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>
#define int long long

signed main(){
	int N, i, j, k, sum;
	scanf("%lld", &N);
	int *a = (int *)malloc(sizeof(int) * (N + 1));
	for(i = 1; i <= N; i++){
		scanf("%lld", &a[i]);
	}
	for(k = 0; k < N; k++){
		for(i = N; i > 0; i--){
			if(a[i] < 0){
				sum = 0;
				for(j = i; j <= N; j += i){
					sum += a[j];
				}
				if(sum < 0){
					for(j = i; j <= N; j += i){
						a[j] = 0;
					}
				}
			}
		}
	}
	sum = 0;
	for(i = 1; i <= N; i++){
		sum += a[i];
	}
	printf("%lld\n", sum);
	return 0;
}