D - An easy problem

We once did a lot of recursional problem . I think some of them is easy for you and some if hard for you.
Now there is a very easy problem . I think you can AC it.
  We can define sum(n) as follow:
  if i can be divided exactly by 3 sum(i) = sum(i-1) + i*i*i;else sum(i) = sum(i-1) + i;
  Is it very easy ? Please begin to program to AC it..-_-

Input  The input file contains multilple cases.
  Every cases contain only ont line, every line contains a integer n (n<=100000).
  when n is a negative indicate the end of file.Output  output the result sum(n).
Sample Input

1
2
3
-1

Sample Output

1
3
30

刚开始看题里面用递归！就用了递归发现超内存了。于是就打表了。注意了，下次可以吸取这次的经验。
#include<iostream>
using namespace std;
#define N 100002
int main()
{
	long long num[N],i;
	num[0] = 0;
	for ( i = 1; i < N; i++)
	{
		if (i % 3) num[i]=num[i - 1] + i;
		else num[i]=num[i - 1] + i*i*i;
	}
	int n;
	while (cin >> n)
	{
		if (n < 0) break;
		cout << num[n] << endl;
	}
	return 0;
}