Problem:
Score : 500 points

Problem Statement

Let S(n) denote the sum of the digits in the decimal notation of n.
For example, S(123) = 1 + 2 + 3 = 6.

We will call an integer n a Snuke number when, for all positive integers m such that m > n, \frac{n}{S(n)} \leq \frac{m}{S(m)} holds.

Given an integer K, list the K smallest Snuke numbers.

Constraints

1 \leq K

The K-th smallest Snuke number is not greater than 10^{15}.

Input

Input is given from Standard Input in the following format:

K

Output

Print K lines. The i-th line should contain the i-th smallest Snuke number.

Sample Input 1

10

Sample Output 1

1
2
3
4
5
6
7
8
9
19

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Text;

namespace D
{
    class Program
    {
        static void Main(string[] args)
        {
            long K = Convert.ToInt64(Console.ReadLine());
            var sb = new StringBuilder();
            long snuke = 0;
            for (var i = 0; i < K; i++)
            {
                snuke = f(snuke + 1);
                sb.AppendFormat($"{snuke}\n");
            }
            Console.Write(sb);
        }

        static long f(long n)
        {
            long ans = n;
            double min = ((double)n / S(n));
            long div = 1;
            for (int i = 0; i < 16; i++)
            {
                // x,xxx,xxx,xxx,xxx,xxx (div == 2のとき)
                long upper = (n / (div * 10)) * (div * 10); // x,xxx,xxx,xxx,xxx,x00
                long num = (n / div) % 10;                  // x,xxx,xxx,xxx,xxx,N00
                for (long j = num + 1; j <= 9; j++)
                {
                    long y = upper + (j * div) + div - 1;
                    double tmp = ((double)y / S(y));
                    if (tmp < min)
                    {
                        min = tmp;
                        ans = y;
                    }
                }
                div *= 10;
            }
            return ans;
        }

        static long S(long x)
        {
            long ans = 0;
            while (x > 0)
            {
                ans += x % 10;
                x /= 10;
            }
            return ans;
        }
    }
}