Codeforces Round #277.5 (Div. 2)-C. Given Length and Sum of Digits...

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

C. 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

解题思路:构造，给你n和s表示你需要构造的数字的位数合个位数字相加的和，你需要找到满足条件的最大和最小值。最小值，从最后一位开始放数字，直到放不了，当然首位不能为零。最大值，从头开始放，没什么要注意的- -。

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 
 6 const int maxn = ;
 7 
 8 int a[maxn], b[maxn], m, s;
 9 
 void solve(){
     int cnt1, cnt2, temp1, temp2;
     int i;
     //特判
     if(m ==  && s == ){
         printf("0 0\n"); return ;
     }
     //无法构造的情况
     if(s > m *  || (m >  && s == )){
         printf("-1 -1\n"); return ;
     }
     memset(a, , sizeof(a));
     memset(b, , sizeof(b));
     temp1 = s; cnt1 = ;
     a[m - ] = ; temp1--;
     for(i = ; i < m - ; i++){
         a[i] = ;
     }
     while(temp1 > ){
         a[cnt1++] = ;
         temp1 -= ;
     }
     if(temp1 > ){
         a[cnt1] = a[cnt1] + temp1;
         cnt1++;
     }
     temp2 = s; cnt2 = ;
     while(temp2 > ){
         b[cnt2++] = ;
         temp2 -= ;
     }
     if(temp2 > ){
         b[cnt2++] = temp2;
     }
     while(cnt2 < m){
         b[cnt2++] = ;
     }
     for(i = m - ; i >= ; i--){
         printf("%d", a[i]);
     }
     printf(" ");
     for(i = ; i < cnt2; i++){
         printf("%d", b[i]);
     }
     printf("\n");
 }
 
 int main(){
     while(scanf("%d %d", &m, &s) != EOF){
         solve();
     }
     return ;

62 }