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

123-1

 

Sample Output

1330

 

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;            }        }    }}