#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.
// 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; }
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