使用两个梯子找到最大无法到达的高度

给定两个高度为h1和h2的梯子。任务是找到两个梯子无法通过任何可能的组合达到的最大高度。如果可以达到所有高度，则打印0 。

例子：

Input: H1 = 2, H2 = 11
Output: 9
We cannot reach heights 1, 3, 5, 7 and 9.
So, the maximum possible height is 9.

Input: H1 = 7, H2 = 5
Output: 23

方法：对于给定的数字a和b ，最大数c不可能使ax + by = c ，其中x≥0并且y≥0是等于(a * b)– a – b的弗罗贝尼斯数。

下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum height
// which can't be reached
int maxHeight(int h1, int h2)
{
    return ((h1 * h2) - h1 - h2);
}
  
// Driver code
int main()
{
    int h1 = 7, h2 = 5;
  
    cout << max(0, maxHeight(h1, h2));
  
    return 0;
}

Java

// Java implementation of the approach 
class GFG
{
  
    // Function to return the maximum height 
    // which can't be reached 
    static int maxHeight(int h1, int h2) 
    { 
        return ((h1 * h2) - h1 - h2); 
    } 
      
    // Driver code 
    public static void main(String args[])
    { 
        int h1 = 7, h2 = 5; 
      
        System.out.println(Math.max(0, maxHeight(h1, h2))); 
    } 
}
  
// This code is contributed by AnkitRai01

Python

# Python3 implementation of the approach
  
# Function to return the maximum height
# which can't be reached
def maxHeight(h1, h2):
    return ((h1 * h2) - h1 - h2)
  
# Driver code
h1 = 7
h2 = 5
  
print(max(0, maxHeight(h1, h2)))
  
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach 
using System;
  
class GFG
{
  
    // Function to return the maximum height 
    // which can't be reached 
    static int maxHeight(int h1, int h2) 
    { 
        return ((h1 * h2) - h1 - h2); 
    } 
      
    // Driver code 
    public static void Main()
    { 
        int h1 = 7, h2 = 5; 
      
        Console.WriteLine(Math.Max(0, maxHeight(h1, h2))); 
    } 
}
  
// This code is contributed by AnkitRai01

输出：

23

时间复杂度： O(1)