Project Euler #1 in Java

Question

I\'m having problems with this code. I don\'t want to look at others, so I\'m wondering what\'s wrong with mine.

If we list all the natural numbers below 10 that are mu

Answer 1

just write this simple java code.

 public static void main(String[] args)
{
int i,sum=0;
for ( i = 3; i <1000; i++)
 {
 if ((i % 3 == 0)||(i%5==0) )
 sum=sum+i;
 }
System.out.print(sum);
}

You will get the output as 233168

Answer 2

Some certain conditions occurs where both conditions satisfies for 3 as well as 5 also. like when i=15 satisfies for both 15%3==0 and 15%5==0. so probably your answer is more than expected because you tried for both 3 and 5 separately. by doing this during these certain conditions you add repeated values. Hence it is better to check in single loop. like this -

    for(int i=0 ; i<1000 ; i++)
    {
        if(i%3==0 || i%5==0)
        {
            temp = temp + i;
        }
        System.out.println(temp);
    }

Answer 3

public static int sumOfMultiples(int i, int j, int limit){
    int s = --limit / i, t = limit / j, u = limit / (i * j);
    return (i*(s*(s+1)/2)) + (j*(t*(t+1)/2)) - ((i*j)*(u*(u+1)/2));
}

Test

actual = Prob1.sumOfMultiples(3, 5, 1000);
assertEquals(233168, actual);

Answer 4

You added all multiples of 15 twice. Using your algorithm, run a third loop and test if the number is divisible by 15, then remove it from the total sum.

Answer 5

Solutions

1) O(n):

A small improvement for the other answers (i can start from 3):

public static void main(String[] args) {
    int sum = 0;
    for (int i = 3; i < 1000; i++) {
        if (i % 3 == 0 || i % 5 == 0) {
            sum += i;
        }
    }
    System.out.println(sum);
}

For a bigger input number ( Integer.MAX_VALUE instead of 1000 ) it takes:

• 195 seconds

2) O(n) = O(n/3) + O(n/5) + O(n/15):

This is more efficient and uses your initial approach (remove numbers that were taken twice):

public static void main(String[] args) {
    long sum = 0 ;
    for ( long i = 3 ; i < 1000 ; i+=3 ){
        sum+=i;
    }
    for ( long i = 5 ; i < 1000 ; i+=5 ){
        sum+=i;
    }       
    for ( long i = 15 ; i < 1000 ; i+=15 ){
        sum-=i;
    }
    System.out.println(sum);
}

In the first case we have about n (1000) values for i, in the second case we have only about n/3 + n/5 + n/15 (600) values for i. The second one is also better because there are fewer comparisons ( no if involved ).

For a bigger input number ( Integer.MAX_VALUE instead of 1000 ) it takes:

• 9 seconds

3) O(1):

This solution is based on the following observation:

1 + 2 + ... + n = n*(n+1)/2

public static void main(String[] args) {
    int nr = 1000;
    nr--;
    int x3 = nr/3;
    int x5 = nr/5;
    int x15 = nr/15;
    
    long sum1 = 3*x3*(x3+1); 
    long sum2 = 5*x5*(x5+1);
    long sum3 = 15*x15*(x15+1);
    
    long sum = (sum1+sum2-sum3)/2;
    System.out.println(sum);
}

In this case, even if the input is Integer.MAX_VALUE, the computation is very fast ( less than 1 ms ).

Answer 6

If a number is a multiplier of both 3 and 5 (e.g.: 15, 30, 45, etc.), you will count it twice. So instead of two for loops, you should have one, with a complex condition:

public class Multiples {
    public static void main (String [] args) {
    int temp = 0;

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

    }

    System.out.println (temp);
   }
}