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最新推荐文章于 2017-08-17 16:38:58 发布

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本文提供了一个关于递归问题的简单实例，通过定义一个函数sum(n)，根据输入n的值来计算特定的数学序列。该问题旨在帮助读者理解并解决类似的递归问题，通过输入整数n（n<100000），程序将输出计算得到的sum(n)。重点在于处理n的边界条件，特别是当n能被3整除时的特殊计算方式。

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An easy problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6143 Accepted Submission(s): 1666

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

Sample Output

Author

Wendell

Source

HDU 2007-11 Programming Contest_WarmUp

Recommend

威士忌

对于这个问题呢一点要注意n的取值哦 n要用_int64来定义否则会报错哦所以一定要小心

#include<iostream>
#include<cstdio>

using namespace std;
int main()
{
    _int64 sum,n;

    while(cin>>n)
    {
        sum=0;
        if(n<0)
        break;

        if(n%3==0)
       {
           n=n/3;
	       sum=(n+1)*(n+1)*n*n/4*27+3*n*n;
       }
        else if(n%3==1)
       {
           n=n/3;
	       sum=(n+1)*(n+1)*n*n/4*27+3*n*n+n*3+1;
       }
        else if(n%3==2)
        {
           n=n/3;
	       sum=(n+1)*(n+1)*n*n/4*27+3*n*n+n*3*2+3;
//这里要小心一点很容易出错哦   呵呵        }
        cout<<sum<<endl;
    }
    return 0;
}