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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <cassert>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const long long MOD = 1000000007;
const long long INF = 1LL << 60;
const int INT_INF = 1000000000;

// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int dx[8] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy[8] = {-1, 0, 1, 1, -1, 1, 0, -1};

void rec(int d, ll val, vector<ll> &all)
{
  all.push_back(val);

  if (d == 10)
    return;

  for (int j = -1; j <= 1; j++)
  {
    int add = (val % 10) + j;
    if (add >= 0 && add <= 9)
      rec(d + 1, val * 10 + add, all);
  }
}
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int K;
  cin >> K;
  vector<ll> all;
  for (int v = 1; v < 10; v++)
    rec(1, v, all);
  sort(all.begin(), all.end());
  cout << all[K - 1] << endl;
  return 0;
}

// 全列挙した後にK番目を取る．サンプル3を見ると最大のKに対する答えが3234566667であることがわかる．つまり，最大でも10桁のlunlun数を列挙すればよさそう．(d桁のlunlun数は大体3^dくらいになりそう．)まず最初に1～9を列挙しておいてそこから次々と新しい数値を付け加えて行くイメージ　DFSでたどれる．