Codeforces 489C 简单构造

传送门：

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

Given Length and Sum of Digits...
[align=center]time limit per test[/align]
[align=center]1 second[/align]
[align=center]memory limit per test[/align]
[align=center]256 megabytes[/align]
[align=center]input[/align]
[align=center]standard input[/align]
[align=center]output[/align]
[align=center]standard output[/align]
You have a positive integer m and a non-negative integers. Your task is to find the smallest and the largest of the numbers that have lengthm
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

题意：给一个数的位数和每一位的和，问这个数最大是多少，最小是多少。不存在输出-1

思路：先检查合法性，是否输出-1 -1；

构造最小数：从尾部开始构造，直接填9（或者剩下的数的和-1），最后剩下的数应该大于等于1，这个填到首位。其他地方填0.

构造最大数：从首位开始构造，直接填9（或者剩下的数的和），其他的地方直接填0.

代码：

#include<cstdio>
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
if(m==0&&n==1)
{
printf("0 0\n");
continue;
}
if(m<1||n*9<m)
{
printf("-1 -1\n");
continue;
}
char s[200]={0};
for(int i=1;i<=n;i++)
{
int tmp=m;
if(tmp>=9)
{
s[i]=9+48;
m-=9;
}
else
{
s[i]=tmp+48;
m-=tmp;
}
}
char temp[105];
sprintf(temp,"%s",s+1);
if(s
=='0')
{
s
++;
for(int i=n-1;i>=1;i--)
if(s[i]>48)
{
s[i]--;
break;
}
}
for(int i=n;i>=1;i--) printf("%c",s[i]);
printf(" ");
printf("%s\n",temp);
}
return 0;
}