All divisors (#8669)

Algorithm Analysis

If $i$ is a divisor of the number $n$ , then $\frac{n}{i}$ will also be a divisor of the number $n$ . In this case, if $n$ is a perfect square, then the divisors $n$ and $\frac{n}{n}$ coincide.

Algorithm Implementation

Read the input value $n$ .

scanf("%d",&n);

Iterate over divisors from 1 to $n$ not inclusively.

for(i = 1; i < sqrt(n); i++)
    if (n % i == 0)
    {
        // If i is a divisor of n, then n / i will also be a divisor of n.
        v.push_back(i);
        v.push_back(n / i);
    }

Check if $n$ is a perfect square. In this case, add the divisor $n$ 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();
    }
}

Python Implementation

import math

n = int(input())
list = []
for i in range(1, int(math.sqrt(n))):
    if n % i == 0:
        list.append(i)
        list.append(n//i)

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

list.sort()
print(*list, sep=" ")