Ice Sculptures



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 ton.
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.

Sample test(s)

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.
n个冰雕，每一个都有吸引度，融化掉一些之后的最大吸引度（融化后依然是正多边形）。暴力求解，首先最少是3个点，判断一下，然后就是找n的因子，再搜每一种情况，与ans比较。这里注意的是ans可能是负的，一开始手残写了0，挂了，然后写了-999999，又挂了- -。只好把ans初始化为所有冰雕都不熔化的值，在与之后比较，这才是正解！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
bool cmp(int a, int b)
{
	return a > b;
}
int max(int a, int b)
{
	return a > b ? a : b;
}
int a[20005];
int main()
{
	int i, j, m, n, ans, pre, temp, k;
	cin >> n;
	ans = 0;
	for (i = 1; i <= n; i++)
	{
		cin >> a[i];
		ans += a[i];
	}
	for (i = 2; i <= n; i++)
	{
		if (n%i == 0&&n/i>=3)
		{
			pre = n / i;
			for (j = 1; j <= i; j++)
			{
				temp = 0;
				for (k = 1; k <= pre; k++)
					temp = temp + a[j + (k - 1)*i];
				ans = max(ans, temp);
			}
		}
	}
	cout << ans << endl;
	return 0;
}