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>

int n;
long long a[101]{};
bool is_broken[101]{};

bool all_broken();

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%lld", a + i);

	while (!all_broken());

	long long sum{};
	for (int i = 1; i <= n; i++)
		if (!is_broken[i])
			sum += a[i];

	printf("%lld\n", sum);
	return 0;
}

bool all_broken()
{
	for (int i = n; i >= 1; i--)
	{
		if (is_broken[i]) continue;
		long long sum{};
		for (int j = i; j <= n; j += i)
			if (!is_broken[j])
				sum += a[j];
		if (sum > 0) continue;
		for (int j = i; j <= n; j += i)
			is_broken[j] = true;
		return false;
	}
	return true;
}