动态规划
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 |
题解:因为数据比较小N+K <= 18,所以int够了,接着当只有一位时,对于k进制,很自然地有k-1种情况,两位时就有(k-1)*k种情况,从有n(n>=3)位开始,我们可以在n-1位的基础上在最后面补非0的数,所以一共有k-1个数,或者在k-2位的情况下先补一个0再补k-1个数,这样得到的数就会和k-1位的不同,以此类推,我们得到递归关系,arr[n]=(arr[n-1]+arr[n-2)*(k-1);
#include <iostream>
using namespace std;
int arr[100000];
int main()
{
int n,k;
cin >> n >> k;
arr[1]=k-1;
arr[2]=k*(k-1);
for(int i=3; i <= n; i++)
arr[i]=(k-1)*(arr[i-1]+arr[i-2]);
cout << arr[n] << endl;
}