Codeforces Round #289 (Div. 2, ACM ICPC Rules)C. Sums of Digits

C. Sums of Digits

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya had a strictly increasing sequence of positive integers a₁, ..., a_n. Vasya used it to build a new sequence b₁, ..., b_n, where b_i is the sum of digits of a_i's decimal representation. Then sequence a_i got lost and all that remained is sequence b_i.

Vasya wonders what the numbers a_i could be like. Of all the possible options he likes the one sequence with the minimum possible last number a_n. Help Vasya restore the initial sequence.

It is guaranteed that such a sequence always exists.

Input

The first line contains a single integer number n (1 ≤ n ≤ 300).

Next n lines contain integer numbers b₁, ..., b_n  — the required sums of digits. All b_i belong to the range 1 ≤ b_i ≤ 300.

Output

Print n integer numbers, one per line — the correct option for numbers a_i, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to b_i.

If there are multiple sequences with least possible number a_n, print any of them. Print the numbers without leading zeroes.

Sample test(s)

Input

3
1
2
3

Output

1
2
3

Input

3
3
2
1

Output

3
11
100

先求出第一个数,这个数低位尽量放9

然后求其他的,先考虑当前的数是否可能和上一个数的位数一样,分两种:可能一样,从低位贪心找比上一个数大的,然后求出这个数;不可能一样:那么构造一个合法的,位数尽量少的数

/*************************************************************************
    > File Name: cf289c.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年02月01日 星期日 19时09分32秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
int pro[300];
string last;

void get_one (int sum)
{
	last = "";
	while (sum)
	{
		if (sum >= 9)
		{
			last += '9';
			sum -= 9;
		}
		else
		{
			last += '0' + sum;
			break;
		}
	}
	int len = last.size();
	for (int i = 0; i < len / 2; ++i)
	{
		swap (last[i], last[len - i - 1]);
	}
}

void get_lag (string now, int sum, int least)
{
	for (int i = 1; i <= least; ++i)
	{
		for (int j = 0; j < 10; ++j)
		{
			if (j == 0 && i == 1)
			{
				continue;
			}
			if (sum <= j + (least - i) * 9)
			{
				now += '0' + j;
				sum -= j;
				break;
			}
		}
	}
	last = now;
	cout << now << endl;
}

void get_equ_sma (string now, int sum)
{
	pro[0] = 0;
	int len = last.length();
	for (int i = 1; i <= len; ++i)
	{
		pro[i] = pro[i - 1] + last[i - 1] - '0';
	}
	int s = -1, ch;
	for (int i = len; i >= 1; --i)
	{
		bool flag = false;
		for (int j = last[i - 1] - '0' + 1; j <= 9; ++j)
		{
			if (sum - j - pro[i - 1] < 0)
			{
				continue;
			}
			int x = sum - j - pro[i - 1];
			int leave = x / 9;
			if (x % 9)
			{
				++leave;
			}
			if (leave <= (len - i))
			{
				flag = true;
				s = i;
				ch = j;
				break;
			}
		}
		if (flag)
		{
			break;
		}
	}
	if (s != -1)
	{
		for (int i = 1; i < s; ++i)
		{
			now += last[i - 1];
		}
		int leave = (sum - ch - pro[s - 1]);
		now += ch + '0';
		for (int i = s + 1; i <= len; ++i)
		{
			for (int j = 0; j < 10; ++j)
			{
				if (leave - j <= (len - i) * 9)
				{
					now += '0' + j;
					leave -= j;
					break;
				}
			}
		}
		cout << now << endl;
		last = now;
	}
	else
	{
		get_lag (now, sum, len + 1);
	}
}

void solve (int sum)
{
	string now = "";
	int least = sum / 9;
	if (sum % 9)
	{
		++least;
	}
	if (least > last.length())
	{
		get_lag (now, sum, least);
	}
	else
	{
		get_equ_sma (now, sum);
	}
}

int main ()
{
	int n;
	while (cin >> n)
	{
		int sum;
		cin >> sum;
		get_one (sum);
		cout << last << endl;
		for (int i = 2; i <= n; ++i)
		{
			cin >> sum;
			solve (sum);
		}
	}
	return 0;
}