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

传送门

Description

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 Input

2 15

0 0

Sample Output

69 96

-1 -1

思路

题意:

求由1-9组成的长度为m且各位数之和为s的最大数和最小数。

+ View Code?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

#include<bits/stdc++.h> using namespace std; const int maxn = 105; int a[maxn];   bool OK(int m, int s) {     return s >= 0 && s <= m * 9; }   int main() {     int m, s;     scanf("%d%d", &m, &s);     if (!OK(m, s) || (m > 1 && s == 0))  printf("-1 -1\n");     else     {         int cnt = 0, sum = s;         for (int i = 0; i < m; i++)         {             for (int j = 0; j < 10; j++)             {                 if (((i > 0 || j > 0) || (m == 1 && j == 0)) && OK(m - i - 1, sum - j))                 {                     sum -= j;                     a[cnt++] = j;                     break;                 }             }         }           for (int i = 0; i < cnt; i++) printf("%d", a[i]);         printf(" ");         cnt = 0, sum = s;         for (int i = 0; i < m; i++)         {             for (int j = 9; j >= 0; j--)             {                 if (((i > 0 || j > 0) || (m == 1 && j == 0)) && OK(m - i - 1, sum - j))                 {                     sum -= j;                     a[cnt++] = j;                     break;                 }             }         }         for (int i = 0; i < cnt; i++) printf("%d", a[i]);         puts("\40");     }     return 0; }