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<cstdio>
#include<vector>
int vis[55];
int t[55];
int a[55];
int b[55];
int all = -1;
long long ans = 0;

int gcd(int a,int b){ return b?gcd(b,a%b):a; }

int min(int a,int b){ return a>b?b:a; }

void getAnswer(int size,int n){
    for(int i = 1; i <= n; i++) t[i] = 0;
    for(int i = 1; i <= size; i++){ 
        //printf("%d ",a[i]);
        for(int j = a[i]; j <= n; j+=a[i]) t[j] = 1;
    }
    //printf("\n");
    //for(int i = 1; i <= n; i++) if(!t[i]) printf("%d ",i);
    //printf("\n");
    for(int i = 1; i <= n; i++){
        if(!t[i] && i>32){
            long long sum = 0;
            for(int j = i; j <= n; j+=i) if(!t[j]) sum += (long long)b[j];
            if(sum<0){
                for(int j = i; j <= n; j+=i) t[j] = 1;
            }
        }
    }
    long long res = 0;
    for(int i = 1; i <= n; i++) if(!t[i]) res += (long long)b[i];
    //printf("res = %lld\n",res);
    if(res>ans) ans = res;
}

void dfs(int last,int size,int n){
    getAnswer(size,all);
    int find = 0;
    for(int i = last+1; i <= n; i++){
        if(!vis[i]){
            find++;
            a[size+1] = i;
            std::vector<int>keep;
            for(int j = i; j <= n; j+=i) if(vis[j]) keep.push_back(j);
            for(int j = i; j <= n; j+=i) vis[j] = 1;
            dfs(i,size+1,n);
            for(int j = i; j <= n; j+=i) vis[j] = 0;
            for(int u: keep) vis[u] = 1;
        }
    }
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) scanf("%d",&b[i]);
    for(int i = 1; i <= n; i++){
        if(b[i]<0){
            if(i*2>n) b[i] = 0;
        }
    }
    all = n;
    int upper = min(n/2,32);
    dfs(0,0,upper);
    printf("%lld\n",ans);
    return 0;
}