Codeforces Round #277.5 (Div. 2)--c greedy--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 has length m and sum of digits s. The required numbers should is 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

/* Greedy for min first to ensure that there are 1 from the back to add, if the addition after the S, indicating no, if not even 1 is not so much for Max, if s 9 is not enough for the other special cases to be sentenced, such as 1 0 pit point should be output 0 0. The other 0 are -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;}

　　

Codeforces Round #277.5 (Div. 2)--c greedy--given Length and Sum of Digits