1 second
256 megabytes
standard input
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.
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.
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).
2 15
69 96
3 0
-1 -1
贪心，考虑最大的时候把数组赋值为9，考虑最小的时候把数组第一位赋值1，其他为赋值0
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int s1[110], s2[110];
int main()
{
int m, s;
while (~scanf("%d%d", &m, &s))
{
if (s == 0 && m == 1)
{
printf("0 0\n");
continue;
}
if (m * 9 < s || s < 1) //每一位全是9都小于s，或者1.....0大于s
{
printf("-1 -1\n");
continue;
}
for (int i = 1; i <= m; ++i)
{
s1[i] = 0;
s2[i] = 9;
}
s1[1] = 1;
int dis = s - 1, cnt = m;
while (1)
{
if (9 - s1[cnt] >= dis)
{
s1[cnt] += dis;
break;
}
dis -= (9 - s1[cnt]);
s1[cnt] = 9;
cnt--;
}
for (int i = 1; i <= m; ++i)
{
printf("%d", s1[i]);
}
printf(" ");
dis = m * 9 - s;
cnt = m;
while (1)
{
if (s2[cnt] >= dis)
{
s2[cnt] -= dis;
break;
}
dis -= (s2[cnt]);
s2[cnt] = 0;
cnt--;
}
for (int i = 1; i <= m; ++i)
{
printf("%d", s2[i]);
}
printf("\n");
}
return 0;
}