Solving with Ruby and Python AtCoder CODE FESTIVAL 2016 qual C B Priority queue

Introduction

• AtCoder Problems * Use Recommendations to solve past problems. Thanks to AtCoder and AtCoder Problems.

This theme

• AtCoder CODE FESTIVAL 2016 qual C B --K cakes * Difficulty: 905

This theme, priority queue

Problems such as the advanced version of * Solved in Ruby, Python and Java AtCoder ABC141 D Priority Queuing *.

Queues the maximum and second maximum of a given array, subtracts 1 and puts it back in the queue. This is repeated until the queue is filled with one or less, and the loop is exited.

There is also a mathematical solution.

Ruby mathematical solution

ruby.rb

k = gets.to_i
a = gets.split.map(&:to_i).max
puts [0, 2 * a - k - 1].max

I will pay attention to the maximum value and solve it, but the refreshing feeling is amazing. Ruby Heap

ruby.rb

class Heap
  attr_reader :size
  def initialize(up: false)
    @up = up
    @heap = []
    @size = 0
  end
  def sum
    x = 0
    @size.times do |i|
      x += @heap[i]
    end
    x
  end
  def push(n)
    n = -n if @up
    i = @size
    while i > 0
      pid = (i - 1) / 2
      break if n >= @heap[pid]
      @heap[i] = @heap[pid]
      i = pid
    end
    @heap[i] = n
    @size += 1
  end
  def pop
    return nil if @size == 0
    top = @heap[0]
    @size -= 1
    n = @heap[@size]
    i = 0
    while i * 2 + 1 < @size
      cid1 = i * 2 + 1
      cid2 = cid1 + 1
      if cid2 < @size && @heap[cid2] < @heap[cid1]
        cid1 = cid2
      end
      break if @heap[cid1] >= n
      @heap[i] = @heap[cid1]
      i = cid1
    end
    @heap[i] = n
    if @up
      -top
    else
      top
    end
  end
end

gets
a = gets.split.map(&:to_i)
h = Heap.new(up: true)
a.each do |x|
  h.push(x)
end
while h.size > 1
  u = h.pop
  v = h.pop
  h.push(u - 1) if u - 1 > 0
  h.push(v - 1) if v - 1 > 0
end
if h.size == 0
  puts "0"
else
  puts h.pop - 1
end

up.rb

  def initialize(up: false)
    @up = up

• Last time * Improved the class so that the maximum value and the minimum value can be switched with a flag.

heap.rb

while h.size > 1
  u = h.pop
  v = h.pop
  h.push(u - 1) if u - 1 > 0
  h.push(v - 1) if v - 1 > 0
end

Take two, subtract them, and if they are 1 or more, put them back in the queue.

Python mathematical solution

python.py

from sys import stdin

def main():
    input = stdin.readline
    k, _ = map(int, input().split())
    a = max(map(int, input().split()))
    print(max(0, 2 * a - k - 1))
main()

Python heapq

python.py

from sys import stdin

def main():
    import heapq
    input = stdin.readline
    input()
    a = [-1 * int(i) for i in input().split()]
    heapq.heapify(a)
    while len(a) > 1:
        u = heapq.heappop(a)
        v = heapq.heappop(a)
        if u + 1 < 0:
            heapq.heappush(a, u + 1)
        if v + 1 < 0:
            heapq.heappush(a, v + 1)
    if len(a) == 0:
        print(0)
    else:
        print(abs(a[0] + 1))
main()

If the operation is complicated, it is better to prepare a function for code conversion.

Summary

• Solved AtCoder CODE FESTIVAL 2016 qual C B
• Become familiar with Ruby
• Become familiar with Python

Referenced site

• [Competitive Pro] Priority Queuing (heapq) Summary *