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 <iostream>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <memory.h>
#include <queue>
#include <random>
#include <time.h>
#include <stack>
#include <bitset>
using namespace std;
const int N = 2e3 + 10;
const int inf = 1e18 + 5;
const int mod = 1e9 + 7;
#define fast ios_base::sync_with_stdio(false)
#define ll long long
int n;
int a[N];
int check(int p){
    if(p == 1) return false;
    for(int i = 2; i * i <= p; ++i){
        if(p % i == 0){
            return false;
        }
    }
    return true;
}
int main(){
    cin >> n;
    ll ans = 0;
    ll sum = 0;
    for(int i = 1; i <= n; ++i){
        cin >> a[i];
        sum += a[i];
    }
    if(n == 1) {
        cout << max(0ll, sum) << endl;
        return 0;
    }
    vector <int > pr;
    for(int i = 1; i <= n; ++i){
        if(check(i)){
            pr.push_back(i);
        }
    }
    for(auto it : pr){
        if(it > 50 && a[it] < 0){
            sum -= a[it];
        }
    }
    while(pr.back() > 50) pr.pop_back();
    for(int i = 1; i <= 10; ++i){
        if(check(i)){
            int cur = i;
            while(cur * i <= n){
                cur *= i;
                pr.push_back(cur);
            }
        }
    }
    
    int sz = (int ) pr.size();
    for(int msk = 1; msk < (1 << sz); ++msk){
        ll cur_sum = 0;
        vector <int > used(n + 1);
        for(int i = 0; i < sz; ++i){
            if(msk & ( 1 << i ) ) {
                for(int j = pr[i]; j <= n; j += pr[i]){
                    used[j] = 1;
                }
            }
        }
        for(int i = 1; i <= n; ++i){
            cur_sum += a[i] * used[i];
        }
        ans = max(ans, sum - cur_sum);
    }
    cout << ans <<endl;
    return 0;
}