HDOJ2132 An easy problem

An easy problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14373    Accepted Submission(s): 3966

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

import java.util.Scanner;

public class Main {
    private static Scanner scanner;
    private static long arr[];
    public static void main(String[] args) {
        scanner = new Scanner(System.in);
        dabiao();
        while (scanner.hasNext()) {
            int n = scanner.nextInt();
            if (n < 0) {
                break;
            }
            System.out.println(arr[n]);
        }
    }
    
    private static void dabiao() {
        arr = new long[100001];
        arr[1] = 1;
        for (long i = 2; i < arr.length; i++) {
            int j = (int)i;
            if(j%3 == 0){
                    //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）
                 arr[j] = arr[j-1]+i*i*i;
            }else {
                 arr[j] = arr[j-1]+j;
            }
        }
    }
}