Codeforces-489C 题解

贪心算法

题意：找到m位数且各位的数字相加为s；

找最大值时，尽可能让9在高位后面尽可能补零；

找最小值时，尽可能让9在个位，最高位为1；

值得注意的是 m*9=2&&s==0时要输出-1 -1  （被坑了QAQ）

#include
#include
#include
#include
#include
using namespace std;
bool compare(int a, int b)
{
	return a>b;
}
int cmp(const void *a, const void *b)
{
	return *(char *)a - *(char *)b;
}
bool composite(int i)
{
	if (i == 1) return false;
	else if (i == 2 || i == 3) return true;
	else for (int j = 2;ja2 ? a1 : a2;
}
int father[1001];
void initial(int n)
{
	for (int i = 1;i <= n;i++)
		father[i] = i;

}
int find(int a)
{
	while (father[a] != a)
		a = father[a];
	return a;
}
void combine(int a, int b)
{
	int temp1 = find(a);
	int temp2 = find(b);
	if (temp1 != temp2)
		father[temp1] = temp2;


}
int a[3];

int main()
{
	ios::sync_with_stdio(false);
	int m, s, a[101] = { 0 }, b[101] = { 0 };//分别是位数和加起来的值
	while (cin >> m >> s)
	{
		int i = 0,j=0,temp1=s,temp2=m;
		if (m == 1 && s == 0) cout << 0 << ' ' << 0 << endl;
		else if (((m == 1) && s > 9)||(s==0)||m*9 9)
		{
			if (m == 1) cout << -1 << ' ' << -1 << endl;
			else {
				while (s > 9 && m > 0)
				{
					a[i] = 9;
					m--;
					s -= 9;
					i++;
				}
				int u = i + m - 1;
				if (m > 1) {
					a[u] = 1;
					a[i] = s - 1;
					for (int k = i + 1;k < u;k++)
						a[k] = 0;
				}
				else a[u] = s;
				for (int k = u;k >= 0;k--)
					cout << a[k];
				printf(" ");
				while (temp1 > 9 && temp2 > 0)
				{
					b[temp2 - 1] = 9;
					temp2--;
					temp1 -= 9;
					j++;
				}
				int p = temp2 - 1;
				b[p] = temp1;
				for (int k = p + j;k >= 0;k--)
					cout << b[k];
				printf("\n");
			}
		}
		else if (s <= 9)
		{
			if (m == 1) printf("%d %d\n", s, s);
			else {
				a[0] = s - 1;
				a[m - 1] = 1;
				for (int k = 1;k < m - 1;k++)
					a[k] = 0;
				b[m - 1] = s;
			
			for (int k = m-1;k >= 0;k--)
				cout << a[k];
			printf(" ");
			for (int k = m - 1;k >= 0;k--)
				cout << b[k];
			printf("\n");}
			
		}
		

	}
}