human
Solve in Python:
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-----
6
1 2 -6 4 5 3

-----Sample Output-----
12

It is optimal to smash Gem 3 and 6.
 ... Give me a Python program to print the maximum amount of money that can be earned?
assistant
#Here is Python program to print the maximum amount of money that can be earned.

#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

#A
def A():

    return

#B
def B():

    return

#C
def C():
    def bfs(s,g,n):
        bfs_map = [-1 for i in range(n)]
        bfs_map[s] = 0
        q = deque()
        q.append(s)
        fin = False
        while q:
            x = q.popleft()
            for y in range(n):
                if c[x][y] > 0 and bfs_map[y] < 0:
                    bfs_map[y] = bfs_map[x]+1
                    if y == g:
                        fin = True
                        break
                    q.append(y)
            if fin:
                break

        if bfs_map[g] == -1:
            return [None,0]
        path = [None for i in range(bfs_map[g]+1)]
        m = float("inf")
        path[bfs_map[g]] = g
        y = g
        for i in range(bfs_map[g])[::-1]:
            for x in range(n+1):
                if c[x][y] > 0 and bfs_map[x] == bfs_map[y]-1:
                    path[i] = x
                    if c[x][y] < m:
                        m = c[x][y]
                    y = x
                    break
        return [path,m]

    def ford_fulkerson(s,g,c,n):
        while 1:
            p,m = bfs(s,g,n)
            if not m:break
            for i in range(len(p)-1):
                c[p[i]][p[i+1]] -= m
                c[p[i+1]][p[i]] += m
        return sum(c[g])

    n = I()
    a = LI()
    e = n+2
    ma = 10**9
    c = [[0]*e for i in range(e)]
 ...