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

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).

Examples

Input

2 15

Output

69 96

Input

3 0

Output

-1 -1

题目大意：

让你找到两个数字串，其中一个是最小的，另外一个是最大的，使得数字串每位数字和等于S.

要求数字串长度必须是n而且不能有前导0.

思路：

对于最小的数字串，我们将第一个数字先置为1.然后从后向前扫尽可能的置9.直到和为S为止，其余位子为0。

对于最大的数字串，我萌从前向后扫，尽可能的置9.直到和为S为止，其余位子为0.

注意前导0的问题，以及注意哪些情况需要输出-1 -1.

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int ans2[150];int ans[150];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        if(n==1&&m==0)        {            printf("0 0\n");            continue;        }        memset(ans2,0,sizeof(ans2));        memset(ans,0,sizeof(ans));        int tmpm=m;        int flag=1;        for(int i=1;i<=n;i++)        {            if(m<=0)break;            if(m>=9)            {                flag=0;                ans[i]=9;                m-=9;            }            else            {                flag=0;                ans[i]=m;                m=0;            }        }        if(m>0||flag==1)        {            printf("-1 -1\n");            continue;        }        m=tmpm;        ans2[1]=1;        m--;        for(int i=n;i>=1;i--)        {            if(m+ans2[i]>=9)            {                m-=9-ans2[i];                ans2[i]=9;            }            else            {                ans2[i]+=m;                m=0;            }        }        if(m<0||m>0)        {            printf("-1 -1\n");            continue;        }        for(int i=1;i<=n;i++)printf("%d",ans2[i]);        printf(" ");        for(int i=1;i<=n;i++)printf("%d",ans[i]);        printf("\n");    }}