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;

long long int N, a[111], f[111], r, l, m, i;
pair<long long, long long> b[111];
vector<long long> d[111];

int main(){
    for(i = 1; i < 111; ++i){
        b[i].second = i;
        for(int j = i; j < 111; j += i){
            d[j].push_back(i);
            ++b[j].first;
        }
    }
    fill(f, f + 111, 0);
    scanf("%lld", &N);
    sort(b + 1, b + N + 1);
    for(i = 1; i <= N; ++i)scanf("%lld", a + i);
    for(i = 1; i <= N; ++i){
        r = b[i].second;
        m = 0;
        for(long long j : d[r]){
            m += (f[j] > 0) ? 0LL : a[j];
        }
        if(m > 0){
            for(long long j : d[r]){
                ++f[j];
            }
            l += m;
        }
    }
    cout << l << endl;
    return 0;
}