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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

// ルンルン数をスタックにpush
void lunlunPush(unsigned long long num, unsigned long long* stack)
{
    static int i = 0;
    // 桁が変わるごとにスタックをリセット
    if (num == 0) {
        i = 0;
        return;
    }
    stack[i] = num;
    i++;
    return;
}


// N桁の数のルンルン数をスタックに入れながら数え上げる
int lunlunCount(int N, int start, int end, unsigned long long num, unsigned long long *stack)
{
    if (N == 0) {
        lunlunPush(num, stack);
        // printf("%d ", num);
        return 1;
    }
    int S = 0;
    int E = 0;
    int total = 0;
    int i;
    for (i = start; i <= end; i++) {
        S = (i - 1) < 0 ? 0 : (i - 1);
        E = (i + 1) > 9 ? 9 : (i + 1);
        total += lunlunCount((N - 1), S, E, (num * 10) + i, stack);
    }
    return total;
}

int main(void)
{
    int K;
    int N, X;
    // 10^5番目のルンルン数は10桁、10桁のルンルン数は130900個
    unsigned long long lunlunStack[130900];

    // input
    scanf("%d", &K);

    // K番目のルンルン数 -> N桁の数のK'番目のルンルン数
    N = 1;
    X = 0;
    while (1) {
        X = lunlunCount(N, 1, 9, 0, lunlunStack);
        if (X >= K) {
            break;
        } else {
            // スタックの添字をリセット
            lunlunPush(0, lunlunStack);
            // printf("\n");
            K -= X;
            N++;
        }
    }
    // printf("%d桁のルンルン数は%d個\n", N, X);
    // printf("答えは%d桁のルンルン数の%d番目の数\n", N, K);

    // output
    // 添字は0から始まるのでスタックのK-1番目
    printf("%lld\n", lunlunStack[K-1]);

    return 0;
}