Project Euler Problem - 1 does not return the value i expected-CodePudding

I did not learn loops using for yet, so I solved this problem using while but I can't find the issue with my problem, sum should be 233168 apparently, and I am getting 234168 I simply cannot identify where I missed it.

Question: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

package net.projecteuler;

public class Problem01 {
    public static void main(String[] args) {
        
        int n, sum;
        boolean notFinished = true;
        n = 0;
        sum = 0;
        
        
        while(notFinished) {
            if(n % 3 == 0 || n % 5 == 0) {
                sum = (sum   n);
            }
            n = (n   1);
            if(n > 1000) {
                notFinished = !true;
            }
        }
        System.out.print("A soma dos números é "   sum);
    }
}

CodePudding user response：

Your condition is wrong. The problem statement is (bolding for emphasis):

Find the sum of all the multiples of 3 or 5 below 1000.

Whereas your program also sums 1000 itself, which is divisible by 5. If you change the condition n > 1000 to n >= 1000, you'll get the correct answer.

Side note: You should really look into for loops. That would be a much better fit for this challenge.

CodePudding user response：

Your condition is wrong. The while loop runs one too many times, because you are comparing n > 1000 instead of n >= 1000.

With that being said, you can also improve your loop. The boolean variable is not needed, as you can straight up check the condition in the while statement, like this - while (n < 1000).

In addition, if you are going to use your boolean variable, I would probably not set it to false by saying isFinished = !true, but by doing isFinished = false or isFinished = !isFinished.

CodePudding user response：

Certainly your issue is with the if condition. It should be n >= 1000

Also you can simplify your algo like this. An alternative solution to your question.

public class Main {
    public static void main(String[] args) {
        System.out.println(sum(1000, 3, 5));
    }

    private static int sum(int i, int a, int b) {
        i = --i;
        int c = a * b;
        int k1 = i / a;
        int k2 = i / b;
        int k3 = i / c;

        return (k1 * (a   (k1 * a)))/2   (k2 * (b   (k2 * b)))/2 - (k3 * (c   (k3 * c)))/2;
    }
}

Note this only works if a and b are prime numbers. If a and b are not prime, then c should be LCM of a and b

int c = lcm(a,b)