dice.empty?
    return 0
  end

  var_score = 0

  conteo = (0..9).to_a.each.map { |x| dice.count(x)} 
  
  #Evaluating 1
  if ( conteo[1] / 3 ) >= 0
    multiplier1 = conteo[1]/3
    var_score += multiplier1 * 1000
  end

  if ( conteo[1] % 3 ) != 0
    var_score += (conteo[1] % 3)*100
  end
  
  #Evaluating 5
  if ( conteo[5] % 3 ) != 0
    var_score += (conteo[5] % 3)* 50
  end

  #Evaluating numbers x 3    
  if (conteo[2..9].count { |x| x >= 3 }) > 0
    triplets = conteo[2..9].map {|x| x / 3}
    array_multiplicator = triplets.each_with_index.select {|num,index| (num > 0)}.map {|x| x[0]}
    product_triplets = triplets.each_with_index.select {|num,index| (num > 0)}.map {|x| x[1]}.map {|x| (x+2)*100}
    var_score += array_multiplicator.zip(product_triplets).map{|x| x.inject(&:*)}.sum
  end

  var_score
end
于 2020-08-05T14:37:37.227 回答
0

花了 29 行，但这是我的第一个 Ruby

    def score(dice)
      return 0 if dice == [] 
      sums = Array.new  # To hold number of occurrences 1 - 6
      for i in 0..6     # Initialize to 0... note [0] is not used 
        sums[i] = 0
      end
      total = 0   # To hold total
      dice.each do |dots|   # Number of dots showing on dice
        sums[dots] += 1     # Increment the array members 1 - 6
      end 
        if sums[1] > 2 then # If 3 1's
          total += 1000 
          sums[1] -= 3   # Remove the 3 you took, in case there's more
        end
        if sums[2] > 2  then total += 200  # If 3 2's
        end
        if sums[3] > 2  then total += 300   #If 3 3's
        end
        if sums[4] > 2  then total += 400    #If 3 4's
        end
        if sums[5] > 2  then total += 500     #If 3 5's
          sums[5] -= 3                        #Remove the 5's you took
        end
        if sums[6] > 2  then total += 600  #If 3 6's
        end
     total += (sums[1] * 100)   # If any ones are left
     total += (sums[5] * 50)    # Same for fives
     return total
end
于 2020-09-14T00:10:45.270 回答
-1 
def score(dice)
    score = 0
    dice.uniq.each do |number|
        count = dice.count number
        weight = if number == 1 then 10 else number end
        if count >= 3
            score += weight * 100
            count -= 3
        end
        if count > 0 and number == 1 or number == 5
            score += count * weight * 10
        end
    end
    score
end
于 2014-09-22T09:44:31.390 回答
-2 
def score(dice)
    ones = fives = rest = 0

    one_count = dice.count(1)
    if one_count > 2
        ones = 1000
        one_count -= 3
    end
    ones += one_count * 100

    five_count = dice.count(5)
    if five_count > 2
        fives = 500
        five_count -= 3
    end
    fives += five_count * 50

    [2,3,4,6].each do |num|
        if dice.count(num) > 2
            rest += num * 100
        end
    end

    return ones + fives + rest
end
于 2017-09-25T11:52:40.940 回答