CodeForces 489C Given Length and Sum of Digits...

题目链接  ：http://codeforces.com/problemset/problem/489/C

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 lengthm and sum of digitss. 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 integersm,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



题意很简单，就是找出最小和最大的满足字符宽为s，和为m的数！！！注意坑点！！！要仔细考虑 0 0 和 -1 -1 的状况

思路：

       由于字符宽度太大，采用字符串，a,b分别记最小的和最大的，要求就是a的头位元素尽量小，b的头位元素尽量大

代码一：

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;#include <string>int main(){    int n, m;    scanf("%d%d", &n, &m);    if(n==1 && m==0)    {        printf("0 0\n");        return 0;    }    if(m<1 || m>n*9)    {        printf("-1 -1\n");        return 0;    }    string a="", b="";    for(int i=0;i<n;i++)    {        int x=min(9, m);        m-=x, a+=(x+'0');    }    b=a;    reverse(a.begin(), a.end());    if(a[0]=='0')        for(int i=0;i<n;i++)            if(a[i]!='0')            {                ++a[0], --a[i];                break;            }    cout<<a<<' '<<b;    return 0;}

代码二：

（与代码一思路差不多，但是显得繁琐很多）

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <string>#include <math.h>#include <stdlib.h>#include <map>#include <time.h>using namespace std;typedef long long LL;const int INF=0x3f3f3f3f;#define pi 3.1415926535897932384626#define eps 1e-8#define N 1005string s1, s2;int main(){    int m,s,s1,s2,i,a[105],b[105];while(~scanf("%d%d",&m,&s))    {        if(m==1&&s==0)        {            printf("0 0\n");            continue;        }        else if(s==0||s>9*m)        {            printf("-1 -1\n");            continue;        }        else        {            s1=s2=s;            for(i=m-1;i>0;i--)            {                if(s1>9)                {                    a[i]=9;                    s1-=9;                }                else if(s1>1)                {                    a[i]=s1-1;                    s1=1;                }                else if(s1==1)                    a[i]=0;            }            a[0]=s1;            for(i=0;i<m;i++)            {                if(s2>9)                {                    b[i]=9;                    s2-=9;                }                else if(s2>0)                {                    b[i]=s2;                    s2=0;                }                else if(s2==0)                    b[i]=0;            }        }        for(i=0;i<m;i++)            printf("%d",a[i]);        printf(" ");        for(i=0;i<m;i++)            printf("%d",b[i]);        printf("\n");    }return 0;}