Java Implementation of Problem 1
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1/*
2Project Euler Problem 1
3
4I know that this could be done faster with a traditional for loop, but I wanted
5to see if iterators were reasonably possible in C, since it makes the prime
6number infrastructure a lot easier to set up.
7
8Problem:
9
10If we list all the natural numbers below 10 that are multiples of 3 or 5, we
11get 3, 5, 6 and 9. The sum of these multiples is 23.
12
13Find the sum of all the multiples of 3 or 5 below 1000.
14*/
15package euler;
16
17public class p0001 implements IEuler {
18 @Override
19 public Object answer() {
20 int answer = 0;
21 for (int i = 0; i < 1000; i += 3)
22 answer += i;
23
24 for (int i = 0; i < 1000; i += 5)
25 answer += i;
26
27 for (int i = 0; i < 1000; i += 15)
28 answer -= i;
29
30 return answer;
31 }
32}