Ural - Timus - 1009 K-based Numbers 题解

Let’s consider K-based numbers, containing exactly N digits. We define a number to be valid if itsK-based notation doesn’t contain two successive zeros. For example:

1010230 is a valid 7-digit number;
1000198 is not a valid number;
0001235 is not a 7-digit number, it is a 4-digit number.

Given two numbers N and K, you are to calculate an amount of valid K based numbers, containing Ndigits.

You may assume that 2 ≤ K ≤ 10; N ≥ 2; N + K ≤ 18.

Input

The numbers N and K in decimal notation separated by the line break.

Output

The result in decimal notation.

Sample

input	output
2 10	90

Problem Source: USU Championship 1997
http://acm.timus.ru/problem.aspx?space=1&num=1009
本题题意：

1 N代表数字的位数， K代表是以什么为基的，如10进制，2进制的10和2

2 判断数值是否合法：开头不能为零， 而不能有两个连续的0

利用动态规划法，可以很好地解决这个问题。

唯一的拐弯处： 要分开处理当前位是0和不是零的情况，所以本程序利用两个数列分别保存这两种情况，最后相加起来就是结果了：

long long calculate(const int n, const int k)
{
	int Nze[16] = {0};
	int Ze[16] = {0};
	Nze[0] = k - 1;
	for (int i = 1; i < n; i++)
	{
		Ze[i] = Nze[i-1];
		Nze[i] = (Ze[i-1] + Nze[i-1]) * (k-1);
	}
	return Nze[n-1] + Ze[n-1];
}

void K_basedNumbers()
{
	int n = 0, k = 0;
	cin>>n>>k;
	cout<<calculate(n, k)<<endl;
}
int main()
{
	K_basedNumbers();
	return 0;
}

当然也可以使用常量保存结果，不过本题的数值都不大，可以认为是空间效率是O(1)了。没有改进的太大必要。