[cfoley]

You should be right, but when I tried it I got the same answer. I wonder if it's a bug with the JVM I'm using. I did search the bug reports but no luck.

[Alex_H]

I tested the code and also got the correct answer....very strange indeed lol

[baavgai]

The goal is smallest code possible now? What an odd aspiration for Java.

Still, silly is fun. For the smallest code, I offer this:

int f(int n) { return (--n<1)?0:f(n)+(n%3*n%5>0?0:n); }

[cfoley]

... but that's not a Java program baavgai



I tried stuff like that (also from recursive constructors!) but calling methods from static main was too many characters.

[cfoley]

93 characters... but you need assertions enabled.

class C {
	public static void main(String[] a) {
		for (int s=0,i=0;;)/> {
			s += ++i%3*i%5 > 0 ? 0:i;
			assert i < 999 : s;
		}
	}

}

[baavgai]

cfoley, on 02 February 2011 - 09:33 AM, said:

... but that's not a Java program baavgai

Agreed, it is clearly a Java function, which is far more useful...

However, I have yet to see some of your programs offer output, which is also fundamentally useless as a program.

Using the logic offered, this is a valid "program." It's 104 chars:

class C {
	static int f(int n) { return (--n<1)?0:f(n)+(n%3*n%5>0?0:n); }
	public static void main(String[] a) { 
		f(1000);
	}
}

A more honest offering would be this, at 124 chars:

class C {
	static int f(int n) { return (--n<1)?0:f(n)+(n%3*n%5>0?0:n); }
	public static void main(String[] a) { 
		System.out.println(f(1000));
	}
}

The declaration in some solutions of the sum in the loop means your really can't print it...

Now that I more fully understand the house rules... 86 chars.

class C {
	public static void main(String[] a) { 
		int s=0,n=1000;
		while(--n>0) { s += n%3*n%5>0?0:n; }
	}
}

Note, I used a very *nixy method of counting chars:

cat C.java | tr '\n' ' ' | sed 's/\s//g' | wc -c

[Dogstopper]

You social hacker you. Read between the line you did, and you managed to pull of a sly victory.

[Alex_H]

Refactored baagvai's code....85 characters

class C {
    public static void main(String[] a) {
        int s=0,n=999;
        while(n-->0) { s += n%3*n%5>0?0:n; }
    }
}

[cfoley]

I did print my answer every time. Sometimes it appeared somewhere in the err output...but it was there

Here's one that calculates it but does not print it (82 chars)

class C {
	public static void main(String[] a) {
		for (int s=0,i=0;i < 999;)
			s += ++i%3*i%5 > 0 ? 0:i;
	}
}

and if you want the declaration outside the loop it's 83 chars.

class C {
	public static void main(String[] a) {
		int s=0,i=0;
		for (;i < 999;)
			s += ++i%3*i%5 > 0 ? 0:i;
	}
}

[itengineer]

class Competition
{
	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 : " + sum);
	}
}

Thanks
itengineer

[Karel-Lodewijk]

A totally different approach.

First some math:

Lets look at the numbers we are trying to add for lets say 3.

n3 = 3, 6, 9, 12, 3n <- this is called mathematically a a sequence with general term 3n

Now we need the running sum of this seqience. The sums of the terms of a sequence is called a series.

s3 = 3, 9, 18, 30, sum from i = 0 to m of 3*i; <- This is the series of sequence n3.

Now for a lot of basic series, calculating that summation is not the only way to find the m-th term.

for this one:
sum from i = 0 to m of 3*i = (m+1)*m*3/2.

But we don't know m at this point or do we ? Well m is the number of elements of the row n3 < 1000, yes or in other workds floor(999/3).

We can do the same for 5.

But we've made a mistake, we've actually counted everything that can be divided by 3 and 5 twice. But no worry we can do the same as we did above for those numbers but this time for 5*3 and subtract that.

Using this we have a closed formula for our problem.

Now how does this translate in a program, something like this:

class C {
    public static void main(String[] a) {
        System.out.println(f(3)+f(5)-f(3*5));
    }
    static int f(int i) {
        int m = 999/i;
        return (m+1)*m*i/2;
    }
}

I couldn't get it short enough though.


P.S.: Is there any way to write math in this forums, any latex support by any chance ?

This post has been edited by Karel-Lodewijk: 03 July 2011 - 05:19 PM

[AdamSpeight2008]

Latex support has be requested before, skyhawk133 and his team are looking at it.

Spoiler

If it come, it most likely be via math tags.

 [math] latex code [/math]

[cfoley]

Karel-Lodewijk, I tried that method a few posts back but ended up sticking with variants of the loop method because they used less characters.

[peterrankings]

This works

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

[nick2price]

I am not that advance at maths, so a problem like this is difficult for me. Therefore, I do not know what formulas you are using. But if you havnt seen it, there is a listing of this problem on wikipedia. It doesnt show you the answer, it does tell you the most effecient formula though. Maybe it can be used to come up with an even shorter answer?
http://en.wikipedia....m_and_solutions