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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define mk make_pair
const int N=300010;

ll n,a[N],bl[N],ans=0;

int main()
{
    memset(bl,0,sizeof(bl));
    scanf("%lld",&n);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    for(int i=n;i>=1;i--)
    {
        if(a[i]>=0||bl[i]) continue;
        ll sum=0;
        for(int j=i;j<=n;j+=i) if(!bl[j]) sum-=a[j];
        if(sum>0) 
        {
            for(int j=i;j<=n;j+=i) bl[j]=1;
        }
    }
    for(int i=1;i<=n;i++) if(!bl[i]) ans+=a[i];
    cout<<ans<<endl;
    return 0;
}