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(void){
  int N,i,j,k,flg;
  long long a[101],sum,minsum;
  scanf("%d",&N);
  for(i=1;i<=N;i++) scanf("%lld",&a[i]);

  for(i=N;i>0;i--){
    sum=flg=0;
    for(j=i;j<=N;j+=i){
      sum+=a[j];
      if(a[j]!=0 && j!=i) flg=1;
    }
    if(sum<0 && flg==0){
      a[i]=0;
    }
  }

  do{
    minsum=0;
    for(i=1;i<=N/2;i++){
      sum=0;
      for(j=i;j<=N;j+=i) sum+=a[j];
      if(sum<minsum){
	minsum=sum;
	k=i;
      }
    }
    if(minsum<0) for(i=k;i<=N;i+=k) a[i]=0;
  }while(minsum<0);
  
  sum=0;
  for(i=1;i<=N;i++) sum+=a[i];
  printf("%lld\n",sum);
  return 0;
}