【ural oj 1009】K-based Numbers(数位dp)_1009. k-based numbers time limit: 0.5 second memor-优快云博客

本文介绍了一种使用动态规划算法解决特定K进制数计数问题的方法，该问题要求计算不含连续两个0的K进制数的数量。通过定义状态转移方程dp[i]来递推计算满足条件的数的个数。

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1009. K-based Numbers

Time limit: 0.5 second
Memory limit: 64 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 ≤ 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

Tags: dynamic programming ( hide tags for unsolved problems)

Difficulty: 96 Printable version Submit solution Discussion (90)
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一个k进制的数，每一位只可能是0到k-1，我们用dp[i]来表示第位数为i的k进制数的个数，显然dp[1]=k-1（0不能出现在首位，位数为2的个数应当是(k-1)*k;由于不能有两个连续的0，因此dp[i](i>2)=dp[i-1]*(k-1)+dp[i-2]*(k-1);

#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 10050
unsigned long long   dp[MAXN]; 
int main()
{
	int n,k;
	cin>>n>>k;
	dp[1]=k-1;
	dp[2]=k*(k-1);
	for(int i=3;i<=n;i++)
	{
		dp[i]=(k-1)*(dp[i-1]+dp[i-2]);
	}
	cout<<dp[n]<<endl;
	return 0;
}