HDOJ(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

Sample Output 1 3 30

水题。。注意范围。！！！java用long型可以AC，只是注意中间计算结果也有可能溢出int型范围，也要转换为long才行。 还有，注意判断条件退出不是输入-1，而是输入小于0的数就是退出了。

import java.util.Scanner;

public class Main{
    static long db[] = new long[100001];
    public static void main(String[] args) {
        dabiao();
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n =sc.nextInt();
            if(n<0){
                return;
            }
            System.out.println(db[n]);
        }
    }
    private static void dabiao() {
        db[1]=1;
        db[2]=3;
        for(int i=3;i<db.length;i++){
            if(i%3==0){
                db[i]=db[i-1]+i*(long)i*i;
                //这里的i*i要强转成long，long*int还是long，否则i*i*i会超int范围
            }else{
                db[i]=db[i-1]+i;
            }
        }

    }

}