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 int max(long long int a,long long int b){return a>b?a:b;}

long long int solve(long long int* x,int N,int end){
    long long tmp[N];
    for(int i=0;i<N;i++)tmp[i]=x[i];
    for(int i=end;i>0;i--){
        long long int sum=0;
        for(int j=i;j<=N;j+=i)sum+=tmp[j-1];
        if(sum<0)for(int j=i;j<=N;j+=i)tmp[j-1]=0;
        if(sum==0)for(int j=i;j<=N;j+=i)if(tmp[j-1]!=0){long long int a=solve(tmp,N,i-1);for(int j=i;j<=N;j+=i)tmp[j-1]=0;return max(a,solve(tmp,N,i-1));}
    }
    long long int ans=0;
    for(int i=0;i<N;i++)ans+=tmp[i];
    return ans;
}

int main(void){
    int N;
    scanf("%d",&N);
    long long int a[N];
    for(int i=0;i<N;i++)scanf("%lld",&a[i]);
    printf("%lld",solve(a,N,N));
    return 0;
}