*DP(2)

K-based Numbers

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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

input	output
2 10	90

Source

Problem Source: 

USU Championship 1997  

 

if(a[i-1] == 0) a[i-2] != 0,a[i-2]就有k-1种选择!

else a[i-1] != 0,此时 a[i-1] 也只有k-1种选择！

所以,a[i] = (k-1)*(a[i-1] + a[i-2]);

 

#include<cstdio>
using namespace std;
long long a[20];
int main(){
    long long n,k;
    while(~scanf("%I64d%I64d",&n,&k)){
        a[1] = k-1;
        a[2] = (k-1)*k;
        for(int i = 3; i <= n; i++)
            a[i] = (k-1)*(a[i-1] + a[i-2]);
        printf("%I64d\n",a[n]);
    }
    return 0;
}