C. Given Length and Sum of Digits...

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

 

大数就是从最高位，能大就大；小数就是从最低位，能小就小，再处理下最高位为0的情况。

 

无结果无非一个sum太小，min全为0，一个sum太大，全为9还有剩

 

public class Main {
    public static void main(String[] args) {
        Scanner io = new Scanner(System.in);
        int len = io.nextInt(), sum = io.nextInt();
        int[] max = new int[len + 1];
        int[] min = new int[len + 1];

        int sum1 = sum, sum2 = sum;
        for (int i = 1; i <= len; sum1 -= max[i++]) max[i] = Math.min(9, sum1);
        for (int i = len; i >= 1; sum2 -= min[i--]) min[i] = Math.min(9, sum2);

        //min最高位最少是1
        if (min[1] == 0) for (int i = 2; i <= len; i++)
            if (min[i] > 0) {
                min[i]--;
                min[1]++;
                break;
            }

        //sum太小：min是否全为0
        int nomin = 1;
        for (int i = 1; i <= len; i++) if (min[i] != 0) nomin = 0;

        //sum太大：sum1、sum2没减完
        if ((nomin == 1 && len != 1) || sum1 != 0 || sum2 != 0) System.out.println("-1 -1");
        else {
            for (int i = 1; i <= len; i++) System.out.print(min[i]);
            System.out.print(" ");
            for (int i = 1; i <= len; i++) System.out.print(max[i]);
        }
    }
}