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..-_-
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.
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
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;
}