找到最小LCM的对(a，b)，使它们的总和等于N

给定数字N ，任务是找到两个数字a和b，使得a + b = N且LCM(a，b)最小。

例子：

Input: N = 15
Output: a = 5, b = 10
Explanation:
The pair 5, 10 has a sum of 15 and their LCM is 10 which is the minimum possible.

Input: N = 4
Output: a = 2, b = 2
Explanation: 
The pair 2, 2 has a sum of 4 and their LCM is 2 which is the minimum possible.

方法：想法是使用GCD和LCM的概念。步骤如下：

• 如果N是素数，则答案为1和N – 1，因为在任何其他情况下， a + b> N或LCM(a，b)都> N – 1 。这是因为如果N为质数，则意味着N为奇数。因此，a和b，它们中的任何一个都必须是奇数而其他偶数。因此，LCM(a，b)必须大于N (如果不是1且N – 1)，因为2始终是一个因数。
• 如果N不是质数，则选择a，b，使它们的GCD最大，因为公式LCM(a，b)= a * b / GCD(a，b) 。因此，为了最小化LCM(a，b)，我们必须 最大化GCD(a，b)。
• 如果x是N的除数，则通过简单的数学运算 a和b可以分别表示为N / x和N / x *(x – 1) 。现在，由于a = N / x和b = N / x *(x – 1) ，因此它们的GCD表示为N / x 。为了最大限度地提高本GCD，取可能的最小x或N个最小可能除数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if number is
// prime or not
bool prime(int n)
{
    // As 1 is neither prime
    // nor composite return false
    if (n == 1)
        return false;
 
    // Check if it is divided by any
    // number then it is not prime,
    // return false
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0)
            return false;
    }
 
    // Check if n is not divided
    // by any number then it is
    // prime and hence return true
    return true;
}
 
// Function to find the pair (a, b)
// such that sum is N & LCM is minimum
void minDivisior(int n)
{
 
    // Check if the number is prime
    if (prime(n)) {
        cout << 1 << " " << n - 1;
    }
 
    // Now, if it is not prime then
    // find the least divisior
    else {
        for (int i = 2; i * i <= n; i++) {
 
            // Check if divides n then
            // it is a factor
            if (n % i == 0) {
 
                // Required output is
                // a = n/i & b = n/i*(n-1)
                cout << n / i << " "
                     << n / i * (i - 1);
                break;
            }
        }
    }
}
 
// Driver Code
int main()
{
    int N = 4;
 
    // Function call
    minDivisior(N);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to check if number is
// prime or not
static boolean prime(int n)
{
    // As 1 is neither prime
    // nor composite return false
    if (n == 1)
        return false;
 
    // Check if it is divided by any
    // number then it is not prime,
    // return false
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
 
    // Check if n is not divided
    // by any number then it is
    // prime and hence return true
    return true;
}
 
// Function to find the pair (a, b)
// such that sum is N & LCM is minimum
static void minDivisior(int n)
{
 
    // Check if the number is prime
    if (prime(n))
    {
        System.out.print(1 + " " +  (n - 1));
    }
 
    // Now, if it is not prime then
    // find the least divisior
    else
    {
        for (int i = 2; i * i <= n; i++)
        {
 
            // Check if divides n then
            // it is a factor
            if (n % i == 0)
            {
 
                // Required output is
                // a = n/i & b = n/i*(n-1)
                System.out.print(n / i + " " +
                                (n / i * (i - 1)));
                break;
            }
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 4;
 
    // Function call
    minDivisior(N);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
 
# Function to check if number is
# prime or not
def prime(n):
     
    # As 1 is neither prime
    # nor composite return false
    if (n == 1):
        return False
 
    # Check if it is divided by any
    # number then it is not prime,
    # return false
    for i in range(2, n + 1):
        if i * i > n:
            break
        if (n % i == 0):
            return False
 
    # Check if n is not divided
    # by any number then it is
    # prime and hence return true
    return True
 
# Function to find the pair (a, b)
# such that sum is N & LCM is minimum
def minDivisior(n):
 
    # Check if the number is prime
    if (prime(n)):
        print(1, n - 1)
 
    # Now, if it is not prime then
    # find the least divisior
    else:
        for i in range(2, n + 1):
            if i * i > n:
                break
 
            # Check if divides n then
            # it is a factor
            if (n % i == 0):
 
                # Required output is
                # a = n/i & b = n/i*(n-1)
                print(n // i, n // i * (i - 1))
                break
 
# Driver Code
N = 4
 
# Function call
minDivisior(N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if number is
// prime or not
static bool prime(int n)
{
     
    // As 1 is neither prime
    // nor composite return false
    if (n == 1)
        return false;
 
    // Check if it is divided by any
    // number then it is not prime,
    // return false
    for(int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
 
    // Check if n is not divided
    // by any number then it is
    // prime and hence return true
    return true;
}
 
// Function to find the pair (a, b)
// such that sum is N & LCM is minimum
static void minDivisior(int n)
{
 
    // Check if the number is prime
    if (prime(n))
    {
        Console.Write(1 + " " + (n - 1));
    }
 
    // Now, if it is not prime then
    // find the least divisior
    else
    {
        for(int i = 2; i * i <= n; i++)
        {
             
            // Check if divides n then
            // it is a factor
            if (n % i == 0)
            {
                 
                // Required output is
                // a = n/i & b = n/i*(n-1)
                Console.Write(n / i + " " +
                             (n / i * (i - 1)));
                break;
            }
        }
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 4;
 
    // Function call
    minDivisior(N);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program for the above approach   
// Function to check if number is
    // prime or not
    function prime(n)
    {
     
        // As 1 is neither prime
        // nor composite return false
        if (n == 1)
            return false;
 
        // Check if it is divided by any
        // number then it is not prime,
        // return false
        for (i = 2; i * i <= n; i++)
        {
            if (n % i == 0)
                return false;
        }
 
        // Check if n is not divided
        // by any number then it is
        // prime and hence return true
        return true;
    }
 
    // Function to find the pair (a, b)
    // such that sum is N & LCM is minimum
    function minDivisior(n)
    {
 
        // Check if the number is prime
        if (prime(n))
        {
            document.write(1 + " " + (n - 1));
        }
 
        // Now, if it is not prime then
        // find the least divisior
        else
        {
            for (i = 2; i * i <= n; i++)
            {
 
                // Check if divides n then
                // it is a factor
                if (n % i == 0)
                {
 
                    // Required output is
                    // a = n/i & b = n/i*(n-1)
                    document.write(n / i + " " + (n / i * (i - 1)));
                    break;
                }
            }
        }
    }
 
    // Driver Code
        var N = 4;
 
        // Function call
        minDivisior(N);
 
// This code is contributed by todaysgaurav

输出：

2 2

时间复杂度： O(平方(N))
辅助空间： O(1)