使用4种颜色为3 * N板着色的方法

给定一块3 X n的木板，找到最多使用4种颜色对其进行着色的方法，以使没有两个相邻的盒子具有相同的颜色。对角邻居不被视为相邻框。
随着答案的迅速增长，请输出％1000000007的方式。
限制条件：
1 <= n <100000
例子 ：

输入：1输出：36我们可以使用3种颜色或2种颜色的组合。现在，从4种颜色中选择3种颜色是${4}\choose{3}$并将它们排列成3个！方式，类似地从4种颜色中选择2种是${4}\choose{2}$而在安排时，我们只能选择其中哪一个位于中心位置，那就是两种方式。答案= ${4}\choose{3}$ * 3！ + ${4}\choose{2}$ * 2！ = 36输入：2输出：588

我们将使用动态方法解决此问题，因为当在板上添加新列时，填充颜色的方式仅取决于当前列中的颜色模式。一列中只能有两种颜色和三种颜色的组合。图像中给出了可以生成的所有可能的新列。请考虑A，B，C和D为4种颜色。

可以从当前列生成的所有可能的颜色组合。

从现在开始，我们将3 * N板第N列的3种颜色组合称为W(n)，将2种颜色称为Y(n)。
我们可以看到每个W可以产生5Y和11W，每个Y可以产生7Y和10W。我们从这里得到两个方程
现在我们有两个方程，

W(n+1) = 10*Y(n)+11*W(n);
Y(n+1) = 7*Y(n)+5*W(n);

C++

// C++ program to find number of ways
// to color a 3 x n grid using 4 colors
// such that no two adjacent have same
// color
#include 
using namespace std;
 
int solve(int A)
{
     
    // When we to fill single column
    long int color3 = 24;
    long int color2 = 12;
    long int temp = 0;
     
    for (int i = 2; i <= A; i++)
    {
        temp = color3;
        color3 = (11 * color3 + 10 *
              color2 ) % 1000000007;
               
        color2 = ( 5 * temp + 7 *
              color2 ) % 1000000007;
    }
     
    long num = (color3 + color2)
                       % 1000000007;
                        
    return (int)num;
}
 
// Driver code
int main()
{
    int num1 = 1;
    cout << solve(num1) << endl;
 
    int num2 = 2;
    cout << solve(num2) << endl;
     
    int num3 = 500;
    cout << solve(num3) << endl;
 
    int num4 = 10000;
    cout << solve(num4);
     
    return 0;
}
 
// This code is contributed by vt_m.

Java

// Java program to find number of ways to color
// a 3 x n grid using 4 colors such that no two
// adjacent have same color.
public class Solution {
    public static int solve(int A) {
        long color3 = 24; // When we to fill single column
        long color2 = 12;
        long temp = 0;
        for (int i = 2; i <= A; i++)       
        {
            long temp = color3;
            color3 = (11 * color3 + 10 * color2 ) % 1000000007;
            color2 = ( 5 * temp + 7 * color2 ) % 1000000007;
        }
        long num = (color3 + color2) % 1000000007;
        return (int)num;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int num1 = 1;
        System.out.println(solve(num1));
 
        int num2 = 2;
        System.out.println(solve(num2));
 
        int num3 = 500;
        System.out.println(solve(num3));
 
        int num4 = 10000;
        System.out.println(solve(num4));
    }
}

Python3

# Python 3 program to find number of ways
# to color a 3 x n grid using 4 colors
# such that no two adjacent have same
# color
 
def solve(A):
     
    # When we to fill single column
    color3 = 24
    color2 = 12
    temp = 0
     
    for i in range(2, A + 1, 1):
        temp = color3
        color3 = (11 * color3 + 10 * color2 ) % 1000000007
             
        color2 = ( 5 * temp + 7 * color2 ) % 1000000007
     
    num = (color3 + color2) % 1000000007
                         
    return num
 
# Driver code
if __name__ == '__main__':
    num1 = 1
    print(solve(num1))
     
    num2 = 2
    print(solve(num2))
     
    num3 = 500
    print(solve(num3))
 
    num4 = 10000
    print(solve(num4))
     
# This code is contributed by
# Shashank_Sharma

C#

// C# program to find number of ways
// to color a 3 x n grid using 4
// colors such that no two adjacent
// have same color.
using System;
 
public class GFG {
     
    public static int solve(int A)
    {
         
        // When we to fill single column
        long color3 = 24;
        long color2 = 12;
        long temp = 0;
         
        for (int i = 2; i <= A; i++)
        {
            temp = color3;
            color3 = (11 * color3 + 10
                 * color2 ) % 1000000007;
                  
            color2 = ( 5 * temp + 7
                 * color2 ) % 1000000007;
        }
        long num = (color3 + color2)
                            % 1000000007;
        return (int)num;
    }
 
    // Driver code
    public static void Main()
    {
        int num1 = 1;
        Console.WriteLine(solve(num1));
 
        int num2 = 2;
        Console.WriteLine(solve(num2));
 
        int num3 = 500;
        Console.WriteLine(solve(num3));
 
        int num4 = 10000;
        Console.WriteLine(solve(num4));
    }
}
 
// This code is contributed by vt_m.

Javascript

// JavaScript program to find number of ways to color
// a 3 x n grid using 4 colors such that no two
// adjacent have same color.
 
    function solve(A) {
        let color3 = 24; // When we to fill single column
        let color2 = 12;
        let temp = 0;
        for (let i = 2; i <= A; i++)       
        {
            let temp = color3;
            color3 = (11 * color3 + 10 * color2 ) % 1000000007;
            color2 = ( 5 * temp + 7 * color2 ) % 1000000007;
        }
        let num = (color3 + color2) % 1000000007;
        return num;
    }
  
// Driver Code
 
        let num1 = 1;
        document.write(solve(num1) + "<br/>");
   
        let num2 = 2;
        document.write(solve(num2) + "<br/>");
   
        let num3 = 500;
        document.write(solve(num3) + "<br/>");
   
        let num4 = 10000;
        document.write(solve(num4));
                 

输出 ：

36
588
178599516
540460643