HDU 2132 An easy problem
http://acm.hdu.edu.cn/showproblem.php?pid=2132
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
代码:
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5 + 10; long long sum[maxn]; int main() { sum[0] = 0; for(long long i = 1; i < maxn; i ++) { if(i % 3 == 0) sum[i] = sum[i - 1] + i * i * i; else sum[i] = sum[i - 1] + i; } int x; while(~scanf("%d", &x)) { if(x < 0) break; else printf("%lld\n", sum[x]); } return 0; }