Codeforces Round #277.5 (Div. 2)-C. Given Length and Sum of Digits （贪心）_html/css_WEB-ITnose

Given Length and Sum of Digits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

2 15

output

69 96

input

3 0

output

-1 -1

题意：给两个数m, s，求出位数等于m 且个位数字之和等于s的最小数和最大数。

解题思路：直接贪心的去构造。先判断输出-1，-1的情况：m*9

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint a[105], b[105];int main(){//        freopen("in.txt","r",stdin);    int m, s;    while(scanf("%d%d", &m, &s)!=EOF)    {        if(!s && m==1) {printf("0 0\n"); continue; }       //m == 1 && s == 0        if(s > m*9 || s == 0) {               //构造不成功的情况            cout0; i--){                       int x = min(ss, 9);            b[i] = x;            ss -= x;        }        if(ss) b[0] += ss;        for(int i=0; i<m i x cout return>  <br>  <br>  <p></p>  <p><br> </p>  <p><br> </p>  <p><br> </p> </m></time.h></stdlib.h></math.h></string></map></set></queue></vector></algorithm></iostream></string.h></stdio.h>