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HDU2132 An easy problem【水题】

2024-11-20 18:12:02   202人阅读

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;
}


HDU2132 An easy problem【水题】


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