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

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

/*
   贪心
   对于min第一位先保证有1从后面开始加，如果加完之后还有s，说明NO，如果连1都没有那么NO
   对于max就从头往后如果s够就9不够就s
   其他特殊情况要判，比如1 0 坑点应该输出0 0。。其他带有0的都是-1 -1
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int n,s;
    int a[110];
    while(~scanf("%d%d", &n, &s)){
        if(n == 1 && s == 0) {
            printf("0 0\n");
            continue;
        }
        if(n == 0 || s == 0 ){
            if( n== 1 && s == 0);
            else {
                printf("-1 -1\n");
                continue;
            }
        }
        memset(a, 0, sizeof(a));
        int temp = s;
        s --;
        a[1] = 1;
        int pos = n;
        int flag = 0;
        while(s > 0){
            if(s <= 8){
                a[pos--] += s;
                s = 0;
            }
            else{ a[pos--] = 9;
                s -= 9;
            }
            if(pos == 0) break;
        }
        int flag1 = 0;
        int sum1 = 0;
        for(int i = 1; i <= n; i++){
            sum1 += a[i];
            if(a[i] > 9)
                flag1 = 1;
        }
        if(sum1 != temp || flag1 == 1) printf("-1 ");
        else {
            for(int i = 1; i <= n ;i++)
                printf("%d", a[i]);
            printf(" ");
        }
        memset(a, 0, sizeof(a));
        s = temp;
        for(int i = 1; i <= n ;i++){
            if(s <= 8) {a[i] = s;s = 0;}
            else {a[i] = 9; s -= 9; }
        }
        int sum = 0;
        for(int i = 1; i <= n ; i++)
            sum += a[i];
        if(sum != temp || a[1] == 0) printf("-1\n");
        else {
            for(int i = 1; i <= n ;i++)
                printf("%d",a[i]);
            puts("");
        }
        
    }
        return 0;
}