How to find divisors of number on specified interval? - c++

Question

I want to write a simple progam using WHILE loop, with which you could get all divisors of the number which you put in.
For example you want all divisors of number 30, which are: 1, 2, 3, 5, 6, 10, 15, 30.
Now you want program to display only numbers on interval (for example) from 5 - 10, which are 5, 6 and 10.
What i tried so far is getting all those divisors using FOR sentence, but without intervals, so I am stucked and don't know how could make it also in WHILE loop.
#include <iostream>
using namespace std;
int main() {
int input_number;
cin >> input_number;
cout << "All numbers are " << input_number << endl;
for (int i = 1; i <= input_number; i++) {
if (input_number % i == 0) {
cout << i << " ";
}
}
return 0;
}
Thanks for your help in addition.

Answer 1

Solving this problem in a while loop is probably not as simple (or brute-force) as solving it in a for loop.
Consider this rendition of the loop using while:
int input_number;
std::cin >> input_number;
int i = 1; //Start trying to divide the input number by 1
int limit = input_number; //Termination condition for the loop
while(i < limit) {
if(input_number % i == 0) { //If divisible by i, both i and input_number/i are factors
std::cout << i << " " << input_number / i << " ";
}
++i; //Try dividing by the next integer
/*Set the limit to our latest input_number/i so we don't get duplicate
results (e.g. (5, 6) and (6, 5) for input_number = 30)*/
limit = input_number / i;
}
For input_number = 30, this loop only runs for 5 iterations, while the for loop version runs for 30 iterations.
Bottom Line
for loops and while loops are interchangeable, but approaching a problem using a for loop might bring you to a different solution faster than approaching the problem with a while loop and vice versa as they help you think about the problem from different perspectives.
Extra Information
The for loop version allows for certain optimization techniques such as parallel accumulators, which are not possible in the while loop version as each iteration depends on the previous.

Answer 2

Why would you want to do this as a while loop? for is definitely the correct choice of loop here. Checking all divisors is done quite clearly via:
for (int i = 1; i <= input_number; i++) { ... }
If you want to add a narrower window to that, you can just change the bounds of the loop:
for (int i = lower_bound; i <= upper_bound; i++) { ... }
Turning that into a while loop would just involve unwrapping those statements into:
int i = lower_bound;
while (i <= upper_bound) {
...
i++;
}
But this is more error-prone - if you had a continue in your loop body, i would not get incremented.