URAL1009——DP——K-based Numbers

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

 大意：和flag那道一样，如果前面是1的话,当前状态是由dp[n-1]转移过来，若果前面是0的话说明前面的前面是1当前的状态是由dp[n-2]转移过来的（0是肯定要的），对于每一种情况都有（k-1）种方法，一开始考虑0的状态想错了。。WA了几发

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    long long dp[200];
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        dp[1] = k - 1;
        dp[2] = k*(k-1);
        for(int i = 3; i <= n ; i ++)
            dp[i] = (k-1)*(dp[i-1] + dp[i-2]);
        printf("%lld
",dp[n]);
    }
    return 0;
}