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;

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<=(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define Eunique(v) v.erase(unique(all(v)),v.end())
#define Eback(s) s.erase(s.end()-1,s.end())
#define rev(v) reverse(all(v))
#define minvec(v) *min_element(all(v))
#define maxvec(v) *max_element(all(v))
#define sumvec(v) accumulate(all(v),0)
#define mapmax(v) v.rbegin()->first
#define mapmin(v) v.begin()->first
#define pb push_back
#define pf push_front
#define mod 1000000007
#define m_p make_pair
#define DOUBLE fixed << setprecision(15)
#define OK puts("OK")
#define OK1 puts("OK1")
#define OK2 puts("OK2")
#define SIZE(s) (int)s.size()
#define INF ((1LL<<62)-(1LL<<31))
#define zero(x,n) setw(x) << setfill('0') << n

typedef long long ll;

typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector<double> vd;
typedef vector<char> vc;
typedef vector<string> vs;

typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef vector<pair<int,int>> vpii;
typedef vector<pair<ll,ll>> vpll;

const double pi = acos(-1.0);

template<class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>&p) {
	ost << "{" << p.first << ", " << p.second << "} ";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost, const vector<T>&v) {
	ost << "{";
	for (int i = 0; i<(int)v.size(); i++) {
		if (i)ost << " ";
		ost << v[i];
	}
	ost << "} \n";
	return ost;
}

template<class A, class B>
ostream& operator<<(ostream& ost, const map<A, B>&v) {
	ost << "{";
	for (auto p:v) {
		ost << "{" << p.first << ", " << p.second << "} ";
	}
	ost << "} ";
	return ost;
}

template<class T>
inline bool chmax(T& a, T b){if(a<b){a=b;return true;} return false;}

template<class T>
inline bool chmin(T& a, T b){if(a>b){a=b;return true;} return false;}

void YES(bool b) {cout << (b?"YES\n":"NO\n");}
void Yes(bool b) {cout << (b?"Yes\n":"No\n");}
void yes(bool b) {cout << (b?"yes\n":"no\n");}
void Yay(bool b) {cout << (b?"Yay!\n":":(\n");}

ll gcd(ll x, ll y) {return __gcd(x,y);}
ll lcm(ll x, ll y) {return x/__gcd(x,y)*y;}

int main()
{
	ll k,cnt=0,x;
	queue<ll> q;
	cin >> k;
	rep(i,9) q.push(i+1);
	while(cnt<k){
		cnt++;
		x=q.front();
		q.pop();
		if(x%10!=0) q.push(10*x+(x%10)-1);
		q.push(10*x+(x%10));
		if(x%10!=9) q.push(10*x+(x%10)+1);
	}
	cout << x << endl;
	return 0;
}