hdu 213

#include<cstdio>int main(){ int n; __int64 a[100001],i; a[0]=0; for(i=1;i<=100000;i++) { if(i%3==0) a[i] = a[i-1] + i*i*i; else a[i] =a[i-1]+i; } while(scanf("%d",&n)!=EOF&&n>=0) { printf("%I64d\n",a[n]); } return 0;}

 

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實習工資）阿里巴巴實習生招聘信息（僅限2013屆本、碩畢業生）

An easy problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4804    Accepted Submission(s): 1309

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