题意：给位数和位数和 求符合要求的最大值和最小值
//想法都有了，可是却没有做粗来（你484傻啊?? ?(?_?)? ??）
Description
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 Input
Input
2 15
Output
69 96
Input
3 0
Output
-1 -1
1 #include<stdio.h>
2 #include<math.h>
3 #include<algorithm>
4 using namespace std;
5 int main()
6 {
7 int m,s,s1,s2,i;
8 int a[101],b[101];
9 while(~scanf("%d %d",&m,&s))
10 {
11 int sum=0;
12 if(m==1&&s==0)
13 puts("0 0");
14 else if(s>9*m||s==0)
15 puts("-1 -1");
16 else
17 {
18 s1=s2=s;
19 for(i=m-1;i>=1;i--)
20 {
21 if(s1>9)
22 {
23 a[i]=9;
24 s1-=9;
25 }
26 else if(s1>1)
27 {
28 a[i]=s1-1;
29 s1=1;
30 }
31 else if(s1==1)
32 a[i]=0;
33 }
34 a[0]=s1;
35 for(i=0;i<m;i++)
36 {
37 if(s2>9)
38 {
39 b[i]=9;
40 s2-=9;
41 }
42 else if(s2>0)
43 {
44 b[i]=s2;
45 s2=0;
46 }
47 else if(s2==0)
48 b[i]=0;
49 }
50 for(i=0;i<m;i++)
51 printf("%d",a[i]);
52 printf(" ");
53 for(i=0;i<m;i++)
54 printf("%d",b[i]);
55 printf("\n");
56 }
57 }
58 return 0;
59 }