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 ms (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;
}

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