给定一个表M * N。总共有M * N个大小为1的正方形。您必须用橙色，蓝色或黑色3种颜色为所有正方形的每一边上色，以使每个正方形具有2种不同的颜色，并且每种颜色必须出现两次。
这意味着每个正方形都有四个边，其中2个是橙色，其中2个是蓝色，其中2个是橙色，其中2个是黑色，其中2个是蓝色，其中2个是黑色。
例子：
Input: N = 1, M = 1
Output: 18
Explanation:
We can fill the upper part of the square and left part of the square with any of the three colours. So the number of ways is 3*3,
Now for filling the right and the lower part, we have only 2 option if upper and left have the same colour.
If the upper and the left part have different colour then we can use those two colours to fill the right and lower part, it has also 2 option. The number of combination to fill right and lower is 2.
A total possible way to colour the square is 3*3*2 = 18
Input: N = 3, M = 2
Output: 15552
方法：
- 查找填充矩形的上部和右侧的方法的数量。上侧有M个单位，右侧有N个单位。因此，为矩形的上，右侧上色的方式是pow(3，M + N)，因为每种方法都有3个选项。
- 现在，为填充每个正方形的下部和左侧，每个正方形有2个选项，则行数为pow(2，M * N) 。
- 最终结果是：
Total count = pow(3, M + N ) * pow(2, M * N)
下面是上述方法的实现：
C++
// C++ program to count the number
// of ways to color boundary of
// each block of M*N table.
#include
using namespace std;
// Function to compute all way to
// fill the boundary of all sides
// of the unit square
int CountWays(int N, int M)
{
int count = 1;
// Count possible ways to fill
// all upper and left side of
// the rectangle M*N
count = pow(3, M + N);
// Count possible ways to fill
// all side of the all squares
// unit size
count *= pow(2, M * N);
return count;
}
// Driver code
int main()
{
// Number of rows
int N = 3;
// Number of columns
int M = 2;
cout << CountWays(N, M);
return 0;
} Java
// Java program to count the number
// of ways to color boundary of
// each block of M*N table.
import java.util.*;
class GFG{
// Function to compute all way to
// fill the boundary of all sides
// of the unit square
static int CountWays(int N, int M)
{
int count = 1;
// Count possible ways to fill
// all upper and left side of
// the rectangle M*N
count = (int)Math.pow(3, M + N);
// Count possible ways to fill
// all side of the all squares
// unit size
count *= (int)Math.pow(2, M * N);
return count;
}
// Driver code
public static void main(String args[])
{
// Number of rows
int N = 3;
// Number of columns
int M = 2;
System.out.println(CountWays(N, M));
}
}
// This code is contributed by ANKITKUMAR34Python3
# Python3 program to count the number
# of ways to color boundary of
# each block of M*N table.
# Function to compute all way to
# fill the boundary of all sides
# of the unit square
def CountWays(N, M):
count = 1
# Count possible ways to fill
# all upper and left side of
# the rectangle M*N
count = pow(3, M + N)
# Count possible ways to fill
# all side of the all squares
# unit size
count *= pow(2, M * N);
return count
# Driver code
# Number of rows
N = 3
# Number of columns
M = 2
print(CountWays(N, M))
# This code is contributed by ANKITKUMAR34C#
// C# program to count the number
// of ways to color boundary of
// each block of M*N table.
using System;
class GFG{
// Function to compute all way to
// fill the boundary of all sides
// of the unit square
static int CountWays(int N, int M)
{
int count = 1;
// Count possible ways to fill
// all upper and left side of
// the rectangle M*N
count = (int)Math.Pow(3, M + N);
// Count possible ways to fill
// all side of the all squares
// unit size
count *= (int)Math.Pow(2, M * N);
return count;
}
// Driver code
static void Main()
{
// Number of rows
int N = 3;
// Number of columns
int M = 2;
Console.Write(CountWays(N, M));
}
}
// This code is contributed by divyeshrabadiya07输出：
15552