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;
using llong = long long int;

#define rep(i, n) for (int i = 0; i < n; ++i)
#define stl_rep(itr, x) for (auto itr = x.begin(); itr != x.end(); ++itr)
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()

const static int MOD = 1000000007;
const static int INF = 1000000000;
const static int dx[4] = {1, 0, -1, 0};
const static int dy[4] = {0, 1, 0, -1};

/* --------------- template start --------------- */
template <class Abel> 
struct BinaryIndexTree {
    vector<Abel> dat[2];
    Abel UNITY_SUM = 0; // to be set
    
    /* [1, n] */
    BinaryIndexTree(int n) { init(n); }
    void init(int n) { for (int iter = 0; iter < 2; ++iter) dat[iter].assign(n + 1, UNITY_SUM); }
    
    /* a, b are 1-indexed, [a, b) */
    inline void sub_add(int p, int a, Abel x) {
        for (int i = a; i < (int)dat[p].size(); i += i & -i)
            dat[p][i] = dat[p][i] + x;
    }
    inline void add(int a, int b, Abel x) {
        sub_add(0, a, x * -(a - 1)); sub_add(1, a, x); sub_add(0, b, x * (b - 1)); sub_add(1, b, x * (-1));
    }
    
    /* a is 1-indexed, [a, b) */
    inline Abel sub_sum(int p, int a) {
        Abel res = UNITY_SUM;
        for (int i = a; i > 0; i -= i & -i) res = res + dat[p][i];
        return res;
    }
    inline Abel sum(int a, int b) {
        return sub_sum(0, b - 1) + sub_sum(1, b - 1) * (b - 1) - sub_sum(0, a - 1) - sub_sum(1, a - 1) * (a - 1);
    }
    
    /* debug */
    void print() {
        for (int i = 1; i < (int)dat[0].size(); ++i) cout << sum(i, i + 1) << ",";
        cout << endl;
    }
};

template<typename T>
struct UnionFind {
    vector<int> par;
    vector<T> sz;

    UnionFind(int n) {
        par.resize(n);
        sz.resize(n);
        for (int i = 0; i < n; ++i) {
            par[i] = i;
            sz[i] = 1;
        }
    }

    int find_root(int x) {
        return par[x] == x ? x : par[x] = find_root(par[x]);
    }

    bool same(int x, int y) {
        return find_root(x) == find_root(y);
    }

    void unite(int x, int y) {
        x = find_root(x);
        y = find_root(y);
        if (x == y) return;
        
        if (sz[x] < sz[y]) swap(x, y);
        sz[x] += sz[y];
        par[y] = x;
    }

    T size_of_tree(int x) {
        return sz[find_root(x)];
    }
};
/* --------------- template end --------------- */

/* --------------- function start --------------- */
llong my_gcd(llong a, llong b) {
    return (a ? my_gcd(b % a, a) : b);
}
llong my_lcm(llong a, llong b) {
    return a * b / my_gcd(a, b);    
}

llong facmod(llong n) {
    if (n == 0) return 1;

    llong res = n;
    for (int i = n - 1; i > 0; --i) {
        res = (res * i) % MOD;
    }

    return res;
}
llong powmod(llong a, llong n) {
    llong res = 1;
    while (n > 0) {
        if (n & 1) res = (res * a) % MOD;
        a = (a * a) % MOD;
        n >>= 1;
    }

    return res;
}
llong modinv(llong n) {
    return powmod(n, MOD - 2);
}
llong combmod(llong n, llong k) {
    llong res = 1;
    for (int i = 0; i < k; ++i) {
        res = (res * (n - i)) % MOD;
    }
    res = (res * modinv(facmod(k))) % MOD;

    return res;
}
/* --------------- function end --------------- */

int main(int argc, char *argv[]) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int k;
    cin >> k;
    if (k < 10) {
        cout << k << endl;
        return 0;
    }

    deque<string> Q;
    for (int i = 1; i < 10; ++i) {
        Q.push_back(to_string(i));
    }
    int cnt = 9;
    while (true) {
        string S = Q.front();
        // cout << cnt << " " << S << endl;
        Q.pop_front();
        int n = (int)S.size();
        if (S[n - 1] != '0') {
            string T = S + (char)(S[n - 1] - 1);
            Q.push_back(T);
        } else {
            string T = S + S[n - 1];
            Q.push_back(T);
        }
        ++cnt;
        if (cnt == k) break;

        if (S[n - 1] != '0') {
            string T = S + S[n - 1];
            Q.push_back(T);
        } else {
            string T = S + (char)(S[n - 1] + 1);
            Q.push_back(T);
        }
        ++cnt;
        if (cnt == k) break;

        if (S[n - 1] != '0' && S[n - 1] != '9') {
            string T = S + (char)(S[n - 1] + 1);
            Q.push_back(T);
            ++cnt;
            if (cnt == k) break;
        }
    }

    string ans = Q.back();

    cout << ans << endl;

    return 0;
}