ural 1009. K-based Numbers dp 高精度

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1009. K-based Numbers

Time limit: 1.0 second

Memory limit: 64 MB

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

2
10

90

Problem Source: USU Championship 1997

dp[i][0]代表第i位不存在0的情况，dp[i][1]代表第i位存在1的情况。

dp[i][1]=dp[i-1][0]表示在前i-1个数中最后一个数不是0 的情况补0

dp[i][0]=(k-1)*(dp[i-1][0]+dp[i-1][1])表示在前i-1位中无论最后一位是不是0都补1~9的情况

//0.125	3 854 KB
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);// 输入
		BigInteger dp[][] = new BigInteger[2007][2];
		int n, k;
		n = cin.nextInt();
		k = cin.nextInt();
		dp[1][0] = (BigInteger.valueOf(k - 1));
		dp[1][1] = (BigInteger.valueOf(0));
		for (int i = 2; i <= n; i++) {
			dp[i][0] = (dp[i - 1][0].add(dp[i-1][1])).multiply(BigInteger.valueOf(k-1));
			dp[i][1] = (dp[i - 1][0]);
		}
		System.out.println(dp[n][0].add(dp[n][1]));
	}
}