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>
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define ld long double
#define rep(i, l, r) for (int i = (l); i < (r); i++)
#define repb(i, r, l) for (int i = (r); i > (l); i--)
#define sz(a) (int)a.size()
#define fi first
#define se second
#define mp(a, b) make_pair(a, b)
#define rank qwertyuio
#define next dfghjk
#define prev fhsgfhjf
#define plus fsghsf
#define minus ytryr
  
using namespace std;
  
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(ll &x, ll y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(ll &x, ll y) { return (y > x) ? x = y, 1 : 0; }
  
const int N = 100000;
const int inf = (int)1e9 + 1;
const ll big = (ll)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const ld eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
 
deque<int> next_good(deque<int> s) {
    int p = sz(s) - 1;
    while (p >= 0 && s[p] == 9) {
        s[p] = 0;
        p--;
    }
    if (p == -1) {
        s.push_front(1);
    } else {
        s[p]++;
    }
    while (true) {
        int b = -1;
        rep(i, 0, sz(s) - 1) {
            if (abs(s[i] - s[i + 1]) > 1) {
                b = i;
                break;
            }
        }
        if (b == -1) {
            break;
        }
        if (s[b + 1] > s[b]) {
            s[b]++;
        }
        int lll = max(s[b] - 1, 0);
        rep(i, b + 1, sz(s)) {
            s[i] = lll;
            lll = max(lll - 1, 0);
        }
    }
    return s;
}
 
int main()
{

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(20);
    cout << fixed;

    int k;
    cin >> k;
    deque<int> ans;
    rep(i, 0, k) {
        ans = next_good(ans);
    }
    for (int i : ans) {
        cout << i;
    }
    cout << "\n";
    return 0;
}