Problem:
Score : 600 points

Problem Statement

There is an infinitely large pond, which we consider as a number line.
In this pond, there are N lotuses floating at coordinates 0, 1, 2, ..., N-2 and N-1.
On the lotus at coordinate i, an integer s_i is written.

You are standing on the lotus at coordinate 0. You will play a game that proceeds as follows:

1. Choose positive integers A and B. Your score is initially 0.

2. Let x be your current coordinate, and y = x+A. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

3. Let x be your current coordinate, and y = x-B. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

4. Go back to step 2.

You want to end the game with as high a score as possible.
What is the score obtained by the optimal choice of A and B?

Constraints

3 \leq N \leq 10^5

-10^9 \leq s_i \leq 10^9

s_0=s_{N-1}=0

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
s_0 s_1 ...... s_{N-1}

Output

Print the score obtained by the optimal choice of A and B.

Sample Input 1

5
0 2 5 1 0

Sample Output 1

3

If you choose A = 3 and B = 2, the game proceeds as follows:

Move to coordinate 0 + 3 = 3. Your score increases by s_3 = 1.

Move to coordinate 3 - 2 = 1. Your score increases by s_1 = 2.

Move to coordinate 1 + 3 = 4. The game ends with a score of 3.

There is no way to end the game with a score of 4 or higher, so the answer is 3. Note that you cannot land the lotus at coordinate 2 without drowning later.

Sample Input 2

6
0 10 -7 -4 -13 0

Sample Output 2

0

The optimal strategy here is to land the final lotus immediately by choosing A = 5 (the value of B does not matter).

Sample Input 3

11
0 -4 0 -99 31 14 -15 -39 43 18 0

Sample Output 3

59

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <fstream>
#include <sstream>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <bitset>     //UWAGA - w czasie kompilacji musi byc znany rozmiar wektora - nie mozna go zmienic
#include <cassert>
#include <iomanip>        //do setprecision
#include <ctime>
#include <complex>
using namespace std;

#define FOR(i,b,e) for(int i=(b);i<(e);++i)
#define FORQ(i,b,e) for(int i=(b);i<=(e);++i)
#define FORD(i,b,e) for(int i=(b)-1;i>=(e);--i)
#define REP(x, n) for(int x = 0; x < (n); ++x)

#define ST first
#define ND second
#define PB push_back
#define PF push_front
#define MP make_pair
#define LL long long
#define ULL unsigned LL
#define LD long double
#define pii pair<int,int>
#define pll pair<LL,LL>

const double pi = 3.14159265358979323846264;
const int mod=1000000007;

int main(){
	 //cin.tie(0);
	//ios::sync_with_stdio(false);
	//std::cout << std::fixed;
	//std::cout << std::setprecision(12);
	//std::cout << std::defaultfloat;
	int n;
	cin>>n;
	vector<LL> s(n);
	LL ans=0;
	FOR(i,0,n)cin>>s[i];
	FOR(i,1,n/2){
		LL t=0;
		int l=i,r=n-1-i;
		set<LL> st;
		FOR(j,0,n){
			if(l>=n-1||r<=0)break;
			if(st.find(l)!=st.end()||st.find(r)!=st.end()||l==r)break;
			if((n-1)%i==0&&l>=r)break;
			if(r-i<=0)break;
			st.insert(l);
			st.insert(r);
			t+=s[l]+s[r];
			ans=max(t,ans);
			l+=i;
			r-=i;
		}
	}

	cout<<ans<<endl;
	return 0;
}