LCM program in C++

Question

#include <iostream>
using std::cout;
using std::cin;

int main()
{

    int num1 = 0;
    int num2 = 0;

    cout << "Please enter two numbers (remember to put 
a space between them):\n";

    cin >> num1 >> num2;

    const int num_limit = num1 * num2;

    for (int i = 1; i <= num_limit; i++)
    {
        int product = num1 * i;
        int product2 = num2 * i;

        // test for when multiples are equal
        if (product == product2)
        {
        cout << "The LCM of " << num1 << " and " << 
num2 << " is: "; 
        }

     }
}

I'm trying to get the LCM from two integers which the user enters. The if statement within the for loop isn't doing what I intended it to do, as nothing prints when product and product2 are equal. What is the solution to this little problem?

*i* is same in both multiplications, so products are equal when *num1 == num2*... — hyde, Aug 22 '18 at 19:32
https://www.geeksforgeeks.org/program-to-find-lcm-of-two-numbers/ — drescherjm, Aug 22 '18 at 19:43

score 1 · Answer 1 · edited Mar 23 '21 at 06:55

1

int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);  
}

int lcm(int a,int b)
{
    //Efficient Solution
    // a*b=gcd(a,b)*lcm(a,b)

    return (a*b)/gcd(a,b);
}

int main()
{
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int a,b;
    cin>>a>>b;
    cout<<lcm(a,b);
    return 0;
}

Output

15 12
60

edited Mar 23 '21 at 06:55

Michael Kotzjan

2,093
2
14
23

answered Mar 23 '21 at 06:44

Amisha Sahu

89
6

This is the correct solution based on the Euclidean algorithm. If anyone is in doubt, try running this and the accepted answer on a couple of 6-digit numbers. – nielsen Mar 23 '21 at 11:42

score 0 · Accepted Answer · edited Mar 23 '21 at 11:17

0

Try this out

//#include "stdafx.h"
#include <iostream>
#include <conio.h>
using std::cout;
using std::cin;

int main()
{

    int num1 = 0;
    int num2 = 0;

    cout << "Please enter two numbers (remember to put a space between them):\n";

    cin >> num1 >> num2;

    int num_limit = num1 * num2,LCM;
    for (int i = 1; i <= num_limit; i++)
    {
        LCM = num1 * i;

        // test for when multiples are equal
        if (LCM%num2==0) break;
    }
    cout << "The LCM of " << num1 << " and " << num2 << " is: " << LCM;
    _getch();// replace with getch(); if didn't work
}

edited Mar 23 '21 at 11:17

Michael Kotzjan

2,093
2
14
23

answered Aug 22 '18 at 19:48

Mostafa Ayaz

480
1
7
16

Seems to work fine. Thank you :) Where does LCM come from? Is this a 'special' variable, as I don't see it declared anywhere? – Njgardner90 Aug 22 '18 at 20:25
You're welcome. I declared it where you declared num_limit variable (at the end of the line)... – Mostafa Ayaz Aug 22 '18 at 20:28

LCM program in C++

2 Answers2