1013. K-based Numbers. Version 2

最新推荐文章于 2024-06-19 15:58:07 发布

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最新推荐文章于 2024-06-19 15:58:07 发布

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分类专栏： Timus 文章标签： numbers c

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本博客探讨了在给定位数和基数的情况下，如何计算满足特定条件的K进制数的数量。通过递归公式和动态规划方法，详细解释了如何计算有效K进制数的总数。

1012. K-based Numbers. Version 2

Time Limit: 1.0 second
Memory Limit: 16 MB

Let’s consider K-based numbers, containing exactly N digits. We define a number to be valid if its K-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 ≤ 180.

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

dp:f(n) = (k-1) * (f(n-1)+f(n-2))

1013. K-based Numbers. Version 3 同秒

#include<cstdio>

int main()
{
	int a[2000] = {0}, b[2000] = {0}, c[2000] = {0};
	int g, i, j, m = 1;
	int n, k;
	scanf("%i %i",&n,&k);
	a[0] = 1; 
	b[0] = k - 1;
	for(i=2; i<=n; i++)
	{
		g = 0;
		for(j=0; j<m; j++)
		{
			c[j] = a[j] + b[j];
			c[j] = c[j]*(k-1) + g;
			if(c[j]>9)
			{
				g=c[j]/10;
				c[j]=c[j]%10;
			}
			else
				g = 0;
			a[j] = b[j]; 
			b[j] = c[j];
		}
		if(g)
			b[m++]=g;
	}
	for(j=m-1; j>=0; j--)
		printf("%d", b[j]);
}