CF 158D Ice Sculptures

来源：互联网发布：淘宝特步鞋编辑：程序博客网时间：2024/06/05 13:25

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 ton.

The site of the University has already conducted a voting that estimated each sculpture's characteristic of t_i — the degree of the sculpture's attractiveness. The values of t_i 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 t_i values of the remaining sculptures is maximized.

Help the Vice Rector to analyze the criticism — find the maximum value of t_i 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, ..., t_n, t_i — the degree of the i-th sculpture's attractiveness ( - 1000 ≤ t_i ≤ 1000). The numbers on the line are separated by spaces.

Output

Print the required maximum sum of the sculptures' attractiveness.

Sample test(s)
input
81 2 -3 4 -5 5 2 3
output
14
input
61 -2 3 -4 5 -6
output
9
input
61 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.

附上有道的翻译

Berland大学正准备庆祝成立256周年!副校长任命的特别庆典准备装修。在校园的中心n冰雕被竖立起来的。雕塑被安排在一个圆在彼此相等的距离,所以他们组建一个正则n角。他们以顺时针顺序编号数字从1到n。

大学的网站已经进行了投票,估计每个雕塑的特点ti -雕塑的程度的吸引力。ti的值可以是正数、负数或零。

当大学校长来评估工作,他说,这可能不是完美的安排。他建议融化一些雕塑,以便:

剩下的雕塑形成一个正多边形(顶点的数量应该是3至n),

ti值之和的雕塑是最大化。

帮助分析批评的副校长——找到ti之和的最大值可以通过这种方式获得的。它允许不融化任何雕塑。这些雕塑不能移动。

#include<iostream>#include<cstdio>#include<map>#include<cstring>#include<string>using namespace std;int array[20002];int n;int solve() {int len;int maxn = -20000000;for(len = 1;len<n;len++){if(n%len==0&&n/len>=3){int b[20002]={0};for(int j = 0;j<n;j++)b[j%len]+=array[j];for(int j=0;j<len;j++)maxn = max(maxn,b[j]);}}return maxn;}int main(){cin >> n;for(int i=0;i<n;i++)cin >> array[i];cout<<solve()<<endl;return 0;}

0 0