找到两个数字 B 和 C 使得它们的乘积是 A 并且它们的 GCD 最大

给定一个正整数A 。任务是找到两个数字B和C 这样他们的产品是A ，他们的GCD应该是最大的。

例子：

Input: A = 72
Output: 12 6
Explanation: The product of 12 and 6 is 72 and GCD(12, 6) is 6 which is maximum possible.  

Input: A = 54
Output: 6 9
Explanation: The product of 6 and 9 is 54, gcd(6, 9) is 3 which is maximum possible. 

方法：这个问题可以通过生成A 的素数来解决。要使GCD最大化，唯一可能的方法是选择不同的素因子，这样它们的乘积就会得到最大的GCD 。请按照以下步骤解决问题。

• 创建一个函数，比如primeFactors()来查找一个数字的所有素因子。
• 首先调用primeFactors()并传递A并通过引用传递一个数组以以排序方式存储所有素数因子。
• 遍历素因子数组，并将A的所有因子交替分布到B和C ，使得，
  • B = primefactor[0] * primefactor[2] * primefactor[4] – – – 以此类推。
  • C = primefactor[1] * primefactor[3] * primefactor[5] – – – 以此类推。
• 输出用空格分隔的数字B和C。

For Example: N = 72
Prime Factorization of 72 = 2 * 2 * 2 * 3 * 3. 
primefactor[] = {2, 2, 2, 3, 3}
B = primefactor[0] * primefactor[2] * primefactor[4] => 2 * 2 * 3 = 12.
C = primefactor[1] * primefactor[3] => 2 * 3 = 6.
Hence, B = 12 and C = 6.

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to stores prime factors
// of a number in a vector.
void primeFactors(int A, vector& prime)
{
    while (A % 2 == 0) {
        prime.push_back(2);
        A = A / 2;
    }
 
    for (int i = 3; i <= sqrt(A); i += 2) {
        while (A % i == 0) {
            prime.push_back(i);
            A = A / i;
        }
    }
 
    if (A > 2)
        prime.push_back(A);
}
 
// Find the numbers B and C
// such that gcd(B, C) is max
// and product is A.
void maxPairGCD(int A)
{
    // Vector to store prime factors
    vector prime;
 
    primeFactors(A, prime);
    int B = 1, C = 1;
    int temp = 0;
 
    for (int i = 0; i < prime.size(); i++) {
        if (temp == 0)
            B *= prime[i];
        else
            C *= prime[i];
        temp = temp ^ 1;
    }
    cout << B << " " << C;
}
 
// Driver code
int main()
{
    int A = 72;
 
    // Function Call
    maxPairGCD(A);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.ArrayList;
 
class GFG
{
 
  // Function to stores prime factors
  // of a number in a vector.
  static void primeFactors(int A, ArrayList prime)
  {
    while (A % 2 == 0)
    {
      prime.add(2);
      A = A / 2;
    }
 
    for (int i = 3; i <= Math.sqrt(A); i += 2)
    {
      while (A % i == 0)
      {
        prime.add(i);
        A = A / i;
      }
    }
 
    if (A > 2)
      prime.add(A);
  }
 
  // Find the numbers B and C
  // such that gcd(B, C) is max
  // and product is A.
  static void maxPairGCD(int A)
  {
 
    // Vector to store prime factors
    ArrayList prime = new ArrayList();
 
    primeFactors(A, prime);
    int B = 1, C = 1;
    int temp = 0;
 
    for (int i = 0; i < prime.size(); i++) {
      if (temp == 0)
        B *= prime.get(i);
      else
        C *= prime.get(i);
      temp = temp ^ 1;
    }
    System.out.println(B + " " + C);
  }
 
  // Driver code
  public static void main(String args[])
  {
    int A = 72;
 
    // Function Call
    maxPairGCD(A);
  }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python 3 program for above approach
import math
 
# Function to stores prime factors
# of a number in a vector.
def primeFactors(A,  prime):
 
    while (A % 2 == 0):
        prime.append(2)
        A = A / 2
 
    for i in range(3, int(math.sqrt(A)) + 1,  2):
        while (A % i == 0):
            prime.append(i)
            A = A / i
 
    if (A > 2):
        prime.append(A)
 
# Find the numbers B and C
# such that gcd(B, C) is max
# and product is A.
def maxPairGCD(A):
 
    # Vector to store prime factors
    prime = []
 
    primeFactors(A, prime)
    B = 1
    C = 1
    temp = 0
 
    for i in range(len(prime)):
        if (temp == 0):
            B *= prime[i]
 
        else:
            C *= prime[i]
        temp = temp ^ 1
 
    print(B, C)
 
# Driver code
if __name__ == "__main__":
 
    A = 72
 
    # Function Call
    maxPairGCD(A)
 
    # This code is contributed by ukasp.

C#

// C# program to implement
// the above approach
using System;
using System.Collections;
 
class GFG
{
 
  // Function to stores prime factors
  // of a number in a vector.
  static void primeFactors(int A, ArrayList prime)
  {
    while (A % 2 == 0) {
      prime.Add(2);
      A = A / 2;
    }
 
    for (int i = 3; i <= Math.Sqrt(A); i += 2) {
      while (A % i == 0) {
        prime.Add(i);
        A = A / i;
      }
    }
 
    if (A > 2)
      prime.Add(A);
  }
 
  // Find the numbers B and C
  // such that gcd(B, C) is max
  // and product is A.
  static void maxPairGCD(int A)
  {
    // Vector to store prime factors
    ArrayList prime = new ArrayList();
 
    primeFactors(A, prime);
    int B = 1, C = 1;
    int temp = 0;
 
    for (int i = 0; i < prime.Count; i++) {
      if (temp == 0)
        B *= (int)prime[i];
      else
        C *= (int)prime[i];
      temp = temp ^ 1;
    }
    Console.Write(B + " " + C);
  }
 
  // Driver code
  public static void Main()
  {
    int A = 72;
 
    // Function Call
    maxPairGCD(A);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

        // JavaScript code for the above approach
 
        // Function to stores prime factors
        // of a number in a vector.
        function primeFactors(A, prime)
        {
         
            while (A % 2 == 0) {
                prime.push(2);
                A = A / 2;
            }
 
            for (let i = 3; i <= Math.sqrt(A); i += 2) {
                while (A % i == 0) {
                    prime.push(i);
                    A = A / i;
                }
            }
 
            if (A > 2)
                prime.push(A);
 
            return prime;
        }
 
        // Find the numbers B and C
        // such that gcd(B, C) is max
        // and product is A.
        function maxPairGCD(A)
        {
         
            // Vector to store prime factors
            let prime = [];
 
            prime = primeFactors(A, prime);
            let B = 1, C = 1;
            let temp = 0;
 
            for (let i = 0; i < prime.length; i++) {
                if (temp == 0)
                    B *= prime[i];
                else
                    C *= prime[i];
                temp = temp ^ 1;
            }
            document.write(B + " " + C)
        }
 
        // Driver code
        let A = 72;
 
        // Function Call
        maxPairGCD(A);
 
  // This code is contributed by Potta Lokesh
    

输出

12 6

时间复杂度： O(Sqrt(A) )
辅助空间： O(log(A))