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;
const int MAX = 700000;
const int MOD = 1000000007;

long long  fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x, long long int y){
	return(x + y - 1) / y;

}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;

	if (number2 > number1) {
		m = number2;
		n = number1;
	}
	long long int s = -1;
	while (s != 0) {
		s = m % n;
		m = n;
		n = s;
	}
	return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;
	return m / lcm(m, n) * n;
}
/*逆元計算*/
long long int  modinv(long long a, long long m) {
	long long int b = m, u = 1, v = 0;
	while (b) {
		long long int t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
// index が条件を満たすかどうか
vector<long long int >meguru;
bool isOK(int index, int key) {
	if (meguru[index] >= key) return true;
	else return false;
}
// 汎用的な二分探索のテンプレ
int binary_search(int key) {
	int left = -1; //「index = 0」が条件を満たすこともあるので、初期値は -1
	int right = (int)meguru.size(); // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()
	/* どんな二分探索でもここの書き方を変えずにできる！ */
	while (right - left > 1) {
		int mid = left + (right - left) / 2;

		if (isOK(mid, key)) right = mid;
		else left = mid;
	}
	/* left は条件を満たさない最大の値、right は条件を満たす最小の値になっている */
	return right;
}
long long modpow(long long a, long long n, long long mod) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}
int dp[3050][8000] = {};
int main() {
	long long int k;
	cin >> k;
	queue<long long int>q;
	vector<long long int>ans;
	q.push(1);
	q.push(2);
	q.push(3);
	q.push(4);
	q.push(5);
	q.push(6);
	q.push(7);
	q.push(8);
	q.push(9);
	while (ans.size() <=100000) {
		long long int now = q.front();
		long long int  c = now % 10;
		if (c == 0) {
			q.push(now * 10 + 1);
			q.push(now * 10);
		}
		else if (c == 9) {
			q.push(now * 10 + 8);
			q.push(now * 10 + 9);
		}
		else {
			q.push(now * 10 +c-1);
			q.push(now * 10 +c  );
			q.push(now * 10 +c+1);
		}
		ans.push_back(now);
		q.pop();
	}
	sort(ans.begin(), ans.end());
	cout << ans[k - 1];
}