CodeForces 489C Given Length and Sum of Digits... （贪心）

Given Length and Sum of Digits...

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121332#problem/F

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

题意：

给出m和s，分别找出最大和最小的非负整数，满足数位长度为m，数位和为s;

题解：

直接贪心枚举每一位就可以了；

最小：能放0就放0; (首位不能是0)

最大：能放9就放9;

WA一次：题目特地强调了非负整数；

那么考虑s为0的情况，并不是全为-1 -1;

当m=1 s=0时应输出0 0;

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 150
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int m,s;

int main(int argc, char const *argv[])
{
//IN;

int n,s;
while(scanf("%d %d", &n, &s) != EOF)
{
if(n*9<s) {printf("-1 -1\n");continue;}

if(!s) {
if(n==1) printf("0 0\n");
else printf("-1 -1\n");
continue;
}

int cur = s;
for(int i=n; i>=1; i--) {
int flag = 0;
if(i==n) flag = 1;
if(9*(i-1) <= cur-flag) {
printf("%d", cur-9*(i-1));
for(int j=1; j<=i-1; j++)
printf("9");
break;
}
if(flag) {
printf("1"); cur-=1;
} else {
printf("0");
}
}

printf(" ");

cur = s;
for(int i=n; i>=1; i--) {
if(9 >= cur) {
printf("%d", cur);
for(int j=1; j<=i-1; j++)
printf("0");
break;
}

printf("9"); cur-=9;
}

printf("\n");
}

return 0;
}