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: Wrong Answer)


#include <cassert>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
using namespace std;

typedef long long ll;

ll vs[100000+10];
int n;

int isNotPrime[100000+10];
int primes[100000+10];
int nPrime;

vector<ll> cs[100000+10];
vector<ll> cs2[100000+10];

inline void calcDs(vector<int> &ds, int v) {
  for(int d = 1; d*d <= v; ++d) {
    if(v % d == 0) {
      ds.push_back(d);
      if(d*d != v) {
        ds.push_back(v/d);
      }
    }
  }
}

ll calc(int a, int b) {
  if(a <= b) {
    // cerr << "HOGE" <<  " " << a << " " << b << endl;
    return -1;
  }
  if((n-1-b) % (a-b) != 0) {
    // cerr << "HOGE2" <<  " " << a << " " << b << endl;
    return -1;
  }
  ll k = (n-1-b) / (a-b);
  if(b % (a-b) == 0) {
    ll x = b / (a-b) + 1;
    if(x <= k-1) {
      // cerr << "HOGE3" <<  " " << a << " " << b << endl;
      return -1;
    }
  }
  // 落ちずに到着できる。
  ll res = 0;
  res += cs2[a-b][k-1];
  res += cs[a-b][k-1];
  // cerr << ">" << a << " " << b << " " << res << endl;
  return res;
}

int main(void) {
  // isNotPrime[0] = isNotPrime[1] = 1;
  // REP(p, 100000+1) {
  //   if(isNotPrime[p] == 0) {
  //     primes[nPrime++] = p;
  //     for(int j = p+p; j <= 100000; j += p) {
  //       isNotPrime[j] = p;
  //     }
  //   }
  // }

  scanf("%d", &n);
  REP(i, n) {
    scanf("%lld", &vs[i]);
  }
  for(int i = 1; i <= n; ++i) {
    ll cur = 0;
    cs[i].push_back(cur);
    for(int j = i; j < n; j += i) {
      cur += vs[j];
      cs[i].push_back(cur);
    }
  }
  for(int i = 1; i < n; ++i) {
    ll cur = 0;
    cs2[i].push_back(cur);
    for(int j = n-1-i; j >= 0; j -= i) {
      cur += vs[j];
      cs2[i].push_back(cur);
    }
  }

  ll res = 0;
  for(int a = 2; a <= n-1; ++a) {
    int rem = (n-1) % a;
    if(rem == 0) {
      continue;
    }
    vector<int> ds;
    calcDs(ds, rem);
    for(int d : ds) {
      int b = a - d;
      ll cur = calc(a, b);
      res = max(res, cur);
    }
  }
  cout << res << endl;
  return 0;
}