📜  计算绘制N x 3网格的独特方法的数量

📅  最后修改于: 2021-06-25 15:31:11             🧑  作者: Mango

给定一个整数N ,任务是使用红色黄色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:

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Red Green Red ,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:

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


输出:
12

时间复杂度: O(N)
辅助空间: O(1)