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>

using namespace std;

typedef long long ll;
ll n, mx, sum, a[105];
bool bo[105];

void g(int p)
{
    for(int i = p; i <= n; i += p)
    {
        bo[i] = true;
    }
}

void f(int p)
{
    if(p > (n + 1) / 2)
    {
        sum = 0;
        ll i;
        for(i = 1; i <= (n + 1) / 2; i++)
        {
            sum += (1 - bo[i]) * a[i];
        }
        for(; i <= n; i++)
        {
            sum += (1 - bo[i]) * max(a[i], 0ll);
        }
        mx = max(mx, sum);
        return;
    }
    f(p + 1);
    if(a[p] < 0) g(p), f(p + 1);
    bo[p] = false;
}

int main()
{
    scanf("%lld", &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
    }
    f(1);
    printf("%lld", mx);
    return 0;
}