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 integers. Your task is to find the smallest and the largest of the numbers that have lengthm 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

#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;}