一对总和为N的整数的最大可能GCD

给定一个整数N ，任务是找到一对整数的最大可能GCD ，以使它们的总和为N。

例子 ：

Input: N = 30 
Output: 15 
Explanation: GCD of (15, 15) is 15, which is the maximum possible GCD

Input: N = 33 
Output: 11 
Explanation: GCD of (11, 22) is 11, which is the maximum possible GCD

天真的方法：
解决此问题的最简单方法是为所有总和为N的整数计算GCD，并在其中找到最大可能的GCD 。
时间复杂度： O(N ² logN)
辅助空间： O(1)

高效方法：
请按照以下给出的步骤优化上述方法：

• 迭代到√N并找到N的最大合适因数。
• 如果N为质数，即无法获得任何因数，则打印1 ，因为所有对都是互质数。
• 否则，打印最大可能的因素作为答案。

下面是上述方法的实现：

C++

// C++ Program to find the maximum
// possible GCD of any pair with
// sum N
#include 
using namespace std;
 
// Function to find the required
// GCD value
int maxGCD(int N)
{
    for (int i = 2; i * i <= N; i++) {
 
        // If i is a factor of N
        if (N % i == 0) {
 
            // Return the largest
            // factor possible
            return N / i;
        }
    }
 
    // If N is a prime number
    return 1;
}
 
// Driver Code
int main()
{
    int N = 33;
    cout << "Maximum Possible GCD value is : "
        << maxGCD(N) << endl;
    return 0;
}

Java

// Java program to find the maximum
// possible GCD of any pair with
// sum N
class GFG{
 
// Function to find the required
// GCD value
static int maxGCD(int N)
{
    for(int i = 2; i * i <= N; i++)
    {
 
        // If i is a factor of N
        if (N % i == 0)
        {
 
            // Return the largest
            // factor possible
            return N / i;
        }
    }
 
    // If N is a prime number
    return 1;
}
     
// Driver Code
public static void main(String[] args)
{
    int N = 33;
    System.out.println("Maximum Possible GCD " +
                       "value is : " + maxGCD(N));
}
}
 
// This code is conhtributed by rutvik_56

Python3

# Python3 program to find the maximum
# possible GCD of any pair with
# sum N
 
# Function to find the required
# GCD value
def maxGCD(N):
     
    i = 2
    while(i * i <= N):
 
        # If i is a factor of N
        if (N % i == 0):
 
            # Return the largest
            # factor possible
            return N // i
         
        i += 1
 
    # If N is a prime number
    return 1
 
# Driver Code
N = 33
 
print("Maximum Possible GCD value is : ",
       maxGCD(N))
 
# This code is contributed by code_hunt

C#

// C# program to find the maximum
// possible GCD of any pair with
// sum N
using System;
 
class GFG{
 
// Function to find the required
// GCD value
static int maxGCD(int N)
{
    for(int i = 2; i * i <= N; i++)
    {
 
        // If i is a factor of N
        if (N % i == 0)
        {
 
            // Return the largest
            // factor possible
            return N / i;
        }
    }
 
    // If N is a prime number
    return 1;
}
     
// Driver Code
public static void Main(String[] args)
{
    int N = 33;
    Console.WriteLine("Maximum Possible GCD " +
                      "value is : " + maxGCD(N));
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript program to find the maximum
// possible GCD of any pair with
// sum N
 
// Function to find the required
// GCD value
function maxGCD(N)
{
    for(var i = 2; i * i <= N; i++)
    {
         
        // If i is a factor of N
        if (N % i == 0)
        {
             
            // Return the largest
            // factor possible
            return N / i;
        }
    }
 
    // If N is a prime number
    return 1;
}
 
// Driver code
var N = 33;
document.write("Maximum Possible GCD " +
               "value is :" + maxGCD(N));
 
// This code is contributed by Ankita saini.
    

输出：

Maximum Possible GCD value is : 11

时间复杂度： O(√N)
辅助空间： O(1)