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..-_-

  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 the result sum(n).

1
2
3
-1

1
3
30

Wendell

HDU 2007-11 Programming Contest_WarmUp

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#include<stdio.h>
#include<string.h>
__int64 sum[100005];
void fun()
{
	__int64 i;//要注意i定义为64位，不懂。。也是醉了
	sum[0]=0;
	for(i=1;i<100005;i++)
	{
		if(i%3==0)
			sum[i]=sum[i-1]+i*i*i;
		else
			sum[i]=sum[i-1]+i;
	}
}
int main()
{
	int n;
	fun();
	while(scanf("%d",&n)!=EOF)
	{
		if(n<0)
			break;
		printf("%I64d\n",sum[n]);
	}
}