使用数字0到K-1组成的最多N个数字的计数，没有相邻的0

给定两个整数N和K ，任务是对最多N个数字进行计数，以使没有两个零相邻且数字范围从0到K-1。
例子：

Input: N = 2, K = 3 
Output: 8 
Explanation: 
There are 8 such numbers such that digits are from 0 to 2 only, without any adjacent 0s: {1, 2, 10, 11, 12, 20, 21, 22}
Input: N = 3, K = 3 
Output: 22 
 

方法：想法是使用动态编程来解决此问题。
令DP [i] [j]为所需数字的数量，直到该数字的第i位，其最后一位为j 。
观察结果：

• 填补一个地方的方法是$(K-1)$
• 众所周知，零不能相邻。因此，当我们的最后一个元素为0时，意味着前一个索引将以1的方式填充，即0。因此，当前位置只能由(K-1)个数字填充。
• 如果最后一位用(K-1)位填充，则当前位可以用0或(K-1)位填充。

基本情况：

• 如果n == 1并且最后一个== K，那么我们可以用(K-1)个数字填充该位置，然后返回(K-1)
• 否则，返回1

递归关系：

When last digit place is not filled by zero then 
 

$dp[i][j] = (K-1)*solve(n-1, K) + (K-1)*solve(n-1, 1)$

When Last digit place is filled by zero then 
 

$dp[i][j] = solve(n-1, K)$

 

下面是上述方法的实现：

C++

// C++ implementation to count the
// numbers upto N digits such that
// no two zeros are adjacent
  
#include 
using namespace std;
  
int dp[15][10];
  
// Function to count the
// numbers upto N digits such that
// no two zeros are adjacent
int solve(int n, int last, int k)
{
    // Condition to check if only
    // one element remains
    if (n == 1) {
  
        // If last element is non
        // zero, return K-1
        if (last == k) {
            return (k - 1);
        }
        // If last element is 0
        else {
            return 1;
        }
    }
  
    // Condition to check if value
    // calculated already
    if (dp[n][last])
        return dp[n][last];
  
    // If last element is non zero,
    // then two cases arise,
    // current element can be either
    // zero or non zero
    if (last == k) {
  
        // Memoize this case
        return dp[n][last]
               = (k - 1)
                     * solve(n - 1, k, k)
                 + (k - 1)
                       * solve(n - 1, 1, k);
    }
  
    // If last is 0, then current
    // can only be non zero
    else {
  
        // Memoize and return
        return dp[n][last]
               = solve(n - 1, k, k);
    }
}
  
// Driver Code
int main()
{
    // Given N and K
    int n = 2, k = 3;
  
    // Function Call
    int x = solve(n, k, k)
            + solve(n, 1, k);
    cout << x;
}

Java

// Java implementation to count the
// numbers upto N digits such that
// no two zeros are adjacent
class GFG{
      
static int[][] dp = new int[15][10];
  
// Function to count the numbers 
// upto N digits such that
// no two zeros are adjacent
static int solve(int n, int last, int k)
{
      
    // Condition to check if only
    // one element remains
    if (n == 1)
    {
          
        // If last element is non
        // zero, return K-1
        if (last == k)
        {
            return (k - 1);
        }
          
        // If last element is 0
        else 
        {
            return 1;
        }
    }
  
    // Condition to check if 
    // value calculated already
    if (dp[n][last] == 1)
        return dp[n][last];
  
    // If last element is non zero,
    // then two cases arise, current 
    // element can be either zero 
    // or non zero
    if (last == k)
    {
          
        // Memoize this case
        return dp[n][last] = (k - 1) *
                        solve(n - 1, k, k) + 
                             (k - 1) * 
                        solve(n - 1, 1, k);
    }
      
    // If last is 0, then current
    // can only be non zero
    else
    {
  
        // Memoize and return
        return dp[n][last] = solve(n - 1, k, k);
    }
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given N and K
    int n = 2, k = 3;
  
    // Function Call
    int x = solve(n, k, k) + 
            solve(n, 1, k);
      
    System.out.print(x);
}
}
  
// This code is contributed by Ritik Bansal

Python3

# Python3 implementation to count the 
# numbers upto N digits such that 
# no two zeros are adjacent 
dp = [[0] * 10 for j in range(15)]
  
# Function to count the numbers
# upto N digits such that no two
# zeros are adjacent 
def solve(n, last, k):
  
    # Condition to check if only 
    # one element remains 
    if (n == 1):
  
        # If last element is non 
        # zero, return K-1 
        if (last == k):
            return (k - 1)
          
        # If last element is 0 
        else:
            return 1
  
    # Condition to check if value 
    # calculated already 
    if (dp[n][last]):
        return dp[n][last]
  
    # If last element is non zero, 
    # then two cases arise, current
    # element can be either zero or
    # non zero
    if (last == k):
  
        # Memoize this case
        dp[n][last] = ((k - 1) * 
                  solve(n - 1, k, k) +
                       (k - 1) * 
                  solve(n - 1, 1, k))
                         
        return dp[n][last]
  
    # If last is 0, then current 
    # can only be non zero
    else:
  
        # Memoize and return
        dp[n][last] = solve(n - 1, k, k)
        return dp[n][last]
  
# Driver code
  
# Given N and K
n = 2
k = 3
  
# Function call
x = solve(n, k, k) + solve(n, 1, k)
  
print(x)
  
# This code is contributed by himanshu77

C#

// C# implementation to count the
// numbers upto N digits such that
// no two zeros are adjacent
using System;
  
class GFG{
      
public static int [,]dp = new int[15, 10];
  
// Function to count the numbers 
// upto N digits such that
// no two zeros are adjacent
public static int solve(int n, int last, int k)
{
      
    // Condition to check if only
    // one element remains
    if (n == 1)
    {
          
        // If last element is non
        // zero, return K-1
        if (last == k)
        {
            return (k - 1);
        }
          
        // If last element is 0
        else
        {
            return 1;
        }
    }
  
    // Condition to check if 
    // value calculated already
    if (dp[n, last] == 1)
        return dp[n, last];
  
    // If last element is non zero,
    // then two cases arise, current 
    // element can be either zero 
    // or non zero
    if (last == k)
    {
          
        // Memoize this case
        return dp[n, last] = (k - 1) *
                        solve(n - 1, k, k) + 
                             (k - 1) * 
                        solve(n - 1, 1, k);
    }
      
    // If last is 0, then current
    // can only be non zero
    else
    {
  
        // Memoize and return
        return dp[n, last] = solve(n - 1, k, k);
    }
}
  
// Driver Code
public static void Main(string[] args)
{
      
    // Given N and K
    int n = 2, k = 3;
  
    // Function Call
    int x = solve(n, k, k) + 
            solve(n, 1, k);
      
    Console.WriteLine(x);
}
}
  
// This code is contributed by SoumikMondal

输出：

8

时间复杂度： O(N)
辅助空间： O(N * 10)