K-based Numbers
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K-based Numbers
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/16384K (Java/Other)
Total Submission(s) : 22 Accepted Submission(s) : 14
Problem 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 21090
题目大意:
给定一个k进制数,求n可以组成多少n位有效的数(没有两个连续的零)
我们设f[i]为k进制i位没有有两个连续零的有效数的个数
则若第i位是0,则前面i-2位就有f[i-2]种选择,第i-1位有k-1种选择(i-1!=0),所以有f[i-2]*(k-1);
若第i位非0则第i位有k-1种选择,i-1有f[i-1]种选择,共有f[i-1]*(k-1)种;
所以状态转移方程为
f[i]=f[i-2]*(k-1)+f[i-1]*(k-1);
代码如下:
#include<iostream>using namespace std;int main(){ int f[19]={0},i,j,n,k; cin>>n>>k; f[1]=k-1; f[0]=1; for (i=2;i<=n;i++) f[i]=f[i-2]*(k-1)+f[i-1]*(k-1); cout<<f[n]; return 0;}
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