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;
int n, a[101], ans;
void dfs(int x) {
	if (x == n + 1) {
		int s = 0;
		for (int i = 1; i <= n; i++) {
			s += a[i];
		}
		ans = max(ans, s);
		return;
	}
	if (a[x] < 0) {
		int b[101] = {0};
		for (int i = 1; i <= n; i++) {
			b[i] = a[i];
		}
		for (int i = x; i <= n; i += x) {
			a[i] = 0;
		}
		dfs(x + 1);
		for (int i = 1; i <= n; i++) {
			a[i] = b[i];
		}
	}
	dfs(x + 1);
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = n; i > n / 6; i--) {
		int s = 0;
		for (int j = i; j <= n; j += i) {
			s += a[j];
		}
		if (s >= 0) {
			continue;
		}
		for (int j = i; j <= n; j += i) {
			a[j] = 0;
		}
	}
	dfs(1);
	printf("%d\n", ans);
	return 0;
}