URAL - 1009 - K-based Numbers （简单DP）

1009. K-based Numbers

Time limit: 1.0 second
Memory limit: 64 MB

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  Ndigits.

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	output
2 10	90

Problem Source: USU Championship 1997

Tags:  dynamic programming   ( hide tags for unsolved problems )

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;

int n, k;
LL dp[55];//dp[i]表示位数为i时可以组成的数的个数，注意0在这里可以看做0位数 

int main() {
	while(scanf("%d %d", &n, &k) != EOF) {
		memset(dp, 0, sizeof(dp));
		
		dp[1] = k - 1;
		dp[2] = (k - 1) * k;
		for(int i = 3; i <= n; i++) {
			dp[i] = (k - 1) * (dp[i - 1] + dp[i - 2]);
		}
		
		cout << dp[n] <<endl;
	}
	return 0;
}