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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define repeat(i, n) for (auto i = decltype(n)(); (i) < (n); (i)++)

int digitSum(int64_t n) {
    int res = 0;
    while (n > 0) {
        res += n % 10;
        n /= 10;
    }
    return res;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int k;
    cin >> k;

    int64_t d = 1;

    vector<int64_t> cand;
    for (int i = 0; i < 14; i++) {
        for (int p = 1; p < 300; p++) {
            int64_t c = p * d - 1;
            if (c == 0) continue;

            cand.push_back(c);
        }
        d *= 10;
    }

    sort(cand.begin(), cand.end());
    cand.erase(unique(cand.begin(), cand.end()), cand.end());

    // 4000個弱

    int next = -1;
    repeat (i, k) {
        bool flag = false;

        while (!flag) {
            next++;
            flag = true;
            for (int j = next + 1; j < cand.size(); j++) {
                if (cand[next] * digitSum(cand[j]) > cand[j] * digitSum(cand[next])) {
                    flag = false;
                    break;
                }
            }
        }
        cout << cand[next] << endl;
    }

    return 0;
}