Problem:
Score : 400 points

Problem Statement

A positive integer X is said to be a lunlun number if and only if the following condition is satisfied:

In the base ten representation of X (without leading zeros), for every pair of two adjacent digits, the absolute difference of those digits is at most 1.

For example, 1234, 1, and 334 are lunlun numbers, while none of 31415, 119, or 13579 is.

You are given a positive integer K. Find the K-th smallest lunlun number.

Constraints

1 \leq K \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K

Output

Print the answer.

Sample Input 1

15

Sample Output 1

23

We will list the 15 smallest lunlun numbers in ascending order:

1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
11,
12,
21,
22,
23.

Thus, the answer is 23.

Sample Input 2

1

Sample Output 2

1

Sample Input 3

13

Sample Output 3

21

Sample Input 4

100000

Sample Output 4

3234566667

Note that the answer may not fit into the 32-bit signed integer type.

Solution in Ruby (2.3.3) (judge verdict: Accepted)

K = gets.chop.to_i

arr = (1..9).to_a

def lulun(num)
  tail = num.split('')[-1]

  prev = (tail.to_i - 1).to_s
  forw = (tail.to_i + 1).to_s

  if tail == '0'
    return [(num + tail).to_i, (num + forw).to_i]
  elsif tail == '9'
    return [(num + prev).to_i, (num + tail).to_i]
  else
    return [(num + prev).to_i, (num + tail).to_i, (num + forw).to_i]
  end
end

def generate(arr1, arr2, count) # 完成した配列と更新したい配列を渡す
  tmp = [] # 新しく作った分を入れる

  arr2.each do |el|
    tmp.concat(lulun(el.to_s))
  end

  arr1.concat(tmp)

  if arr1.count > count
    return arr1
  else
    generate(arr1, tmp, count)
  end
end

luluns = generate(arr, arr, K)

puts luluns[K - 1]