Codeforces 489C 简单构造

传送门：



http://codeforces.com/contest/489/problem/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).

Sample test(s)

Input

2 15

Output

69 96

Input

3 0

Output

-1 -1

题意：给一个数的位数和每一位的和，问这个数最大是多少，最小是多少。不存在输出-1

思路：先检查合法性，是否输出-1 -1；

构造最小数：从尾部开始构造，直接填9（或者剩下的数的和-1），最后剩下的数应该大于等于1，这个填到首位。其他地方填0.

构造最大数：从首位开始构造，直接填9（或者剩下的数的和），其他的地方直接填0.

代码：

#include
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        if(m==0&&n==1)
        {
            printf("0 0\n");
            continue;
        }
        if(m<1||n*9=9)
            {
                s[i]=9+48;
                m-=9;
            }
            else
            {
                s[i]=tmp+48;
                m-=tmp;
            }
        }
        char temp[105];
        sprintf(temp,"%s",s+1);
        if(s[n]=='0')
        {
            s[n]++;
            for(int i=n-1;i>=1;i--)
            if(s[i]>48)
            {
                s[i]--;
                break;
            }
        }
        for(int i=n;i>=1;i--) printf("%c",s[i]);
        printf(" ");
         printf("%s\n",temp);
    }
    return 0;
}