D - An easy problem
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.Output output the result sum(n).
Sample Input
1 2 3 -1
Sample Output
1
3
30
刚开始看题里面用递归!就用了递归发现超内存了。于是就打表了。注意了,下次可以吸取这次的经验。
#include<iostream>
using namespace std;
#define N 100002
int main()
{
long long num[N],i;
num[0] = 0;
for ( i = 1; i < N; i++)
{
if (i % 3) num[i]=num[i - 1] + i;
else num[i]=num[i - 1] + i*i*i;
}
int n;
while (cin >> n)
{
if (n < 0) break;
cout << num[n] << endl;
}
return 0;
}