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>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <math.h>
#include <limits.h>
#define swap(type,a,b) do{type t=a;a=b;b=t;}while(0);
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define ll long long
#define INF 100000000
#define MOD 1000000007
#define SIZE 100005
#define FOR(i,a,n) for(i=(a);i<(n);i++)
int comp(const void* a,const void* b){
	return *(int*)a-*(int*)b;
}
ll a[101];
int used[101];
int prime[]={2,3,5,7,11,13,17,19,23,29,31,
37,41,43,47,53,59,61,67,71,73,79,83,89,97,101};
int main(void)
{
	int n,id,i,j;
	ll res=0,sum=0,min;
	scanf("%d",&n);
	FOR(i,1,n+1){
		scanf("%lld",a+i);
		res+=a[i];
	}
	for(i=0;prime[i]<=n;i++){
		sum=min=0; id=-1;
		for(j=1;j*prime[i]<=n;j++); j--;
		for(;j>=1;j--){
			if(used[j*prime[i]]) continue;
			sum+=a[j*prime[i]];
			if(min>sum){
				min=sum; id=j;
			}
		}
		if(id>0){
			res-=min;
			for(j=id;j*prime[i]<=n;j++){
				if(used[j*prime[i]]) continue;
				used[j*prime[i]]=1;
			}
		}
	}
	printf("%lld\n",MAX(res,0));
	return 0;
}