打印所有 N 的除数且与除数商互质的数

给定一个正整数N ，任务是打印所有数字，比如K ，使得K是N的除数，并且K和N / K互质。

例子：

Input: N = 12  
Output: 1 3 4 12  
Explanation:
All numbers K such that it is divisor of N(= 12) and K and N/K are coprime:

1. 1: Since 1 is a divisor of 12 and 1 and 12(= 12/1) are coprime.
2. 3: Since 3 is a divisor of 12 and 3 and 4( = 12/3) are coprime.
3. 4: Since 4 is a divisor of 12 and 4 and 3( = 12/4) are coprime.
4. 12: Since 12 is a divisor of 12 and 12 and 1( = 12/12) are coprime.

Input: N = 100  
Output: 1 4 25 100

朴素方法：解决给定问题的最简单方法是迭代范围[1, N] 并检查每个数字K是否K是N的除数 K 和 N/K 的 GCD 是否为1 。如果发现是true ，则打印K 。否则，检查下一个数字。

时间复杂度： O(N*log N)
辅助空间： O(1)

有效方法：上述方法也可以通过使用所有除数成对存在的观察来优化。例如，如果N是100 ，那么所有的除数对是： (1, 100), (2, 50), (4, 25), (5, 20), (10, 10) 。

因此，想法是在[1, √N]范围内进行迭代并检查是否满足两个给定条件，即任何数K是否是N的除数 K和N/K的 GCD 是否为1 。如果发现是true ，则打印两个数字。在两个相等的除数的情况下，只打印其中一个。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print all numbers
// that are divisors of N and are
// co-prime with the quotient
// of their division
void printUnitaryDivisors(int n)
{
    // Iterate upto square root of N
    for (int i = 1; i <= sqrt(n); i++) {
 
        if (n % i == 0) {
 
            // If divisors are equal and gcd is
            // 1, then print only one of them
            if (n / i == i && __gcd(i, n / i) == 1) {
                printf("%d ", i);
            }
 
            // Otherwise print both
            else {
                if (__gcd(i, n / i) == 1) {
                    printf("%d %d ", i, n / i);
                }
            }
        }
    }
}
 
// Driver Code
int main()
{
    int N = 12;
    printUnitaryDivisors(N);
 
    return 0;
}

Python3

# python 3 program for the above approach
from math import sqrt, gcd
 
# Function to print all numbers
# that are divisors of N and are
# co-prime with the quotient
# of their division
def printUnitaryDivisors(n):
   
    # Iterate upto square root of N
    for i in range(1,int(sqrt(n)) + 1, 1):
        if (n % i == 0):
           
            # If divisors are equal and gcd is
            # 1, then print only one of them
            if (n // i == i and gcd(i, n // i) == 1):
                print(i)
 
            # Otherwise print both
            else:
                if (gcd(i, n // i) == 1):
                    print(i, n // i,end = " ")
                 
# Driver Code
if __name__ == '__main__':
    N = 12
    printUnitaryDivisors(N)
 
    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
  
static int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}
// Function to print all numbers
// that are divisors of N and are
// co-prime with the quotient
// of their division
static void printUnitaryDivisors(int n)
{
    // Iterate upto square root of N
    for (int i = 1; i <= (int)Math.Sqrt(n); i++) {
 
        if (n % i == 0) {
 
            // If divisors are equal and gcd is
            // 1, then print only one of them
            if (n / i == i && gcd(i, n / i) == 1) {
                Console.Write(i+" ");
            }
 
            // Otherwise print both
            else {
                if (gcd(i, n / i) == 1) {
                    Console.Write(i + " " +n / i+ " ");
                }
            }
        }
    }
}
 
// Driver Code
public static void Main()
{
    int N = 12;
    printUnitaryDivisors(N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    static int gcd(int a, int b)
    {
        return b == 0 ? a : gcd(b, a % b);
    }
    // Function to print all numbers
    // that are divisors of N and are
    // co-prime with the quotient
    // of their division
    static void printUnitaryDivisors(int n)
    {
        // Iterate upto square root of N
        for (int i = 1; i <= (int)Math.sqrt(n); i++) {
 
            if (n % i == 0) {
 
                // If divisors are equal and gcd is
                // 1, then print only one of them
                if (n / i == i && gcd(i, n / i) == 1) {
                    System.out.print(i + " ");
                }
 
                // Otherwise print both
                else {
                    if (gcd(i, n / i) == 1) {
                        System.out.print(i + " " + n / i
                                         + " ");
                    }
                }
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 12;
        printUnitaryDivisors(N);
    }
}

Javascript

 
// JavaScript program for the above approach
function gcd( a,  b)
    {
        return b == 0 ? a : gcd(b, a % b);
    }
    // Function to print all numbers
    // that are divisors of N and are
    // co-prime with the quotient
    // of their division
    function printUnitaryDivisors( n)
    {
        // Iterate upto square root of N
        for (var i = 1; i <= Math.sqrt(n); i++) {
 
            if (n % i == 0) {
 
                // If divisors are equal and gcd is
                // 1, then print only one of them
                if (n / i == i && gcd(i, n / i) == 1) {
                    document.write(i + " ");
                }
 
                // Otherwise print both
                else {
                    if (gcd(i, n / i) == 1) {
                        document.write(i + " " + n / i
                                         + " ");
                    }
                }
            }
        }
    }
 
    // Driver Code
        var N = 12;
        printUnitaryDivisors(N);
         
// This code is contributed by shivanisingh2110  
 
 

输出：

1 12 3 4

时间复杂度： O(√N*log N)
辅助空间： O(1)