计算绘制N x 3网格的独特方法的数量

给定一个整数N ，任务是使用红色，黄色或绿色绘制一个大小为N x 3的网格，同时使相邻的像元对没有相同的 颜色。打印各种可能的不同方式

例子：

Input: N = 1
Output: 12
Explanation: 
Following 12 possible ways to paint the grid exists:

1. Red, Yellow, Red
2. Yellow, Red, Yellow
3. Green, Red, Yellow
4. Red, Yellow, Green
5. Yellow, Red, Green
6. Green, Red, Green
7. Red, Green, Red
8. Yellow, Green, Red
9. Green, Yellow, Red
10. Red, Green, Yellow
11. Yellow, Green, Yellow
12. Green, Yellow, Green

Input: N = 2
Output: 102

方法：请按照以下步骤解决问题：

• 为行着色的方法可以分为以下两类：
  • 最左边和最右边的单元格具有相同的颜色。
  • 最左边和最右边的单元格具有不同的颜色。
• 考虑第一种情况：
  • 存在六种可能的方式来绘制行，以使最左边和最右边的颜色相同。
  • 对于同时占据最左边和最右边单元格的每种颜色，都有两种不同的颜色可用来对中间行进行着色。
• 考虑第二种情况：
  • 存在六种可能的方式来绘制最左边和最右边的颜色是不同的。
  • 左边的单元格有三个选择，中间的单元格有两个选择，并用唯一剩余的颜色填充最右边的单元格。因此，可能性的总数为3 * 2 * 1 = 6 。
• 现在，对于随后的单元格，请看以下示例：
  • 如果将前一行绘制为红色绿色红色，则可以通过以下五种有效方法为当前行着色：
    • {绿色红色绿色}
    • {绿色红色黄色}
    • {绿色黄绿色}
    • {黄色红色绿色}
    • {黄色红色黄色}
  • 从以上观察可知，三种可能性的末端具有相同的颜色，而两种可能性的末端具有不同的颜色。
  • 如果前一行的颜色为红色绿色黄色，则存在以下四种为当前行着色的可能性：
    • {绿色红色绿色}
    • {绿色黄色红色}
    • {绿色黄绿色}
    • {黄色红色绿色}
  • 从上面的观察中可以明显看出，可能性的末端具有相同的颜色，而两种可能性的末端具有不同的颜色。
• 因此，基于上述观察，可以为绘制N行的方式数量定义以下递归关系：

Count of ways to color current row having ends of same color S_N+1 = 3 * S_N + 2D_N

Count of ways to color current row having ends of different colors D_N+1 = 2 * S_N + 2D_N

Total number of ways to paint all N rows is equal to the sum of S_N and D_N.

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the number
// of ways to paint  N * 3 grid
// based on given conditions
void waysToPaint(int n)
{
 
    // Count of ways to pain a
    // row with same colored ends
    int same = 6;
 
    // Count of ways to pain a row
    // with different colored ends
    int diff = 6;
 
    // Traverse up to (N - 1)th row
    for (int i = 0; i < n - 1; i++) {
 
        // Calculate the count of ways
        // to paint the current row
 
        // For same colored ends
        same = 3 * same + 2 * diff;
 
        // For different colored ends
        diff = 2 * same + 2 * diff;
    }
   
    // Print the total number of ways
    cout << (same + diff);
}
 
// Driver Code
int main()
{
    int N = 1;
    waysToPaint(N);
}
 
// This code is contributed by ukasp.

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
    // Function to count the number
    // of ways to paint  N * 3 grid
    // based on given conditions
    static void waysToPaint(int n)
    {
 
        // Count of ways to pain a
        // row with same colored ends
        long same = 6;
 
        // Count of ways to pain a row
        // with different colored ends
        long diff = 6;
 
        // Traverse up to (N - 1)th row
        for (int i = 0; i < n - 1; i++) {
 
            // Calculate the count of ways
            // to paint the current row
 
            // For same colored ends
            same = 3 * same + 2 * diff;
 
            // For different colored ends
            diff = 2 * same + 2 * diff;
        }
 
        // Print the total number of ways
        System.out.println(same + diff);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int N = 1;
 
        // Function call
        waysToPaint(N);
    }
}

Python3

# Python3 program for the above approach
 
# Function to count the number
# of ways to paint  N * 3 grid
# based on given conditions
def waysToPaint(n):
 
    # Count of ways to pain a
    # row with same colored ends
    same = 6
     
    # Count of ways to pain a row
    # with different colored ends
    diff = 6
 
    # Traverse up to (N - 1)th row
    for _ in range(n - 1):
 
        # Calculate the count of ways
        # to paint the current row
         
        # For same colored ends
        same = 3 * same + 2 * diff
         
        # For different colored ends
        diff = 2 * same + 2 * diff
 
    # Print the total number of ways
    print(same + diff)
 
 
# Driver Code
 
N = 1
waysToPaint(N)

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to count the number
// of ways to paint N * 3 grid
// based on given conditions
static void waysToPaint(int n)
{
     
    // Count of ways to pain a
    // row with same colored ends
    int same = 6;
 
    // Count of ways to pain a row
    // with different colored ends
    int diff = 6;
 
    // Traverse up to (N - 1)th row
    for(int i = 0; i < n - 1; i++)
    {
         
        // Calculate the count of ways
        // to paint the current row
 
        // For same colored ends
        same = 3 * same + 2 * diff;
 
        // For different colored ends
        diff = 2 * same + 2 * diff;
    }
     
    // Print the total number of ways
    Console.WriteLine(same + diff);
}
 
// Driver code
static public void Main()
{
    int N = 1;
     
    waysToPaint(N);
}
}
 
// This code is contributed by offbeat

Javascript

 
// Javascript program for the above approach
 
// Function to count the number
// of ways to paint  N * 3 grid
// based on given conditions
function waysToPaint(n)
{
     
    // Count of ways to pain a
    // row with same colored ends
    var same = 6;
 
    // Count of ways to pain a row
    // with different colored ends
    var diff = 6;
 
    // Traverse up to (N - 1)th row
    for(var i = 0; i < n - 1; i++)
    {
         
        // Calculate the count of ways
        // to paint the current row
 
        // For same colored ends
        same = 3 * same + 2 * diff;
 
        // For different colored ends
        diff = 2 * same + 2 * diff;
    }
 
    // Print the total number of ways
    document.write(same + diff);
}
 
// Driver code
var N = 1;
 
// Function call
waysToPaint(N);
         
// This code is contributed by Khushboogoyal499
 

输出：

12

时间复杂度： O(N)
辅助空间： O(1)