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\n",dp[n]); } return 0; }
时间: 2024-08-10 01:44:45