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)

/*
    飲んだ魔剤で家が建つ。
    created at: 2020-04-05 18:37:07
*/

#include <bits/stdc++.h>
using namespace std;

#define all(x) (x).begin(),(x).end()
#define rev(x) greater<x>()
#define MOD 1000000007
#define INF 1000000000

typedef long long LL;
typedef long double LD;

bool compare_by_b(pair<int, int> a, pair<int, int> b)
{
    if (a.second != b.second) {
        return a.second < b.second;
    } else {
        return a.first < b.first;
    }
}

LL modpower(LL n, LL m, LL mod) {
    if (m == 0) {
        return 1;
    } else if (m % 2 == 0) {
        LL p = modpower(n, m/2, mod);
        return p*p % mod;
    } else {
        return n*modpower(n, m-1, mod) % mod;
    }
}

LL modinv(LL n, LL mod) {
    return modpower(n, mod-2, mod);
}

LL modcombination(LL n, LL r, LL mod) {
    LL ret = 1;

    for (int i = 0; i < r; i++) {
        ret *= (n-i);
        ret %= mod;
        ret *= modinv(i+1, mod);
        ret % mod;
    }

    return ret % mod;
}

int gcd(int k, int l) {
    if (l > 0) return gcd(l, k%l);
    else return k;
}

struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n) {
        for (int i = 0; i < n; i++) par[i] = i;
    }

    int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        if (rx == ry) return;
        par[rx] = ry;
    }

    bool same(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

int main()
{
    int k;
    cin >> k;

    queue<LL> que;
    for (int i = 1; i <= 9; i++)
        que.push(i);
    
    int cnt = 0;
    LL ans;
    while (!que.empty()) {
        LL x = que.front();
        que.pop();

        if (++cnt == k) {
            ans = x;
            break;
        }

        if (x % 10 > 0) que.push(10*x + (x%10) - 1);

        que.push(10*x + (x%10));

        if (x % 10 < 9) que.push(10*x + (x%10) + 1);
    }

    cout << ans << endl;

    return 0;
}