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"
#include "ext/pb_ds/assoc_container.hpp"
#include "ext/pb_ds/tree_policy.hpp"
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
#define int long long
#define pb push_back
#define fi first
#define se second
#define fr(i, a, b) for(int i = a; i <= b; i++)
#define all(x) x.begin(), x.end()
#define IO ios :: sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define pii pair<int,int>
#define sz(x) (int)x.size()
const int mod = 1e9 + 7;
// const int mod1 = 998244353;
typedef long double f80;
 
#ifndef LOCAL
#pragma GCC optimize ("O2")
#define endl '\n'
#endif
 
template<typename T>
using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r){
    uniform_int_distribution<int> uid(l, r);
    return uid(rng);
}
 
ll pwr(ll a,ll b, ll mod){
    a %= mod;
    int ans = 1;
    while(b){
        if(b & 1) ans = (ans * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return ans;
}
 
string to_string(string s) {
  return '"' + s + '"';
}
 
string to_string(const char* s) {
  return to_string((string) s);
}
 
string to_string(bool b) {
  return (b ? "true" : "false");
}
 
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
 
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
 
void debug_out() { cerr << endl; }
 
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
 
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

void debug1(int a[], int l,int r) {
	#ifdef LOCAL
		vector<int> temp(a + l, a + r + 1);
		debug(temp);
	#endif
}

const int N = 105;
int a[N], b[N];

void solve(){
	int n;
	cin >> n;
	fr(i, 1, n) {
		cin >> a[i];
	}
	n = 100;
	int ans = 0;
	for(int mask = 0; mask < (1 << 10); mask++) {
		fr(j, 1, n) {
			b[j] = a[j];
		}
		fr(j, 1, 10) {
			if(mask & (1 << (j - 1))) {
				for(int k = j; k <= n; k += j) {
					b[k] = 0;
				}
			}
		}
		for(int j = n; j >= 11; j--) {
			int s = 0;
			for(int k = j; k <= n; k += j) {
				s += b[k];
			}
			if(s < 0) {
				for(int k = j; k <= n; k += j) {
					b[k] = 0;
				}
			}
		}
		int tans = 0;
		fr(j, 1, n) {
			tans += b[j];
		}
		ans = max(ans, tans);
	}
	cout << ans;
}

signed main(){
    IO;
    #ifdef LOCAL
        freopen("inp.txt","r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    cout << setprecision(10) << fixed;
    int t = 1;
    // cin >> t;
    fr(tt, 1, t){
        solve();
    }
    return 0;
}