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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MOD1 1000000007
#define MOD2 998244353
#define LIMIT1 200002
#define LIMIT2 500002
typedef long long ll;
typedef long double ld;
typedef const void cv;
#define rep(i,n) for(i=0;i<n;i++)
#define max(a,b) ((a)>(b) ? (a) : (b))
#define min(a,b) ((a)<(b) ? (a) : (b))
#define zt(a,b) (max((a),(b))-min((a),(b)))
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll nPr(int n,int r){ll i,result=1;rep(i,r){result*=(ll)(n-i);}return result;}
ll nCr(int n,int r){ll i,result=1;if(n<r) return 0;rep(i,min(r,n-r)){result*=(ll)(n-i);result/=(ll)(i+1);}return result;}
#define fact(n) nPr((int)(n),(int)(n))
#define nHr(n,r) nCr((int)((n)+(r)+1),(int)(r))
#define sankaku(x) (((x)*((x)+1))/2)
int dx[8]={1,0,-1,0,1,-1,-1,1};
int dy[8]={0,1,0,-1,1,1,-1,-1};
int upint(const void *a, const void *b) { return *(int *)a < *(int *)b ? -1 : *(int *)a > *(int *)b ? 1 : 0; }
int downint(const void *a, const void *b) { return *(int *)a < *(int *)b ? 1 : *(int *)a > *(int *)b ? -1 : 0; }
int upchar(const void* left, const void* right) {return strcmp((char *)left,(char *)right);}
int downchar(const void* left, const void* right) {return strcmp((char *)right,(char *)left);}
ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
void initialize(){
    
}
int main(void){
  initialize();
  ll n,m,i,j,k,result=0;
  ll a[LIMIT2]={0};
  int aCount=0;
  
  scanf("%lld",&n);
  for(i=1;i<=min(9,n);i++){
    a[aCount]=i;
    aCount++;
  }
  j=0;
  for(i=10;i<=n;i++){
    if(a[j]%10!=0){
      a[aCount]=a[j]*10+a[j]%10-1;
      aCount++;
    }
    a[aCount]=a[j]*10+a[j]%10;
    aCount++;
    if(a[j]%10!=9){
      a[aCount]=a[j]*10+a[j]%10+1;
      aCount++;
    }
    j++;
  }
  printf("%lld",a[n-1]);
  return 0;
}