题目链接 ：http://codeforces.com/contest/158/problem/D

D. Ice Sculptures
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

The Berland University is preparing to celebrate the 256-th anniversary of its founding! A specially appointed Vice Rector for the celebration prepares to decorate the campus. In the center of the campus n ice sculptures were erected. The sculptures are arranged in a circle at equal distances from each other, so they form a regular n-gon. They are numbered in clockwise order with numbers from 1 to n.

The site of the University has already conducted a voting that estimated each sculpture's characteristic of ti — the degree of the sculpture's attractiveness. The values of ti can be positive, negative or zero.

When the university rector came to evaluate the work, he said that this might be not the perfect arrangement. He suggested to melt some of the sculptures so that:

  • the remaining sculptures form a regular polygon (the number of vertices should be between 3 and n),
  • the sum of the ti values of the remaining sculptures is maximized.

Help the Vice Rector to analyze the criticism — find the maximum value of ti sum which can be obtained in this way. It is allowed not to melt any sculptures at all. The sculptures can not be moved.

Input

The first input line contains an integer n (3 ≤ n ≤ 20000) — the initial number of sculptures. The second line contains a sequence of integers t1, t2, ..., tn, ti — the degree of the i-th sculpture's attractiveness ( - 1000 ≤ ti ≤ 1000). The numbers on the line are separated by spaces.

Output

Print the required maximum sum of the sculptures' attractiveness.

Examples
input
8
1 2 -3 4 -5 5 2 3
output
14
input
6
1 -2 3 -4 5 -6
output
9
input
6
1 2 3 4 5 6
output
21
Note

In the first sample it is best to leave every second sculpture, that is, leave sculptures with attractivenesses: 2, 4, 5 и 3.

 

题目大意 ：

题目大意 ：

　　很多个雕像围在一起构成一个多边形，每一个占一个点并有个分数。现在需要移除一些雕像，使分数和最大，并且还是能构成多边形。

枚举每次删除的步长l，然后维护一个最大值就好了。枚举步长的上界是l * 3 <= n。

 代码 ：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
const int MaxN = 2e4;
using namespace std;
int n;
int ans;
int a[MaxN + 5];
int main()
{
　　scanf("%d",&n);
　　for(int i = 1;i <= n;i++){
　　　　scanf("%d",&a[i]);
　　　　ans += a[i];
　　}
　　for(int i = 1;i <= n / 3;i++){ // 枚举步长到n / 3
　　　　if(n % i ) continue;
　　for(int j = 1;j <= i;j++){    // 每次枚举的起始位置小于当前枚举步长
　　　　int s = 0;
　　for(int k = j;k <= n;k += i){
　　　　s += a[k];　　　　　　　　//求每次枚举的和
　　}
　　ans = max(ans,s);　　　　　　//维护最大值
　　}
　　}
　　printf("%d\n",ans);
}