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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define repi(i, a, b) for(int i=(a); i>(b); i--)
#define db(x) (cerr << #x << ": " << (x) << '\n')
#define sync ios_base::sync_with_stdio(false), cin.tie(NULL)
#define cps CLOCKS_PER_SEC
#define tests(t) int t; cin >> t; while(t--)
#define iceil(n, x) (((n) + (x) - 1) / (x))
#define ll long long
#define gcd __gcd
#define eb emplace_back
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define sz size()
#define all(v) (v).begin(), (v).end()
#define uni(v) sort(all(v)), (v).erase(unique(all(v)), (v).end())
#define pii pair<int, int>
#define vi vector<int>
#define vpii vector<pii>
#define vvi vector<vi>
#define fi first
#define se second
#define mt make_tuple
#define pqueue priority_queue
#define bitcount(x) __builtin_popcount(x)
#define PI acos(-1.0)
#define EPS 1e-9
#define mod 1000000007
using namespace std;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ','); cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}

template<typename T1, typename T2>
ostream& operator << (ostream& os, const pair<T1, T2>& p) { return os << '(' << p.fi << ", " << p.se << ')'; }

template<typename T>
void printv(const T& v) { for(auto i : v) cerr << i << ' '; cerr << '\n'; }

template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;

template<typename T>
using maxpq = priority_queue<T>;

//All indexing is 0-based
using namespace __gnu_pbds;
template<class key, class cmp = std::less<key>>
using ordered_set = tree<key, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
//methods: find_by_order(k); & order_of_key(k);
//To make it an ordered_multiset, use pairs of (value, time_of_insertion)
//to distinguish values which are similar

template<class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;

//Returns no. of values x for which ceil(n / x) == y (y must be > 1).
inline ll CC(ll n, ll y) { return iceil(n, y-1) - iceil(n, y); }

//Returns no. of values x for which floor(n / x) == y
inline ll FF(ll n, ll y) { return n / y - n / (y+1); }

//a and b are assumed to be taken modulo p
inline int add(int a, int b, int p = mod){ int c = a + b; if(c >= p) c -= p; return c; }
inline int sub(int a, int b, int p = mod){ int c = a - b; if(c < 0) c += p; return c; }
inline int mul(int a, int b, int p = mod){ return (a * 1ll * b) % p; }

#define N 105
#define int ll
// #define trace(...) 42

vi v;

void bt(int len, int n) {
    if(!len) {
        v.pb(n);
        return;
    }

    int d = n % 10;
    if(d) bt(len-1, 10*n + d-1);
    bt(len-1, 10*n + d);
    if(d < 9) bt(len-1, 10*n + d+1);
}

//Generate nos. for a fixed length len
void generate(int len) {
    rep(d, 1, 10) {
        bt(len-1, d);
    }
}

main()
{   
    #ifdef CP
        freopen("/home/tarun/Desktop/input.txt", "r", stdin);
     // freopen("/home/tarun/Desktop/output.txt", "w", stdout);
    #endif
    sync;
    clock_t clk = clock();
    cerr << "I will return...\n";

    int k; cin >> k;
    for(int len = 1; v.size() < k; len++) {
        generate(len);
    }

    cout << v[k-1] << '\n';

    cerr << "...don't you ever hang your head.\n";
    cerr << "Time (in ms): " << (double)(clock() - clk) * 1000.0 / cps << '\n';
}

//Compile using:
//g++ -o filename.exe --std=c++11 filename.cpp
//Use -D CP for defining a macro CP in the local environment