Codeforces Round #277.5 (Div. 2) C

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：给你m,s两个数,m代表位数,s代表这个数各个位上的数加起来的和，让我们求这样能组成的最大数和最小数，如果不存在则输出-1 -1。 
          
 
          

            做法：先判断是否存在这样的数，首先判断m*9>s或者(m>1&&s=0)是否成立，如果成立则不存在这样的数。最大值比较好找，从最前位开始每一位取min(s,9),取完之后s-取得值。最小值则要注意一下最前位，从最后一位开始，如果不是最前位则取min（s-1,9）,如果是最前位则取min(s,9)。 
          
 
          
 
           #include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#define esp 1e-6
#define LL long long
#define inf 0x0f0f0f0f
using namespace std;
int main()
{
    int m,s,s1;
    int i,j,sum;
    int num[105],num2[105];
    while(scanf("%d%d",&m,&s)!=EOF)
    {
        s1=s;
        memset(num,0,sizeof(num));
        memset(num2,0,sizeof(num2));
        if(m==1&&s==0)
        {
            printf("0 0\n");
            continue;
        }
        if(m*91&&s==0))
        {
            printf("-1 -1\n");
            continue;
        }
        if(m==1)
        {
            printf("%d %d\n",s,s);
            continue;
        }
        for(i=0;i=9)
            {
                num[i]=9;
                s-=9;
            }
            else
            {
                num[i]=s;
                s-=num[i];
            }
        }
        for(i=m-1;i>=0;i--)
        {
            if(s1>=10&&i!=0)
            {
                num2[i]=9;
                s1-=9;
            }
            else if(s1>=9&&i==0)
                {
                     num2[i]=9;
                     s1-=9;
                }
            else if(s1>=2&&s1<=9&&i!=0)
            {
                num2[i]=s1-1;
                s1-=num2[i];
            }
            else if(s1>=1&&s1<=9&&i==0)
            {
                num2[i]=s1;
                s1-=num2[i];
            }
            else
                num2[i]=0;
        }
        for(i=0;i 
           
 
          
 
          
 
         
 
        
 
       

Codeforces Round #277.5 (Div. 2) C

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