codeforces#277.5 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 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

题意：要求出两个m位数，他们m位上数字的和为s，求出最小和最大的这样的数，如果不存在，就输出-1 -1

贪心的策略，希望数字最小，那么除了第m位，其他位尽可能为9，希望数字最大，那么尽量让前几位为9

#include<cstdio>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
int m,s;
int main()
{
	while(~scanf("%d%d",&m,&s))
	{
	    if((s<1&&m>1)||s>m*9)
	    {
	        cout<<-1<<" "<<-1<<endl;
	        continue;
	    }

	    for(int i=m-1,k=s;i>=0;i--)
	    {
	        int j=max(0,k-9*i);
	        if(j==0&&i==m-1&&k) j=1;//k!=0很关键，当数据为1 0时，若k=0则此时要输出0
	        cout<<j;
	        k-=j;
	    }
	    cout<<" ";
	    for(int i=m-1,k=s;i>=0;i--)
	    {
	        int j=min(9,k);
	        cout<<j;
	        k-=j;

	    }

	}
	return 0;
}