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>

long long score(int i, int n, long long* a, int* flag) {
  int m = n/i;
  int ans = 0;
  for (int j=1; j <= m; j++) {
    if (flag[i*j] == 0) {
      ans += a[i*j];
    }
  }
  return ans;
}

int main() {
  int n;
  int i;
  scanf("%d", &n);
  long long a[110] = {};
  int flag[110] = {};
  for (i = 1; i < n+1; i++) {
    scanf("%lld", &a[i]);
  }
  for (i = n; i >= 1; i--) {
    /*printf("%d\n", score(i, n, a, flag));*/
    if (score(i, n, a, flag) < 0) {
      int m = n/i;
      for (int j=1; j<=m; j++) {
	flag[i*j] = 1;
      }
    }
    /*printf("%d\n", flag[i]);*/
  }
  long long ans = 0;
  for (i = 1; i < n+1; i++) {
    if (flag[i] == 0) {
      ans += a[i];
    }
  }
  printf("%lld", ans);
  return 0;
}