矩形网格中的相同分割

给定一个NxM尺寸的矩形网格，任务是找到将给定的矩形网格分成大小为1×1的正方形所需的最小切割次数。

例子：

Input: N = 4, M = 4 
Output: 15

Input: N = 2, M = 1 
Output: 1 

方法：

上图显示了矩形网格的分裂。我们可以观察到，每一次切割都会将不同维度的矩形数量增加1 。我们将进行分裂，直到我们达到维度1×1的平方。
因此，对于给定的 NxM 矩形尺寸，1×1 尺寸的正方形总数为 N*M。因此，我们需要N*M – 1次切割才能将 NxM 的给定矩形尺寸分解为尺寸为 1×1 的正方形。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to find the minimum cuts
void minimumCuts(int N, int M)
{
 
    // Print the minimum cuts using
    // the formula
    cout << (N * M - 1);
}
 
// Driver Code
int main()
{
 
    // Given dimensions
    int N = 4, M = 4;
 
    // Function call
    minimumCuts(N, M);
 
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
 
class GFG{
 
// Function to find the minimum cuts
static void minimumCuts(int N, int M)
{
     
    // Print the minimum cuts using
    // the formula
    System.out.print(N * M - 1);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given dimensions
    int N = 4, M = 4;
 
    // Function call
    minimumCuts(N, M);
}
}
 
// This code is contributed by Rohit_ranjan

Python3

# Python3 program of the above approach
 
# Function to find the minimum cuts
def minimumCuts(N, M):
     
    # Print the minimum cuts using
    # the formula
    print(N * M - 1)
 
# Driver Code
if __name__ == "__main__":
     
    # Given dimensions
    N = 4
    M = 4
     
    # Function call
    minimumCuts(N, M)
     
# This code is contributed by coder001

C#

// C# program of the above approach
using System;
 
class GFG{
 
// Function to find the minimum cuts
static void minimumCuts(int N, int M)
{
     
    // Print the minimum cuts using
    // the formula
    Console.Write(N * M - 1);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given dimensions
    int N = 4, M = 4;
 
    // Function call
    minimumCuts(N, M);
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript program of the above approach
 
// Function to find the minimum cuts
function minimumCuts(N, M)
{
 
    // Print the minimum cuts using
    // the formula
    document.write(N * M - 1);
}
 
// Driver Code
// Given dimensions
var N = 4, M = 4;
 
// Function call
minimumCuts(N, M);
 
// This code is contributed by noob2000.

输出：

15

时间复杂度： O(1)