URAL 1009K-based Numbers dp练习

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.

 题意问的是在不出现连续0的条件下 k进制的N位数共有多少个

                     对于第N个数来说    如果之前的数即N-1位为0 则 dp[N] 加上 dp[N-2] 的k-1倍

                      如果N-1不为0 则加上dp[N-1]的 k-1倍

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;ll dp[50];int main(){    int n ,  k ;    while(scanf("%d%d",&n,&k)!=EOF)    {        dp[1] = k - 1;        dp[2] = k * dp[1];        for(int i = 3; i <= n; i++)        {            dp[i] = (k-1)*(dp[i-1]+dp[i-2]);        }        printf("%I64d\n",dp[n]);    }    return 0;}