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

123-1

 

Sample Output

1330

题目没什么好说的，思路很简单，打表。注意一个问题：i*i*i 可能会超int ，所以 i 必须定义 long long 。

#include<stdio.h>#include<string.h>__int64 f[100005],i,n;int main(){    memset(f,0,sizeof(f));    f[1]=1;    for(i=2; i<=100002; i++)        if(i%3==0) f[i]=f[i-1]+i*i*i;    //此处 i 必须是__int64，不然会WA        else f[i]=f[i-1]+i;    while(~scanf("%I64d",&n) && n>=0) {        printf("%I64d\n",f[n]);    }    return 0;}