K-based Numbers

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Description

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 ≤ 18.

Input

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

Output

The result in decimal notation.

Sample Input

inputoutput

210

90

题解：因为数据比较小N+K <= 18,所以int够了，接着当只有一位时，对于k进制，很自然地有k-1种情况，两位时就有（k-1）*k种情况，从有n（n>=3）位开始，我们可以在n-1位的基础上在最后面补非0的数，所以一共有k-1个数，或者在k-2位的情况下先补一个0再补k-1个数，这样得到的数就会和k-1位的不同，以此类推，我们得到递归关系，arr[n]=(arr[n-1]+arr[n-2)*(k-1);

#include <iostream>using namespace std;int arr[100000];int main(){int n,k;cin >> n >> k; arr[1]=k-1;arr[2]=k*(k-1);for(int i=3; i <= n; i++)arr[i]=(k-1)*(arr[i-1]+arr[i-2]);cout << arr[n] << endl;}