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

 1 public class Main {
 2     public static void main(String[] args) {
 3         Scanner io = new Scanner(System.in);
 4         int len = io.nextInt(), sum = io.nextInt();
 5         int[] max = new int[len + 1];
 6         int[] min = new int[len + 1];
 7
 8         int sum1 = sum, sum2 = sum;
 9         for (int i = 1; i <= len; sum1 -= max[i++]) max[i] = Math.min(9, sum1);
10         for (int i = len; i >= 1; sum2 -= min[i--]) min[i] = Math.min(9, sum2);
11
12         //min最高位最少是1
13         if (min[1] == 0) for (int i = 2; i <= len; i++)
14             if (min[i] > 0) {
15                 min[i]--;
16                 min[1]++;
17                 break;
18             }
19
20         //sum太小：min是否全为0
21         int nomin = 1;
22         for (int i = 1; i <= len; i++) if (min[i] != 0) nomin = 0;
23
24         //sum太大：sum1、sum2没减完
25         if ((nomin == 1 && len != 1) || sum1 != 0 || sum2 != 0) System.out.println("-1 -1");
26         else {
27             for (int i = 1; i <= len; i++) System.out.print(min[i]);
28             System.out.print(" ");
29             for (int i = 1; i <= len; i++) System.out.print(max[i]);
30         }
31     }
32 }

原文地址：https://www.cnblogs.com/towerbird/p/11257227.html