All divisors#8669
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All divisors (Editorial)

If is a divisor of , then is also a divisor of . Moreover, if is a perfect square, then the divisors \sqrt{n} and n /\sqrt{n} are the same.

Algorithm implementation

Read the input value of .

scanf("%d",&n);

Iterate over divisors from to \sqrt{n} not inclusive.

for(i = 1; i < sqrt(n); i++)
  if (n % i == 0)
  {

If is a divisor of , then is also a divisor of .

    v.push_back(i);
    v.push_back(n / i);
  }

Check if is a perfect square. In this case, add the divisor \sqrt{n} to the vector.

i = sqrt(n);
if (i * i == n) v.push_back(i);

Sort the divisors in ascending order.

sort(v.begin(),v.end());

Print all divisors in one line.

for(i = 0; i < v.size(); i++)
  printf("%d ",v[i]);
printf("\n");

Java implementation

import java.util.*;

public class Main
{
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    int n = con.nextInt();
    ArrayList<Integer> v = new ArrayList<Integer>(); 
    for(int i = 1; i < Math.sqrt(n); i++)
      if (n % i == 0)
      {
        v.add(i);
        v.add(n / i);
      }

    int i = (int)Math.sqrt(n);
    if (n % i == 0) v.add(i);

    Collections.sort(v);

    for(i = 0; i < v.size(); i++)
      System.out.print(v.get(i) + " ");
    System.out.println();
    con.close();
  }
}
Java

Python implementation

import math

Read the input value of .

n = int(input())

We'll add the divisors of to the list .

lst = []

Iterate over divisors from to \sqrt{n} not inclusive.

for i in range(1,int(math.sqrt(n))):

If is a divisor of , then is also a divisor of .

  if n % i == 0:
    lst.append(i)
    lst.append(n//i)

Check if is a perfect square. In this case, add the divisor \sqrt{n} to the vector.

i = int(math.sqrt(n));
if n % i == 0: lst.append(i);

Sort the divisors in ascending order.

lst.sort()

Print all divisors in one line.

print(*lst)