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 <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <cassert>
#include <queue>
#include <ctime>
#include <string>
#include <cstring>
#include <random>
#define mp make_pair
#define pb push_back
#define NAME ""
#define y1 y1_423
#define list lista

using namespace std;

typedef long long ll;
typedef long double ld;

random_device gen;
mt19937 rnd(gen());

const int nmax = 2000 * 1000;
const int inf = 2000 * 1000 * 1000;
const ll infl = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 1000ll;
const int mod = 1000 * 1000 * 1000 + 7;
const ld pi = acos(-1.0);

int n, a[nmax];

ll get(int ind) {
    ll answer = 0;
    for (int i = ind; i <= n; i += ind) {
        answer += a[i];
    }
    return answer;
}

int main() {
    //freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
    //freopen(NAME".in", "r", stdin);freopen(NAME".out", "w", stdout);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
        int ind = 1;
        for (int j = 1; j <= n; j++) {
            if (get(j) < get(ind)) {
                ind = j;
            }
        }
        if (get(ind) >= 0) {
            break;
        }
        for (int j = ind; j <= n; j += ind) {
            a[j] = 0;
        }
    }
    ll answer = 0;
    for (int i = 1; i <= n; i++) {
        answer += a[i];
    }
    cout << answer << endl;
    return 0;
}