URAL 1009 K-based Numbers dp练习

原创于 2015-08-26 20:14:12 发布 · 283 阅读

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本文介绍了如何计算在给定的进制和位数条件下，形成的有效数字总数，有效数字定义为不包含连续两个零的数字。通过动态规划的方法，逐步构建解，最终得出所需答案。

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;
}