生成给定数目的K个互素因数对

给定两个整数N和K ，任务是找到K个对数为N的因数，以使每对因数的GCD为1。
注意：对于给定的数字， K个素数因子始终存在
例子：

Input: N = 6, K = 1 
Output: 2 3 
Explanation: 
Since 2 and 3 are both factors of 6 and gcd(2, 3) = 1.
Input: N = 120, K = 4 
Output: 
2 3 
3 4 
3 5 
4 5 
 

天真的方法：
最简单的方法是检查直到N的所有数字，并检查该对的GCD是否为1。
时间复杂度： O(N ² )
空间复杂度： O(1)
线性进近：
查找N的所有可能的除数，并将其存储在另一个数组中。遍历数组以从数组中搜索所有可能的互质对，并打印它们。
时间复杂度： O(N)
空间复杂度： O(N)
高效方法：
请按照以下步骤解决问题：

• 可以观察到，如果任何数量的GCD，说x，其中1始终为1，即GCD(1，X)= 1。
• 由于1将始终是一个因子N，简单地打印的N中的任K系数与1作为互质对。

下面是上述方法的实现。

C++

// C++ implementation of
// the above approach
#include 
using namespace std;
  
// Function prints the
// required pairs
void FindPairs(int n, int k)
{
    // First co-prime pair
    cout << 1 << " " << n << endl;
  
    // As a pair (1 n) has
    // already been Printed
    k--;
  
    for (long long i = 2;
         i <= sqrt(n); i++) {
  
        // If i is a factor of N
        if (n % i == 0) {
  
            cout << 1 << " "
                 << i << endl;
            k--;
            if (k == 0)
                break;
  
            // Since (i, i) won't form
            // a coprime pair
            if (i != n / i) {
                cout << 1 << " "
                     << n / i << endl;
                k--;
            }
            if (k == 0)
                break;
        }
    }
}
  
// Driver Code
int main()
{
  
    int N = 100;
    int K = 5;
  
    FindPairs(N, K);
  
    return 0;
}

Java

// Java implementation of 
// the above approach 
import java.util.*;
  
class GFG{ 
  
// Function prints the 
// required pairs 
static void FindPairs(int n, int k) 
{ 
      
    // First co-prime pair 
    System.out.print(1 + " " + n + "\n"); 
  
    // As a pair (1 n) has 
    // already been Printed 
    k--; 
  
    for(long i = 2; i <= Math.sqrt(n); i++)
    { 
          
        // If i is a factor of N 
        if (n % i == 0)
        { 
            System.out.print(1 + " " + i + "\n"); 
            k--; 
            if (k == 0) 
                break; 
  
            // Since (i, i) won't form 
            // a coprime pair 
            if (i != n / i)
            { 
                System.out.print(1 + " " + 
                             n / i + "\n"); 
                k--; 
            } 
            if (k == 0) 
                break; 
        } 
    } 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int N = 100; 
    int K = 5; 
  
    FindPairs(N, K); 
}
} 
  
// This code is contributed by princiraj1992

Python3

# Python3 implementation of
# the above approach
from math import sqrt
  
# Function prints the
# required pairs
def FindPairs(n, k):
  
    # First co-prime pair
    print(1, n)
  
    # As a pair (1 n) has
    # already been Printed
    k -= 1
  
    for i in range(2, int(sqrt(n)) + 1):
  
        # If i is a factor of N
        if(n % i == 0):
            print(1, i)
  
            k -= 1
            if(k == 0):
                break
  
            # Since (i, i) won't form
            # a coprime pair
            if(i != n // i):
                print(1, n // i)
                k -= 1
  
            if(k == 0):
                break
  
# Driver Code
if __name__ == '__main__':
  
    N = 100
    K = 5
  
    FindPairs(N, K)
  
# This code is contributed by Shivam Singh

C#

// C# implementation of 
// the above approach 
using System;
  
class GFG{ 
  
// Function prints the 
// required pairs 
static void FindPairs(int n, int k) 
{ 
      
    // First co-prime pair 
    Console.Write(1 + " " + n + "\n"); 
  
    // As a pair (1 n) has 
    // already been Printed 
    k--; 
  
    for(long i = 2; i <= Math.Sqrt(n); i++)
    { 
          
        // If i is a factor of N 
        if (n % i == 0)
        { 
            Console.Write(1 + " " + i + "\n"); 
            k--; 
            if (k == 0) 
                break; 
  
            // Since (i, i) won't form 
            // a coprime pair 
            if (i != n / i)
            { 
                Console.Write(1 + " " + 
                          n / i + "\n"); 
                k--; 
            } 
            if (k == 0) 
                break; 
        } 
    } 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
    int N = 100; 
    int K = 5; 
  
    FindPairs(N, K); 
}
} 
  
// This code is contributed by Rajput-Ji

输出：

1 100
1 2
1 50
1 4
1 25

时间复杂度： O(sqrt(N))
辅助空间： O(1)