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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(int argc, char const *argv[]){
	int n,i,b,a[101],sum=0,as=0,s[100];
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&a[i]);
		as+=a[i];
		s[i]=1;
	}
	b=0;
	while(b++<n){
		sum=0;
		i=b;
		while(1){
			if(a[i]%b==0&&s[i]==1)
				sum+=a[i];
			i+=b;
			if(i>n)
				break;
		}
		if(sum<0){
			i=b;
			while(1){
				if(s[i]==1){
					as-=a[i];
					s[i]=0;
				}
				i+=b;
				if(i>n)
					break;
			}
		}
	}
	printf("%d",as);
	return 0;
}