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; } } }

}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/51306358