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.



Sample input:
15


Sample output:
23

Submission status: Runtime Error

Solution code:
K = int(input())

count = 0
num = 0
while count <= K:
    num += 1
    flag = True
    num_str = str(num)
    if len(num_str) >= 2:
        for i in range(len(num_str) - 1):
            if ads(int(num_str[i]) - int(num_str[i + 1])) > 1:
                flag = False

    if flag:
        count += 1

print(num)

Submission status: Runtime Error

Solution code:
k = str(input())
list = []
def lunlun(k):
    for i in range(len(k) - 1):
        if abs(int(k[i]) - int(k[i+1])) > 1:
            return False
            break
    return True

i = 1
while len(list) == int(k):
    if lunlun(i):
        list.append(i)
    i += 1

print(lunlun(len(k)))

Submission status: Runtime Error

Solution code:
k = int(input())
x =[0]*(k*k)
lis=[]
count=0
for a in range(k*k):
    n = str(a)
    if(len(n))==1:
        x[int(n)]=1
        lis.append(n)
        count+=1
    elif x[int(n[:-1])]==1 and abs(int(n[-1])-int(n[-2]))<2:
        x[int(n)]=1
        lis.append(n)
        count+=1
    if count>k:
        break
print(lis[k])

Submission status: Runtime Error

Solution code:
K = int(input())
 

    h = 10
    cnt = 9
    while h <= 10 ** 100:
        X = str(h)
        N = len(X)
 
        i = 0
        ans = 'Yes'
        while i < N-1:
            if (int(X[i]) - int(X[i+1]) == 1) or (int(X[i+1]) - int(X[i]) == 1) or (int(X[i]) == int(X[i+1])):
                i += 1
            else:
                ans = 'No'
                break
        if ans == 'Yes':
            cnt += 1
    
        if cnt == K:
            print(h)
            break
        else:
            h += 1

Submission status: Runtime Error

Solution code:
K = int(input())
ans = []
for i in range(10**5):
    a = list(map(int,str(i).split()))
    for k in range(len(a)):
        if abs(a[k] - a[k+1]) <= 1:
            ans.append(i)
print(ans[K])

Submission status: Runtime Error

Solution code:
k = int(input())

if k<=10:
  print(k)
else:
  a = [1,2,3,4,5,6,7,8,9]
  a1 = [1,2,3,4,5,6,7,8,9]
  a2 = []
  while True:
    for i in a1:
      if i%10 != 0:
        a.append(int(str(i)+str(i%10-1)))
        a2.append(int(str(i)+str(i%10-1)))
      if i%10 != 9:
        a.append(int(str(i)+str(i%10+1))) 
        a2.append(int(str(i)+str(i%10+1)))
      a.append(int(str(i)+str(i%10)))
      a2.append(int(str(i)+str(i%10)))
    if len(a)>=k:
      break
    a1 = a2
    a2 = []
a.sort()
print(a[k-1])

Submission status: Runtime Error

Solution code:
k=int(input())

keta=1
total=9
while total<=k:
  keta+=1
  total*=3

if keta==1:
  print(k)
  exit()
ans=''
num=1
k-=9*3**(keta-2)
while k-(3**(keta-1))>0:
  k-=3**(keta-1)
  num+=1
else:
  ans+=str(num)
  keta-=1

while keta>0:
  san=ans[-1]
  for i in range(4):
    if k>=3**(keta-1):
      k-=3**(keta-1)
    else:
      ans+=str(int(san)+i-2)
      break

  keta-=1

print(int(ans))

Submission status: Runtime Error

Solution code:
K=int(input())
S=1
i=1
ans=0
while S<=K: 
	a=str(i)
	N=len(a)
	x=a[0:N//2]
	y=a[N//2 +1:N-1]
	flag=0
	sam=int(a[N//2])-int(a[N//2 +1])
	bm=abs(sam)
	if bm>=2:
		flag=1
		continue
	for j in range(N//2-1):
		sax=int(x[j])-int(x[j+1])
		bx=abs(sax)
		if bx>=2:
			flag=1
			break
	for j in range(len(y)):
		say=int(y[j])-int(y[j+1])
		by=abs(say)
		if by>=2:
			flag=1
			break
	if flag==0:
		ans=i
		S=S+1
	i=i+1
print(ans)

Submission status: Runtime Error

Solution code:
#include <math.h>
#include <bits/stdc++.h>
#include<iostream>
using namespace std;
int main()
{
    int K;
    cin >> K;

    if (12 >= K) {
	    cout << K << endl;
    } else {
	    int runrun_count = 12;
	    unsigned int value = 12;
	    while (runrun_count < K) {
		    value += 1;
		    unsigned int num = value;
		    int i = 0;
		    int mae_value = 0;
		    bool is_runrun = true;
		    while( num >= 1 )   {
			    int mod = num % 10; // 一の位の余り
			    num = num / 10; // 取得済みの余りを"num"から除去
			    if (i > 0) {
				    if (abs(mae_value - mod) > 1) {
					    is_runrun = false;
					    break;
				    }
			    }
			    i += 1;
			    mae_value = mod;
		    }

		    if (is_runrun) {
			    runrun_count += 1;
		    }
	    }

	    cout << value << endl;
    }



    return 0;
}