結果

問題 No.33 アメーバがたくさん
ユーザー ciel
提出日時 2020-02-08 22:16:27
言語 Crystal
(1.14.0)
結果
AC  
実行時間 2 ms / 5,000 ms
コード長 2,272 bytes
コンパイル時間 11,839 ms
コンパイル使用メモリ 296,676 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-30 19:59:24
合計ジャッジ時間 12,598 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
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ファイルパターン 結果
other AC * 11
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ソースコード

diff #
プレゼンテーションモードにする

struct Range(B, E)
  #Perform binary search
  #Based on Marc-André Lafortune's Ruby backports implementation, ported by @cielavenir
  def bsearch
    #return to_enum(:bsearch) unless block_given?
    from = self.begin
    to   = self.end
    unless from.is_a?(Int) && to.is_a?(Int)
      # Float support is currently dropped
      raise "can't do binary search for #{from.class}"
    end
    midpoint = 0 # placeholder
    #convert = ->{ (pointerof(midpoint) as Pointer(Int64)).value }
    convert = ->{ midpoint }
    to -= 1 if excludes_end?
    satisfied = nil
    while from <= to
      midpoint = (from + to)//2
      result = yield(cur = convert.call)
      case result
      when Int
        return cur if result == 0
        result = result < 0
      when Float
        return cur if result == 0
        result = result < 0
      when true
        satisfied = cur
      when nil, false
        # nothing to do
      else
        raise "wrong argument type #{result.class} (must be numeric, true, false or nil)"
      end
      if result
        to = midpoint - 1
      else
        from = midpoint + 1
      end
    end
    satisfied
  end
end
class Array(T)
  def bsearch
    idx=(0...self.size).bsearch{|i|yield self[i]}
    idx ? self[idx] : nil
  end
  def bsearch_index
    (0...self.size).bsearch{|i|yield self[i]}
  end
end
def checkio(data)
    result=0
    se=[] of Tuple(Int64,Int64)
    data.each{|e|
        l,r=e
        right_idx=(0...se.size).bsearch{|i|
            x={l,0}<=>se[i]
            x<=0
        }||se.size #l <= se[right][0]
        if right_idx>0
            left_idx=right_idx-1
            if l<=se[left_idx][1]+1 # overlap with left
                l=se[left_idx][0]
                r=[r,se[left_idx][1]].max
                result-=se[left_idx][1]-se[left_idx][0]+1
                se.delete_at(left_idx)
                right_idx-=1
            end
        end
        while right_idx<se.size && se[right_idx][0]<=r # overlap with right
            r=[r,se[right_idx][1]].max
            result-=se[right_idx][1]-se[right_idx][0]+1
            se.delete_at(right_idx)
        end
        result+=r-l+1
        se.insert(right_idx,{l,r})
    }
    return result
end
dic={} of Int64 => Array(Tuple(Int64,Int64))
n,d,t=gets.not_nil!.split.map(&.to_i64)
gets.not_nil!.split.map(&.to_i64).each{|e|
    mod=e%d
    dic[mod]||=[] of Tuple(Int64,Int64)
    dic[mod]<<{(e-mod)//d-t,(e-mod)//d+t}
}
r=0_i64
dic.each{|k,v|r+=checkio(v)}
puts r
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