K-based Numbers. Version 2 URAL - 1012 [dp + java大数]_1012. k-based numbers. version 2-优快云博客

本文探讨了如何计算特定长度的K进制数中不含连续两个0的有效组合数量。使用动态规划方法解决这一问题，并提供了Java实现代码。

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1012. K-based Numbers. Version 2

Time limit: 0.5 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 N digits.

You may assume that 2 ≤ K ≤ 10; N ≥ 2; N + K ≤ 1800.

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

Tags: dynamic programming ( hide tags for unsolved problems)

Difficulty: 172 Printable version Submit solution Discussion (53)
All submissions (51204) All accepted submissions (13281) Solutions rating (7492)

题目我已经在之前解释过了，可以看这里;

就是数据有点大，需要用大数。。。我用java写;

代码:

import java.math.BigDecimal;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.util.Scanner;


public class howell {
     public static void main(String[] args)
     {
         Scanner scanf = new Scanner(System.in);
         int n = scanf.nextInt();
         int k = scanf.nextInt();
         BigInteger dp[][] = new BigInteger[1800][2];
         dp[1][0] = BigInteger.ONE;
         dp[1][1] = BigInteger.valueOf(k - 1);
         for(int i = 2; i <= n+1; i++)
         {
             dp[i][1] = dp[i - 1][0].add(dp[i - 1][1]);
             dp[i][1] = dp[i][1].multiply(BigInteger.valueOf(k - 1));
             dp[i][0] = dp[i - 1][1];
         }
         dp[n][1].add(dp[n][0]);
         System.out.println(dp[n][1]);
     }
}

发现不能加包名。。。