hdu 2132... 被基本问题考住了。。

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

 

Sample Output

1
3
30

#include<iostream>
using namespace std;
int main()
{
    __int64 i,n,a[100001];
    a[0]=0;
    for(i=1;i!=100001;++i)
    {
        if(i%3==0)
            a[i]=a[i-1]+i*i*i;//////  i*i*i的过程中 当I等于99999的时候 用int去存的话 有溢出（这里是吧i*i*i的值放在一个int的空间里面 然后再进行赋值运算  要尤其注意）
        else
            a[i]=a[i-1]+i;
    }
    while(cin>>n&&n>=0)
    {
        cout<<a[n]<<endl;
    }
    return 0;
}