HDU 2132 An easy problem

 Problem Description 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

1
3
30

#include
#include
__int64 f[100005],i,n;  //i也要64位
int main()
{
    memset(f,0,sizeof(f));
    f[1]=1;
    for(i=2; i<=100002; i++)
        if(i%3==0) f[i]=f[i-1]+i*i*i;
        else f[i]=f[i-1]+i;
    while(~scanf("%I64d",&n) && n>=0) {
        printf("%I64d\n",f[n]);
    }
    return 0;
}