N以下3或5的所有倍数的总和。

Question

So I'm doing the Project Euler challenge and I'm stuck at the first one, I'm using Java as pl. for example if we have to 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. We have to find the sum of all the multiples of 3 or 5 below N.

My code works on Eclipse but I get "Nice try, but you did not pass this test case." with stdout : No Response and when I submit the code I get Wrong Answer on all test cases, here's the code:

public class Solution {
    public static void main(String[] args) {
        for (int j = 0; j < args.length; j++) {
            int N = Integer.parseInt(args[j]);
            if (Somme(N) != 0) {
                System.out.println(Somme(N));
            }
        }
    }

    public static int Somme(int Nn) {
        int s = 0;
        for (int i = 0; i < Nn; i++) {
            if (((i % 3) == 0) || ((i % 5) == 0)
                && !(((i % 3) == 0) && ((i % 5) == 0))) {
                s = s + i;
            }
        }
        return (s);
    }
}

UPDATE : So, I looked more and it turns out that this is how it's supposed to be done:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Solution{
public static void main(String[] args) throws IOException {


    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String line = br.readLine();
    int Nbr = Integer.parseInt(line);


        for(int j=0; j<Nbr;j++)
        {
            BufferedReader br2 = new BufferedReader(new   InputStreamReader(System.in));
            String line2 = br2.readLine();
            String[] numbers = new String[Nbr];
            numbers[j]= line2;
            System.out.println(Somme(Long.parseLong(numbers[j])));
        }

        }


public static long Somme(long Nn) {
    long s = 0;
    for (int i = 0; i < Nn; i++) {
        if (((i % 3) == 0) || ((i % 5) == 0)) {
            s = s + i;
        }
    }
    return (s);
}

}

Now the only problem left is that I want it to be able to read ALL the numbers THEN display the sum, for now it reads one number and display the sum right after it, any ideas?

Answer 1

You are skipping some numbers that should not be skipped.

if (((i % 3) == 0) || ((i % 5) == 0)
    && !(((i % 3) == 0) && ((i % 5) == 0)))

This statement says: i must be divisible by 3 or 5 AND is must not be divisible by 3 and 5 . Rephrased: i must be divisible by 3 or 5 , but not both of them. Just delete the second line and it should work.

Answer 2

I believe it is a combination of what Turing85 said and wazaaaap. The examples for Project Euler all show it does not take different inputs. You just have to produce the correct output. So replace Integer.parseInt(args[j]); with Integer.parseInt(1000); To add to what Turing said, the solution should follow the following psuedocode:

target=999
sum=0
for i=1 to target do
if (i mod 3=0) or (i mod 5)=0 then sum:=sum+i
output sum

Answer 3

Using a for loop, you can get all the numbers from 0 to 1000 , and by using an if condition, you get the required numbers ie, multiples of 3 , 5 , adding them gives the final output, like the following:

public class SUM_3_5{

    public static void main(String []args) {
        int sum=0;
        int i;

        for(i=0;i<1000;i++)
        {
             if(i%3==0||i%5==0)
                 sum=sum+i;
        }

        System.out.println(sum);
    }
}