SDUT——K-based Numbers

原题：

题目描述

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:
(1)1010230 is a valid 7-digit number;
(2)1000198 is not a valid number;
(3)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.

输入

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

输出

The result in decimal notation.

示例输入

2 10

示例输出

90

 

分析：

递推，哎呀，如果我自己递推，肯定哇。

原码；

#include <iostream>

using namespace std;

const int maxn=20;

long long f[maxn],n,k;

int main()

{



    while(cin>>n>>k)

    {

        if(n==1)

        {

            cout<<k<<endl;

            return 0;

        }

        f[1]=k-1;

        f[2]=k*(k-1);

        for( int i=3; i<=n; i++)

        {

            f[i]=(f[i-1]+f[i-2])*(k-1);

        }

        cout<<f[n]<<endl;

    }

    return 0;

}