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<algorithm>
#include<functional>
#include<string>
#include<string.h>
#include<limits.h>
#include<vector>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#define int long long
using namespace std;

int n;
int MAX;
int a[101];
bool k[101];//false 使わない true 使う
void saiki(int b) {
	if (b == 20||b==n+1) {
		int t = 0;
		for (int i = 1; i <= 19; i++) { if (k[i])t += a[i]; }
		for (int i = 20; i <= n; i++) {
			for (int j = 1; j <= 19; j++) {
				if (i%j == 0 && k[j]) {
					t += a[i];
					k[i] = true;
					break;
				}
				else {
					k[i] = false;
				}
			}
		}
		for (int i = n; i > 19; i--) {
			if (k[i])continue;
			bool OK = true;//自由に使えるか？
			for (int j = i*2; j <= n; j+=i) {
				if (!k[j])OK = false;
			}
			if (OK) {
				if (a[i] >= 0) {
					k[i] = true;
					t += a[i];
				}
			}
			else {
				int o = 0;
				if (a[i] >= 0) {
					for (int j = i; j <= n; j+=i) {
						if(!k[j])o += a[j];
					}
					if (o >= 0) {
						for (int j = i; j <= n; j += i) {
							k[j] = true;
						}
						t += o;
					}
				}
			}
		}
		MAX = max(MAX, t);
		return;
	}
	for (int i = 1; i < b; i++) {
		if (b%i == 0 && k[i]) {
			k[b] = true;
			saiki(b + 1);
			return;
		}
	}
	k[b] = false;
	saiki(b + 1);
	k[b] = true;
	saiki(b + 1);
}
signed main() {
	cin >> n;
	int sum = 0;
	for (int i = 0; i < n; i++) {
		cin >> a[i + 1];
		sum += a[i + 1];
		a[i + 1] *= -1;
	}
	saiki(1);
	cout << sum + MAX << endl;
}