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;

using ll = long long;

ll a[105];
bool b[105];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    ll ans = 0;
    int m = 0;
    while (true) {
        if (m * m > n) {
            break;
        }
        m++;
    }
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        ans += a[i];
    }
    /*
    for (int x = n + 1; x > 0; x--) {
        for (int i = n; i > 0; i--) {
            b[i] = (i % x == 0);
        }
        for (int i = n; i > 0; i--) {
            ll tmp = 0;
            for (int j = i; j <= n; j += i) {
                if (!b[j]) tmp += a[j];
            }
            if (tmp <= 0) {
                for (int j = i; j <= n; j += i) {
                    b[j] = true;
                }
            }
        }
        ll tmp = 0;
        for (int i = 1; i <= n; i++) {
            if (!b[i]) tmp += a[i];
        }
        ans = max(ans, tmp);
    }*/
    for (int s = 0; s < (1 << m); s++) {
        for (int i = n; i > 0; i--) {
            b[i] = false;
            for (int j = 0; j < m; j++) {
                if (s & (1 << j)) {
                    if (i % (j + 1) == 0) b[i] = true;
                }
            }
        }
        for (int i = n; i > 0; i--) {
            ll tmp = 0;
            for (int j = i; j <= n; j += i) {
                if (!b[j]) tmp += a[j];
            }
            if (tmp <= 0) {
                for (int j = i; j <= n; j += i) {
                    b[j] = true;
                }
            }
        }
        ll tmp = 0;
        for (int i = 1; i <= n; i++) {
            if (!b[i]) tmp += a[i];
        }
        ans = max(ans, tmp);
    }

    cout << ans << endl;
    return 0;
}