Question

Need help programing this in C.

rinteivsors Print the proper divisors of an integer value The program should read a single integer input value, which you can assume will be positive. It should then print a single line of output with all of the proper divisors of the input value in order from least to greatest. A proper divisor d of an integer n is an integer that evenly divides n: i.e., nld is an integer For example, if the input is 12, the output of the program should be 1 2 3 4 6 Hints All of the proper divisors of n are less than or equal to n/2 Use a for loop to iterate through possible divisors. divisors. least one space between each divisor. . Think about how to determine which possible divisors are actual . Make sure all of the divisors are printed on the same line, with at

Answer 1

#include<stdio.h>
int main()
{
   int divisor, input; // Declare variables
   printf("Enter an integer :"); // Ask user to enter input
   scanf("%d", &input); // Read input from terminal
   for(divisor=1; divisor<=input/2; divisor++) //Loop from 1 to input/2
   {
       if(input%divisor == 0) //Check if it is a proper divisor
           printf("%d ", divisor); //If yes, then print it
   }
   printf("\b\n"); //Remove the extra whitespace at end and print newline
   return 0; //return success
}

Note : The program is done under linux environment. Please note that the program won't display any output for numbers which don't have a proper divisor. And I am assuming the input to be positive. For negative input, add a check after taking input, to return if number is negative. Thanks. Ping me back for any queries/help.

Answer 2

// A O(sqrt(n)) program that prints all divisors
// in sorted order
#include <bits/stdc++.h>
using namespace std;

// function to print the divisors
void printDivisors(int n)
{
	// Vector to store half of the divisors
	vector<int> v;
	for (int i = 1; i <= sqrt(n); i++) {
		if (n % i == 0) {

			// check if divisors are equal
			if (n / i == i)
				printf("%d ", i);
			else {
				printf("%d ", i);

				// push the second divisor in the vector
				v.push_back(n / i);
			}
		}
	}

	// The vector will be printed in reverse
	for (int i = v.size() - 1; i >= 0; i--)
		printf("%d ", v[i]);
}

/* Driver program to test above function */
int main()
{
	printf("The divisors of 100 are: \n");
	printDivisors(100);
	return 0;
}