HDU2132 An easy problem【水题】

An easy problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10149    Accepted Submission(s): 2689

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

Author

Wendell

Source

HDU 2007-11 Programming Contest_WarmUp

题目大意：给了递推公式，如果当前i%3==0，则sum(i) = sum(i-1) + i*i*i；否则

sum(i) = sum(i-1) + i。

思路：因为数据略大一些，所以用__int64整型来存储结果。

#include<iostream>
#include<algorithm>
using namespace std;

__int64 ans[100010];

int main()
{
    for(__int64 i = 1; i <= 100000; ++i)
    {
        if(i % 3 == 0)
            ans[i] = ans[i-1] + i*i*i;
        else
            ans[i] = ans[i-1] + i;
    }
    __int64 N;
    while(cin >> N && N >= 0)
    {
        cout << ans[N] << endl;
    }
    return 0;
}