找出两个数的最大公约数

Question

I want to find the greatest common divisor between two input numbers, and I bumped into a problem.

I am not sure if the method I used to actually find the divisor is right. I did it by dividing both of the numbers with every number until the number reaches itself.

#include <iostream>
#include <string>

using namespace std;
    
int main() {

    int num1, num2;
    int large = 0;
    int gcd = 0;

    cout << "this program finds the greatest common divisor" << endl;
    cout << "input first  number > "; cin >> num1; 
    cout << "input second number > "; cin >> num2;

    if (num1 > num2)
        large = num1;
    else
        large = num2;

    cout << "larger number  > " << large << endl;
    cout << endl;

    for (int i = 0; i < large + 1; i++) {
        if (num1 % i == 0 && num2 % i == 0) {
            gcd = i;
        }
    }

    cout << "The gcd of " << num1 << " and " << num2 << " is " << gcd << endl;
    cout << endl;
}

Answer 1

Yes it makes sense. Basically it does the job. You do few things unnecessary. Start loop:

for (int i = 0; i < large + 1; i++)

with i=1, as you cant div by 0. Also you are performing unnecessary modulo operations when i is between larger and smaller number. Ofc it can be optimized a lot. Obviously on numbers smaller/2 up to smaller you also cant find common divisor(excluding border numbers), and than earlier smaller/3 to smaller/2(-#-). However logic seems solid: last found divisor by the loop will be the gratest common divisor.

Answer 2

You need some modifications to make it correct.
First, check whether one of them is zero, if so, return the other.
Second, if this didn't happen, start dividing from 1, not zero, up to the smallest, not the largest.

int gcd, small;
if (num1 == 0)
    gcd = num2;
else if (num2 == 0)
    gcd = num1;
else
{
    if (num1 < num2)
        small = num1;
    else
        small = num2;
    for (gcd = 1; gcd < small + 1; gcd++)
        if (num1 % gcd == 0 && num2 % gcd == 0)
            break;
}
cout << "The gcd of " << num1 << " and " << num2 << " is " << gcd << endl;

But you should now that there are easier and faster ways, like my answer to this question .