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 pb push_back
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define sz(x)  (long long)(x).size()
#define sqr(x) (x)*(x)
#define sof sizeof
#define ins insert
#define lowerB lower_bound
#define upperB upper_bound
#define mkp make_pair
#define mkt make_tuple
#define act(a,v) (get<(long long)a>(v))
#define mod 1000000007
#define mod1 998244353
#define N 100003
#define inf 1e18
#define minf -1e18
#define min3(a,b,c) min(a,min(b,c));
#define max3(a,b,c) max(a,max(b,c));
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define sumall(v) accumulate(all(v),(ll)0)
#define sum accumulate
#define FastIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define PRECISION std::cout.unsetf ( std::ios::floatfield );std::cout.precision(15)
#define PI M_PI
#define fi first
#define se second
#define here cout<<"RSSB"<<endl
#define showzero cout<<std::showpoint
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*b)/gcd(a,b)
#define br "\n"
#define sp " "
#define zero (ll)0
using namespace  std ;
typedef double db;
typedef long long ll ;
typedef vector<int>vi ;
//typedef vector<l>vl ;
typedef vector<ll>vll ;
typedef vector<vector<ll> >vvl ;
typedef pair<int,int> pii ;
typedef pair<ll,ll> pll ;
typedef vector<pll>vpll;
typedef vector<string>vstr;
typedef vector<bool>vbool ;
void READ(vll &v, ll n)
{
    ll a ;
    for(ll i=0 ; i<n ; i=i+1)
    {
        cin>>a ;
        v.pb(a) ;
    }
}
void PRINT(vll &v , ll a = 0)
{
    //ll a ;
    for(ll i=a ; i<v.size() ; i=i+1)
    {
        v[i] = max((ll)0,v[i]);
        cout<<v[i]<<" " ;
    }
    cout<<"\n" ;
}
/*vbool isprime(N,true);
vll spf(N,0);
vll prime;
void sieve()
{
    isprime[0] = isprime[1] = false;
    for(ll i = 2 ; i<N ; i++)
    {
        if(isprime[i])
        {
            for(ll j=2*i ; j<N ; j+=i)
            {
                isprime[j] = false;
                if(spf[j]==0)
                {
                    spf[j] = i;
                }
            }
            prime.pb(i);
        }
    }
}*//*
ll nck(ll n , ll k ,ll m=1e18+7)
{
    ll i , j,b=1,c=1 ;
    if(n-k<k)
    {
        k = n-k ;
    }
    while(k)
    {
        b*=n ;
        c*=k ;
        ll g = __gcd(b,c) ;
        b/=g ;
        c/=g ;
        n--;
        k-- ;
    }
    return(b) ;
}*//*
long long power(long long k , long long n , long long m=mod)
{
    long long res = 1;
    while(n)
    {
        if(n%2!=0)
        {
            res = (res*k)%m ;
        }
        k = (k*k)%m ;
        n = n/2 ;
    }
    return(res) ;
}*/
ll n,m,k,l,r;
bool ok = false;
vll dp;
map<ll,ll>mp;
void solve(ll x , ll y , ll j)
{
    if(y==m)
    {
        if(mp.count(x))
        {
            return;
        }
        mp[x]++;
        dp.pb(x);
        return;
    }
    if(j)
    {
        solve(x*10+j-1,y+1,j-1);
    }
    if(sz(dp)>=N)
    {
        return;
    }
    solve(x*10+j,y+1,j);
    if(sz(dp)>=N)
    {
        return;
    }
    if(j!=9)
    {
        solve(x*10+j+1,y+1,j+1);
        if(sz(dp)>=N)
        {
            return;
        }
    }
}
int main()
{
    FastIO ;
    PRECISION ;
    ll a,b,c,d,i,j,t,x,y,z;//,x1,y1,z1,e,f;
    ll p,q,r,u,w ;
    char ch,chr ;
    string str,st,str1,str2,str3,str4;
    //ll a1=0,a2=0,b1=0,b2=0,c1=0 ;
    //cin>>str>>str1;
    for(i=1 ; i<=9 ; i++)
    {
        dp.pb(i);
    }
    i = 2;
    while(true)
    {
        m = i;
        for(j=1 ; j<=9 ; j++)
        {
            x = j;
            y = 1;
            solve(x,y,j);
            if(sz(dp)>=N)
            {
                break;
            }
        }
        i++;
        if(sz(dp)>=N)
        {
            break;
        }
    }
    sort(all(dp));
    //PRINT(dp);
    //cout<<sz(dp);
    cin>>k;
    cout<<dp[k-1];
}