最大化(a + b)的值，使得(a * ab * b = N)

给定一个奇数值N ，任务是找到(a + b)的最大值(其中a和b是整数)，使得(a ² – b ² = N)
例子：

Input: N = 1
Output: 1
Since a*a - b*b = 1 
The maximum value occurs when a = 1 and b = 0. 
Thus, a + b = 1.

Input: N = 3
Output: 3
Since a*a - b*b = 3 
The maximum value occurs when a = 2 and b = 1. 
Thus, a + b = 3.

方法：

• 给定

a*a - b*b = N
=> (a+b)*(a-b) = N
=> (a+b) = N/(a-b)

• 现在上述方程式成立

We know that 
|a - b| ≥ 1

Therefore,
a + b ≤ N

• 现在为了最大化(a + b)的值，

Maximising a + b ≤ N

=> a + b = N

• 因此，当我们需要最大化(a + b)的值使得(a * ab * b = N)时，N是(a + b)的最大值。

下面是上述方法的实现：

C++

// C++ program to maximize the value
// of (a+b) such that (a*a-b*b = N)
 
#include 
using namespace std;
 
// Function to maximize the value
// of (a+b) such that (a*a-b*b = n)
int maxValue(int n)
{
    return n;
}
 
// Driver code
int main()
{
    int n = 1;
 
    cout << maxValue(n);
 
    return 0;
}

Java

// Java program to maximize the value
// of (a+b) such that (a*a-b*b = N)
class GFG
{
 
// Function to maximize the value
// of (a+b) such that (a*a-b*b = n)
static int maxValue(int n)
{
    return n;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 1;
 
    System.out.print(maxValue(n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to maximize the value
# of (a+b) such that (a*a-b*b = N)
 
# Function to maximize the value
# of (a+b) such that (a*a-b*b = n)
def maxValue(n) :
 
    return n;
 
# Driver code
if __name__ == "__main__" :
 
    n = 1;
 
    print(maxValue(n));
     
# This code is contributed by AnkitRai01

C#

// C# program to maximize the value
// of (a+b) such that (a*a-b*b = N)
using System;
 
class GFG
{
 
    // Function to maximize the value
    // of (a+b) such that (a*a-b*b = n)
    static int maxValue(int n)
    {
        return n;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int n = 1;
     
        Console.Write(maxValue(n));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript program to maximize the value
// of (a+b) such that (a*a-b*b = N)
 
// Function to maximize the value
// of (a+b) such that (a*a-b*b = n)
function maxValue(n)
{
    return n;
}
 
// Driver code
var n = 1;
document.write(maxValue(n));
 

输出：

1

时间复杂度： $O(1)$