Codeforces Round #277.5 (Div. 2)C——Given Length and Sum of Digits...

C. Given Length and Sum of Digits...

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a positive integer m and a non-negative integer
s. Your task is to find the smallest and the largest of the numbers that have length
m and sum of digits
s. The required numbers should be non-negative integers written in the decimal base without leading zeroes.

Input

The single line of the input contains a pair of integers
m, s (1 ≤ m ≤ 100, 0 ≤ s ≤ 900) — the length and the sum of the digits of the required numbers.

Output

In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1
-1" (without the quotes).

Sample test(s)

Input

2 15

Output

69 96

Input

3 0

Output

-1 -1

贪心，考虑最大的时候把数组赋值为9，考虑最小的时候把数组第一位赋值1。其它为赋值0

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int s1[110], s2[110];

int main()
{
	int m, s;
	while (~scanf("%d%d", &m, &s))
	{
		if (s == 0 && m == 1)
		{
			printf("0 0\n");
			continue;
		}
		if (m * 9 < s || s < 1) //每一位全是9都小于s，或者1.....0大于s
		{
			printf("-1 -1\n");
			continue;
		}
		for (int i = 1; i <= m; ++i)
		{
			s1[i] = 0;
			s2[i] = 9;
		}
		s1[1] = 1;
		int dis = s - 1, cnt = m;
		while (1)
		{
			if (9 - s1[cnt] >= dis)
			{
				s1[cnt] += dis;
				break;
			}
			dis -= (9 - s1[cnt]);
			s1[cnt] = 9;
			cnt--;
		}
		for (int i = 1; i <= m; ++i)
		{
			printf("%d", s1[i]);
		}
		printf(" ");
		dis = m * 9 - s;
		cnt = m;
		while (1)
		{
			if (s2[cnt] >= dis)
			{
				s2[cnt] -= dis;
				break;
			}
			dis -= (s2[cnt]);
			s2[cnt] = 0;
			cnt--;
		}
		for (int i = 1; i <= m; ++i)
		{
			printf("%d", s2[i]);
		}
		printf("\n");
	}
	return 0;
}