Problem:
Score : 700 points

Problem Statement

We have N gemstones labeled 1 through N.

You can perform the following operation any number of times (possibly zero).

Select a positive integer x, and smash all the gems labeled with multiples of x.

Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan).
However, a_i may be negative, in which case you will be charged money.

By optimally performing the operation, how much yen can you earn?

Constraints

All input values are integers.

1 \leq N \leq 100

|a_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the maximum amount of money that can be earned.

Sample Input 1

6
1 2 -6 4 5 3

Sample Output 1

12

It is optimal to smash Gem 3 and 6.

Sample Input 2

6
100 -100 -100 -100 100 -100

Sample Output 2

200

Sample Input 3

5
-1 -2 -3 -4 -5

Sample Output 3

0

It is optimal to smash all the gems.

Sample Input 4

2
-1000 100000

Sample Output 4

99000

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

#include<bits/stdc++.h>
 
using namespace std;
using ll = long long;


int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    int n;
    cin >> n;
    ll sum = 0;
    int k = 1;
    while (k * 2 <= n) ++k;
    vector<ll> p, a(n + 1);
    vector<bool> f(n + 1, true);
    for (int i = 2; i <= n; ++i) {
        if (f[i]) {
            if (i < k) p.push_back(i);
            for (int j = i; j <= n; j += i) {
                f[j] = false;
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        sum += a[i];
    }
    for (int i = k; i <= n; ++i) {
        if (a[i] < 0) {
            sum -= a[i];
            a[i] = 0;
        }
    }
    int sz = p.size();
    ll res = max(res, sum);
    for (int mask = 0; mask < (1 << sz); ++mask) {
        ll qwe = 0;
        set<int> s;
        for (int j = 0; j < sz; ++j) {
            if ((mask >> j) & 1) {
                for (int c = p[j]; c < k; c += p[j]) {
                    s.insert(c);
                }
            }
        }
        for (int x : s) {
            qwe += a[x];
        }
        res = max(res, sum - qwe);
    }
    cout << res << '\n';
}