ruby完成一路百度的测试题目

ruby完成一道百度的测试题目
题目：
第一题：连续正整数

题目描述：一个正整数有可能可以被表示为n(n>=2)个连续正整数之和，如：

15=1+2+3+4+5
15=4+5+6
15=7+8

请编写程序，根据输入的任何一个正整数，找出符合这种要求的所有连续正整数序列。

输入数据：一个正整数，以命令行参数的形式提供给程序。

puts "Input the integer:"
number = gets

int_number = Integer(number)

hasResult = false;
(1..int_number/2).each  do |number_low|
  ((number_low+1)..int_number).each do |number_high|
    #The equation is (a+b)*(b-a+1)/2 数列求和公式
    result = (number_low+number_high) * (number_high-number_low+1) / 2
    if(result == int_number)
      puts (number_low..number_high).to_a.inspect
      hasResult = true;
      break;
    elsif(result > int_number)
      break
    end
  end
end

if(!hasResult)
  puts "NONE"
end

1 楼 bitlanghen 2011-11-07

改进了一些的Ruby代码 ruby完成一路百度的测试题目

puts "Input the integer:"
number = gets

def sub_seq(int_number)
  seqs = Array.new;
  (1..int_number/2).each  do |number_low|
    ((number_low+1)..int_number).each do |number_high|
      result = (number_low+number_high) * (number_high-number_low+1) / 2
      if(result == int_number)
        #The equation is (a+b)*(b-a+1)/2
        seqs << (number_low..number_high)
        break;
      elsif(result > int_number)
        #this can reduce some loops
        break
      end
    end
  end
  seqs
end

sub_seq(Integer(number)).each do |ran|
  puts ran.to_a.inspect
end

ruby完成一路百度的测试题目

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