Codeforces Round #277.5 (Div. 2)(C题)

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

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;bool can(int m, int s){    if(s >= 0 && 9*m >= s) return true;    else return false;}int main(){    int m,s;    cin>>m>>s;    if(!can(m,s))    {        cout<<"-1"<<" "<<"-1"<<endl;        return 0;    }    if(m == 1)    {        if(s >= 10)        {            cout<<"-1"<<" "<<"-1"<<endl;        }        else cout<<s<<" "<<s<<endl;    }    else {        if(s == 0) cout<<"-1"<<" "<<"-1"<<endl;        else {            string minn, maxn;            int sum = s;            for(int i = 1; i <= m; i++)                for(int j = 0; j < 10; j++)            {                if((j > 0 || (j == 0 && i > 1) ) && can(m - i, sum - j))                   {                       minn += char('0' + j);                       sum -= j;                       break;                   }            }            sum = s;               for(int i = 1; i <= m; i++)                for(int j = 9; j >= 0; j--)            {                if(can(m - i, sum - j))                   {                       maxn += char('0' + j);                       sum -= j;                       break;                   }            }            cout<<minn<<" "<<maxn<<endl;        }    }    return 0;}