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 <iostream>
#include <cstdio>
#include <ctime>
#include <cstdlib>
using namespace std;

typedef long long ll;
ll n, num, s1, s2, md, en, mx = 0, Q =200000, a[105], s[105], z[105];
int main()
{
	ll i, j;
	cin >> n;
	for (i = 1; i <= n; i++) {
		scanf ("%lld", &a[i]);
	}
	if (n == 1) {
		cout << max(a[1], 0ll);
		return 0;
	}
	md = n / 2;
	for (i =1; i <= md; i++) s1 += a[i];
	for (i = md + 1; i <= n; i++) {
		if (a[i] > 0) s2 += a[i];
	}
	mx = max(mx, s1 + s2);
//	srand(time(NULL));
	en = 0;
	while (Q--) {
//		en = num;
		en = 0;
		for (j = 0; j < 50; j++) {
			if (md == en) break;
			num = rand() % (md-en) + 1 + en;
			z[num] = 1 - z[num];
			en = num;
			if (z[num] == 1) {
				for (i = num; i <= n; i += num) {
					s[i]++;
					if (i <= md) {
						if (s[i] == 1) s1 -= a[i];
					} else {
						if (s[i] == 1 && a[i] > 0) s2-= a[i];
					}
				}
			} else {
				for (i = num; i <= n; i += num) {
					s[i]--;
					if (i <= md) {
						if (s[i] == 0) s1 += a[i];
					} else {
						if (s[i] == 0 && a[i] > 0) s2 += a[i];
					}
				}
			}
			if (s1 + s2 > mx) {
				j = -1;
				en = 0;
				mx = s1 + s2;
			}
		}
//		cout <<num <<endl;
	}
	cout << mx;
    return 0;
}