题目：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..-_-
output the result sum(n).
1 2 3 -1
1 3 30
题解：对于这道题我有两种解法:1、题目已经给出1 2 3的sum[i]，对于公式sum(i) = sum(i-1) + i*i*i，只要知道知道前一个数，就可以算出后一个数，就可以用一个for循环，算出后面的数，存成一个数组，在知道i值后，直接在数组里面调出就可以了。不过这个数组要给的很大。2、用函数来做，用递归思想，也涉及到循环，不过不用开数组。
#include<stdio.h>
int main()
{
__int64 a[];
__int64 n,i;
a[]=;
a[]=;
a[]=;
for(i=;i<=;i++)
{
if(i%==) a[i]=a[i-]+i*i*i;
else a[i]=a[i-]+i;
}
while(scanf("%I64d",&n)!=EOF)
{
if(n<) break;
printf("%I64d\n",a[n]);
} return;
}