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>

using namespace std ;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll ;
typedef vector<int> vint;
typedef vector<double> vdbl ;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<vector<int>> vvint;
typedef vector<pair<int,int>> vpii ;
typedef vector<pair<ll,ll>> vpll ;
typedef priority_queue<int, vector<int>, greater<int>> spqint ; //小さい順に取り出し
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll ; //小さい順に取り出し
typedef priority_queue<int, vector<int>, less<int>> bpqint ; //大きい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll ; //大きい順に取り出し


#define REP(i, n) for (int (i) = 0; (i) < (int)(n); i++) 
#define FOR(i, a, b) for(int (i) = a; (i) < (int)b; i++)
#define IREP(i,v) for(auto i=v.begin(); i !=v.end(); i++)
#define MOD 1000000007
#define NIL -1
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
#define PF push_front
#define TS to_string
#define BS binary_search
#define LB lower_bound
#define UB upper_bound
#define NP next_permutation
#define ALL(v) (v).begin(),(v).end()
#define SZ(x)  (ll)x.size()
#define SP(x)  setprecision((ll)x)

const int INF = 1e9;
const ll LINF = 1e18;
const double EPS = 1e-9;
const double PI = M_PI;
const int dx[8] = {1, 0 -1, 0, -1, 1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} //最小公倍数

void yes(){
  cout <<"Yes"<<endl ;
}
void no(){
  cout <<"No"<<endl ;
}
//-----------------------------------------


const int dz[3] = {-1,0,1} ;
vll vec ;

void dfs(int x,string s){
  if(SZ(s)==11){
    return ;
  }
  vec.PB(stol(s)) ;
  REP(i,3){
    if(x+dz[i]>=0 && x+dz[i]<=9){
      dfs(x+dz[i],s+TS(x+dz[i])) ;
    }
  }
  return ;
}
//-----------------------------------------
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  FOR(i,1,10){
    dfs(i,TS(i)) ;
  }
  sort(ALL(vec)) ;
  int n ;
  cin >> n ;
  cout << vec[n-1] <<endl ;
  return 0 ;
}