Codeforces Round #277.5 (Div. 2)C??Given Length and Sum of Digits._html/css_WEB-ITnose

C. Given Length and Sum of Digits...

time limit per test

memory limit per test

input

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，其他为赋值0

div+css3阶梯分页样式

div+css3阶梯分页样式

84

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#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;}