K-based Numbers-优快云博客

本文探讨了如何计算特定条件下K进制数的有效数量，避免连续的零出现。使用递归函数解决这一问题，通过C语言实现算法，并提供了一个示例程序。该程序能够接收用户输入的N和K值，计算并输出所有N位K进制数中有效数的数量。

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K-based Numbers

Time Limit: 1000MS Memory limit: 65536K

题目描述

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: (1)1010230 is a valid 7-digit number; (2)1000198 is not a valid number; (3)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.

输入

The numbers N and K in decimal notation separated by the line break.

输出

The result in decimal notation.

示例输入

2 10

示例输出

#include<stdio.h>
#include<string.h>
int f[101];
int k;
int D(int n,int k)
{
    if(n==1)
        return f[1]=k-1;
    if(n==2)
        return f[2]=(k-1)*k;
    else
        return f[n]=(k-1)*(D(n-1,k)+D(n-2,k));
}
int main()
{
    int n;
    while(~scanf("%d %d",&n,&k))
    {
        memset(f,0,sizeof(f));
        D(n,k);
        printf("%d\n",f[n]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/guoyongzhi/p/3232992.html