求 N 的两个真因数，使得它们的和与 N 互质

给定一个整数N ，您必须找到N的两个适当因子，使得它们的和与给定整数N互质。如果不存在此类因素，则打印 -1。

例子：

Input: N = 15
Output: 3, 5
Explanation: 3 and 5 are the proper factors of 15 and 3+5 -> 8 is coprime with 15.

Input: N = 4
Output: -1
Explanation: there are no proper factors that satisfy the required conditions

朴素方法：生成 N 的所有适当因子的列表，并为每个可能的对检查它们的总和是否与 N 互质，即 GCD(整数对之和，N) = 1。这里 GCD 表示最大公约数。

有效方法：如果两个数字A和B互质，那么它们的和与它们的乘积互质。牢记这一点，找到 N 的所有因子，对于每个因子d1 ，计算与d1互质的 N 的最大因子d2 。要计算 d2，只需将N除以d1直到N%d1 != 0 。最后，检查d1和d2是否是N的真因数（即 d1>1 和 d2>1）。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
 
// Function to find two proper
// factors of N such that their
// sum is coprime with N
void printFactors(int n)
{
 
    // Find factors in sorted order
    for (int i = 2; i <= sqrt(n); i++) {
 
        if (n % i == 0) {
            int d1 = i, d2 = n;
 
            // Find largest value of d2 such
            // that d1 and d2 are co-prime
            while (d2 % d1 == 0) {
                d2 = d2 / d1;
            }
 
            // Check if d1 and d2 are proper
            // factors of N
            if (d1 > 1 && d2 > 1) {
                // Print answer
                cout << d1 << ", " << d2;
                return;
            }
        }
    }
 
    // No such factors exist
    cout << -1;
}
 
// Driver code
int main()
{
    int N = 10;
 
    // Function Call
    printFactors(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find two proper
// factors of N such that their
// sum is coprime with N   
static void printFactors(int n)
{
     
    // Find factors in sorted order
    for(int i = 2; i <= (int)Math.sqrt(n); i++)
    {
        if (n % i == 0)
        {
            int d1 = i, d2 = n;
 
            // Find largest value of d2 such
            // that d1 and d2 are co-prime
            while (d2 % d1 == 0)
            {
                d2 = d2 / d1;
            }
 
            // Check if d1 and d2 are proper
            // factors of N
            if (d1 > 1 && d2 > 1)
            {
                 
                // Print answer
                System.out.print(d1 + ", " + d2);
                return;
            }
        }
    }
 
    // No such factors exist
    System.out.print(-1);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 10;
     
    // Function Call
    printFactors(N);
}
}
 
// This code is contributed by Potta Lokesh

Python3

# Python Program for the above approach
import math
 
# Function to find two proper
# factors of N such that their
# sum is coprime with N
def printFactors(n):
    # Find factors in sorted order
    for i in range(2, int(math.sqrt(n))+1):
        if (n % i == 0):
            d1 = i
            d2 = n
             
            # Find largest value of d2 such
            # that d1 and d2 are co-prime
            while (d2 % d1 == 0):
                d2 = d2 // d1
             
            # Check if d1 and d2 are proper
            # factors of N
            if (d1 > 1 and d2 > 1):
                 
                # Print answer
                print(d1, d2, sep=", ")
                return
    # No such factors exist
    print(-1)
# Driver code
N = 10
 
# Function Call
printFactors(N)
     
# This code is contributed by Shivani

C#

// C# Program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find two proper
// factors of N such that their
// sum is coprime with N
static void printFactors(int n)
{
 
    // Find factors in sorted order
    for (int i = 2; i <= (int)Math.Sqrt(n); i++) {
 
        if (n % i == 0) {
            int d1 = i, d2 = n;
 
            // Find largest value of d2 such
            // that d1 and d2 are co-prime
            while (d2 % d1 == 0) {
                d2 = d2 / d1;
            }
 
            // Check if d1 and d2 are proper
            // factors of N
            if (d1 > 1 && d2 > 1)
            {
               
                // Print answer
                Console.Write(d1 + ", "+d2);
                return;
            }
        }
    }
 
    // No such factors exist
    Console.Write(-1);
}
 
// Driver code
public static void Main()
{
    int N = 10;
   
    // Function Call
    printFactors(N);
}
}
 
// This code is contributed by ipg2016107.

Javascript

// Javascript Program for the above approach
 
// Function to find two proper
// factors of N such that their
// sum is coprime with N
function printFactors(n) {
  // Find factors in sorted order
  for (let i = 2; i <= Math.sqrt(n); i++) {
    if (n % i == 0) {
      let d1 = i,
        d2 = n;
 
      // Find largest value of d2 such
      // that d1 and d2 are co-prime
      while (d2 % d1 == 0) {
        d2 = Math.floor(d2 / d1);
      }
 
      // Check if d1 and d2 are proper
      // factors of N
      if (d1 > 1 && d2 > 1) {
        // Print answer
        document.write(d1 + ", " + d2);
        return;
      }
    }
  }
 
  // No such factors exist
  document.write(-1);
}
 
// Driver code
 
let N = 10;
 
// Function Call
printFactors(N);
 
// This code is contributed by _saurabh_jaiswal.

输出：

2, 5

时间复杂度： O(√N)
辅助空间： O(1)