最后剩余的地块面积

给定两个整数N和M ，代表矩形图的长度和宽度，另一个整数K代表人数。每个人将情节分成两部分，使其成为情节中最大的正方形，并将第二部分留给其他人，一直持续到情节结束或每个人都得到情节。现在，任务是确定最后留下的地块面积。
例子：

Input: N = 5, M = 3, K = 2
Output: 2
1st person divides the 5×3 plot into 2 parts i.e 3×3 and 2×3
and will get the plot having dimension 3×3. The other person divides the 2×3 plot again into
two parts i.e 2×2 and 1×2 and will get the plot having dimension 2×2. Now, the remaining
part of plot is having dimension 1×2 and area as 2 units.
Input: N = 4, M = 8, K = 4
Output: 0

编程需要懂一点英语

方法：

如果长度大于宽度，则从长度中减去宽度。
如果宽度大于长度，则从宽度中减去长度。
当剩余地块的面积大于 0 时，对所有人员重复上述步骤。
最后打印剩余图的面积，即长度 * 宽度。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the area
// of the remaining plot
int remainingArea(int N, int M, int K)
{
 
    // Continue while plot has positive area
    // and there are persons left
    while (K-- && N && M) {
 
        // If length > breadth
        // then subtract breadth from length
        if (N > M)
            N = N - M;
 
        // Else subtract length from breadth
        else
            M = M - N;
    }
 
    if (N > 0 && M > 0)
        return N * M;
    else
        return 0;
}
 
// Driver code
int main()
{
    int N = 5, M = 3, K = 2;
 
    cout << remainingArea(N, M, K);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the area
    // of the remaining plot
    static int remainingArea(int N, int M, int K)
    {
 
        // Continue while plot has positive area
        // and there are persons left
        while (K-- > 0 && N > 0 && M > 0) {
 
            // If length > breadth
            // then subtract breadth from length
            if (N > M)
                N = N - M;
 
            // Else subtract length from breadth
            else
                M = M - N;
        }
 
        if (N > 0 && M > 0)
            return N * M;
        else
            return 0;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 5, M = 3, K = 2;
 
        System.out.println(remainingArea(N, M, K));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
 
# Function to return the area
# of the remaining plot
def remainingArea(N, M, K):
 
    # Continue while plot has positive area
    # and there are persons left
    while (K > 0 and N > 0 and M > 0):
 
        # If length > breadth
        # then subtract breadth from length
        if (N > M):
            N = N - M;
 
        # Else subtract length from breadth
        else:
            M = M - N;
        K = K - 1;
    if (N > 0 and M > 0):
        return N * M;
    else:
        return 0;
 
# Driver code
N = 5;
M = 3;
K = 2;
 
print(remainingArea(N, M, K));
 
# This code contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the area
    // of the remaining plot
    static int remainingArea(int N, int M, int K)
    {
 
        // Continue while plot has positive area
        // and there are persons left
        while (K-- > 0 && N > 0 && M > 0) {
 
            // If length > breadth
            // then subtract breadth from length
            if (N > M)
                N = N - M;
 
            // Else subtract length from breadth
            else
                M = M - N;
        }
 
        if (N > 0 && M > 0)
            return N * M;
        else
            return 0;
    }
 
    // Driver code
    static public void Main()
    {
        int N = 5, M = 3, K = 2;
 
        Console.WriteLine(remainingArea(N, M, K));
    }
}
 
/* This code contributed by ajit */

PHP

 breadth
        // then subtract breadth from length
        if ($N > $M)
            $N = $N - $M;
 
        // Else subtract length from breadth
        else
            $M = $M - $N;
    }
 
    if ($N > 0 && $M > 0)
        return $N * $M;
    else
        return 0;
}
 
// Driver code
$N = 5;
$M = 3;
$K = 2;
 
echo remainingArea($N, $M, $K);
 
// This code is contributed by AnkitRai01
 
?>

Javascript

输出：