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)

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<cmath>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<numeric>
#include<functional>
#include<algorithm>
#include<bitset>
#include<tuple>
#include<unordered_set>
#include<random>
#include<array>
#include<cassert>
using namespace std;
#define INF (1<<29)
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(v) v.begin(),v.end()
#define uniq(v) v.erase(unique(all(v)),v.end())



int n;
int a[101];
long long dp[101][2];
bool root[101];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	cin>>n;
	rep(i, n)cin >> a[i+1];

	rep(i, 101)rep(j, 2)dp[i][j] = 1LL<<60;

	for (int i=1;i<=n;i++){
		dp[i][0] = 0;
		dp[i][1] = a[i];
		root[i] = true;
		for (int j=2; j <= i ; j++) if(i%j==0){
			int k = i/j;
			root[k] = false;
			dp[i][1] += min(dp[k][0], dp[k][1]);
			dp[k][1] = 0;
		}
	}

	long long ans = 0;
	rep(i, n)ans += a[i+1];
	for (int i=1;i<=n;i++)if(root[i]){
		ans -= min(dp[i][0], dp[i][1]);
	}
	cout<<ans<<endl;


	return 0;
}