生成矩阵的方法计数，每行和每列的乘积为 1 或 -1

给定两个整数N和M ，任务是找到形成大小为N * M的矩阵的方法数，该矩阵仅由 1 或 -1 组成，使得每行和每列中整数的乘积等于 1 或-1.

例子：

Input: N = 2, M = 2 
Output: 4 
Explanation: 
Possible ways to get product of each row and column as 1 are, 
{{1, 1}, {1, 1}} and {{-1, -1}, {-1, -1}} 
Possible ways to get product of each row and column as -1 are, 
{{1, -1}, {-1, 1}} and {{-1, 1}, {1, -1}} 
Hence, number of ways = 2 + 2 = 4
Input: N = 3, M = 3 
Output: 32 
Explanation: 
There are 16 ways to get product as 1 and 16 ways to get product as -1. 
Hence, number of ways = 16 + 16 = 32

天真的方法：
解决这个问题的最简单方法是生成大小为N * M 的所有可能矩阵，并为每个矩阵计算所有行和列的乘积并检查它是 1 还是 -1。
时间复杂度： O(2 ^N*M )
辅助空间： O(M*N)
有效的方法：
假设前N-1行和前M-1列由 1 或 -1 填充。现在，每行最多N-1行和每列最多M-1列的乘积将是1或-1 。总共有2 ^{(N-1) * (M-1)}种方式可以组成一个大小为(N-1)*(M-1)的矩阵，填充 1 或 -1。根据需要作为 N 行和 M 列的乘积，可以相应地填充最后一行和最后一列。
请按照以下步骤解决问题：

• 如果N + M是偶数，
  获得乘积的可能矩阵数为 1 = 2 ^{(N-1) * (M-1)}
  获得乘积的可能矩阵数为 -1 = 2 ^{(N-1) * (M-1)}
• 如果N + M是奇数，
  获得乘积的可能矩阵数为 1 = 2 ^{(N-1) * (M-1)}
  获得乘积为 -1 = 0 的可能矩阵数

下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
 
#include
using namespace std;
 
// Function to return the
// number of possible ways
void Solve(int N, int M)
{
 
    int temp = (N - 1) * (M - 1);
    int ans = pow(2, temp);
 
    // Check if product can be -1
    if ((N + M) % 2 != 0)
        cout << ans;
    else
        cout << 2 * ans;
 
    cout << endl;
}
// Driver Code
int main()
{
    int N = 3;
    int M = 3;
 
    Solve(N, M);
    return 0;
}

Java

// Java implementation of the above approach
import java.util.Arrays;
 
class GFG{
     
// Function to return the
// number of possible ways
static void Solve(int N, int M)
{
    int temp = (N - 1) * (M - 1);
    int ans = (int)(Math.pow(2, temp));
 
    // Check if product can be -1
    if ((N + M) % 2 != 0)
        System.out.print(ans);
    else
        System.out.print(2 * ans);
}
 
// Driver code
public static void main (String[] args)
{
    int N = 3;
    int M = 3;
     
    Solve(N, M);
}
}
 
// This code is contributed by Shubham Prakash

Python3

# Python3 program to implement
# the above approach
# Function to return
# possible number of ways
def Solve(N, M):
  temp = (N - 1) * (M - 1)
  ans = pow(2, temp)
 
  # Check if product can be -1
  if ((N + M) % 2 != 0):
    print(ans)
    else:
      print(2 * ans)
 
      # driver code
      if __name__ == '__main__':
        N, M = 3, 3
        Solve(N, M)
 
# This code is contributed by Sri_srajit

C#

// C# implementation of the above approach
using System;
 
class GFG{
     
// Function to return the
// number of possible ways
static void Solve(int N, int M)
{
    int temp = (N - 1) * (M - 1);
    int ans = (int)(Math.Pow(2, temp));
 
    // Check if product can be -1
    if ((N + M) % 2 != 0)
        Console.Write(ans);
    else
        Console.Write(2 * ans);
}
 
// Driver Code
public static void Main(string[] args)
{
    int N = 3;
    int M = 3;
     
    Solve(N, M);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript program implementation
// of the approach
 
// Function to return the
// number of possible ways
function Solve(N, M)
{
    let temp = (N - 1) * (M - 1);
    let ans = (Math.pow(2, temp));
   
    // Check if product can be -1
    if ((N + M) % 2 != 0)
        document.write(ans);
    else
        document.write(2 * ans);
}
 
// Driver Code
     
       let N = 3;
    let M = 3;
       
    Solve(N, M);
          

输出：

32

时间复杂度： O(log(N*M))
辅助空间： O(1)

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