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).
Examples
input
Copy
2 15
output
Copy
69 96
input
Copy
3 0
output
Copy
-1 -1
我刚开始以为是要把所有和等于s的都写出来，心态都炸穿了。。。
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
//#include <xfunctional>#define ll long long
#define mod 998244353
usingnamespace std;
int dir[4][2] = { {0,1},{0,-1},{-1,0},{1,0} };
constlonglong inf = 0x7f7f7f7f7f7f7f7f;
constint INT = 0x3f3f3f3f;
bool can(int m, int s)
{
return s >= 0 && s <= 9 * m;
}
int main()
{
int m, s;
cin >> m >> s;
if ((s==0 && m>1) || can(m,s)==false)
{
cout << "-1 -1";
return0;
}
if (m == 1 && s == 0)
{
cout << "0 0";
return0;
}
string minn="";
int sum = s;
for (int i = 1; i <= m; i++)
{
for (int d = 0; d < 10; d++)
{
if ((i > 1 || d > 0 || (m == 1 && d == 0)) && can(m - i, sum - d))
{
minn += to_string(d);
sum -= d;
break;
}
}
}
sum = s;
string maxn;
for (int i = 1; i <= m; i++)
{
for (int d = 9; d >= 0; d--)
{
if (can(m - i, sum - d))
{
maxn += to_string(d);
sum -= d;
break;
}
}
}
cout << minn << "" << maxn;
return0;
}
