I was going through your blog and found the problem interesting.

An alternative solution to your constant time algorithm could be the use of sum of an arithmetic progression series. I think its pretty similar to the solution you suggested.

Finding the sum is like series of two numbers:

A (multiples of 3)= 3,6,9,12,15………An; where n is the number of terms in the series

B (multiples of 5)= 5,10,15,20,25,30…….Bn

where Sum of a finite series= (n/2)*(2*first_term+(n-1)diff.)

So, considering an example;

Case 1: Series A(multiples of 3 between 1-10): 3,6,9

No. of terms=3

Sum = (3/2)* [2*3+(3-1)*3]
= 18

Case 2: Series B: 5, 10, 15, 20

No. of terms = 4

Sum = (4/2) * [2*5+(4-1)*5]
= 2 * 25
= 50

This way, the modification looks like;

limit=999
threes=sum_ap(limit,3);
fives=sum_ap(limit,5);

where
int sum_ap(limit, n)
{ fact=limit/n;
Sum=(fact/2)* [2*n + (fact-1)*n];
return Sum;
}

And, the final output would be
threes+fives-fifteen

Thanks,

Shirish

I’m far more proud of a polished, well-rounded product than a piece of nice code.

Oh someone already said that.. programmers mind it tell ya!

I wish there were groupies too, but unfortunately programming is not that kind of business ;(