K-based Numbers

Accept:141??	??Submit:314
Time Limit:1000MS??	??Memory Limit:65536KB

Description

Let’s considerK-based numbers, containing exactlyN 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-digitnumber;

• 1000198 is not a validnumber;

• 0001235 is not a 7-digitnumber, it is a 4-digit number.

Given two numbersNandK,you are to calculate an amount of validK based numbers, containingN digits.

You may assume that 2 ≤K≤ 10;N≥ 2;N+K≤ 18.

Input

The numbersN andK in decimal notation separated by the line break.

Output

The result in decimal notation.

SampleInput

2

10

SampleOutput

90

1 N代表數字的位數， K代表是以什么為基的，如10進制，2進制的10和22 判斷數值是否合法：開頭不能為零， 而不能有兩個連續的0設f(n)表示符合題目條件的n位K進制的數的總數。則有：f(n)=(k-1)*(f(n-1)+f(n-2)),且f(1)=k-1,f(2)=k*(k-1)。當n==1時，直接輸出k,進行特判！

#include <stdio.h> int main(){int f[20];int n,k;scanf("%d%d",&n,&k);if(n==1) {printf("%d\n",k);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);printf("%d\n",f[n]);return 0; }

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