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++14 (Clang 3.8.0) (judge verdict: Accepted)


#include <vector>
#include <list>
#include <iostream>
#include <numeric>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>


// C++ implementation of Dinic's Algorithm
using namespace std;
//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);

//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))

//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;




 
// A structure to represent a edge between
// two vertex
struct Edge
{
    int v ;  // Vertex v (or "to" vertex)
             // of a directed edge u-v. "From"
             // vertex u can be obtained using
             // index in adjacent array.
 
    int flow ; // flow of data in edge
 
    int C;    // capacity
 
    int rev ; // To store index of reverse
              // edge in adjacency list so that
              // we can quickly find it.
};
 
// Residual Graph
class Graph
{
    int V; // number of vertex
    int *level ; // stores level of a node
    vector< Edge > *adj;
public :
    Graph(int V)
    {
        adj = new vector<Edge>[V];
        this->V = V;
        level = new int[V];
    }
 
    // add edge to the graph
    void addEdge(int u, int v, int C)
    {
        // Forward edge : 0 flow and C capacity
        Edge a = {v, 0, C, static_cast<int>(adj[v].size())};
 
        // Back edge : 0 flow and 0 capacity
        Edge b = {u, 0, 0, static_cast<int>(adj[u].size())};
 
        adj[u].push_back(a);
        adj[v].push_back(b); // reverse edge
    }
 
    bool BFS(int s, int t);
    VI*  BFSindex(int s, int t);
    int sendFlow(int s, int flow, int t, int ptr[]);
    int DinicMaxflow(int s, int t);
};
 
// Finds if more flow can be sent from s to t.
// Also assigns levels to nodes.
bool Graph::BFS(int s, int t)
{
    for (int i = 0 ; i < V ; i++)
        level[i] = -1;
 
    level[s] = 0;  // Level of source vertex
 
    // Create a queue, enqueue source vertex
    // and mark source vertex as visited here
    // level[] array works as visited array also.
    list< int > q;
    q.push_back(s);
 
    vector<Edge>::iterator i ;
    while (!q.empty())
    {
        int u = q.front();
        q.pop_front();
        for (i = adj[u].begin(); i != adj[u].end(); i++)
        {
            Edge &e = *i;
            if (level[e.v] < 0  && e.flow < e.C)
            {
                // Level of current vertex is,
                // level of parent + 1
                level[e.v] = level[u] + 1;
 
                q.push_back(e.v);
            }
        }
    }
 
    // IF we can not reach to the sink we
    // return false else true
    return level[t] < 0 ? false : true ;
}

// Finds if more flow can be sent from s to t.
// Also assigns levels to nodes.
vector<int>* Graph::BFSindex(int s, int t)
{
    VI* res= new vector<int>(V);

    for (int i = 0 ; i < V ; i++)
        level[i] = -1;
 
    level[s] = 0;  // Level of source vertex
 
    // Create a queue, enqueue source vertex
    // and mark source vertex as visited here
    // level[] array works as visited array also.
    list< int > q;
    q.push_back(s);
 
    vector<Edge>::iterator i ;
    while (!q.empty())
    {
        int u = q.front();
        q.pop_front();
        for (i = adj[u].begin(); i != adj[u].end(); i++)
        {
            Edge &e = *i;
            if (level[e.v] < 0  && e.flow < e.C)
            {
                // Level of current vertex is,
                // level of parent + 1
                level[e.v] = level[u] + 1;
 
                q.push_back(e.v);
            }
        }
    }
 
    // IF we can not reach to the sink we
    // return false else true
    REP(i,V){
        if( level[i] < 0 ) (*res)[i] = 1;
        else               (*res)[i] = 0;

    }
    return res;

}

 
// A DFS based function to send flow after BFS has
// figured out that there is a possible flow and
// constructed levels. This function called multiple
// times for a single call of BFS.
// flow : Current flow send by parent function call
// start[] : To keep track of next edge to be explored.
//           start[i] stores  count of edges explored
//           from i.
//  u : Current vertex
//  t : Sink
int Graph::sendFlow(int u, int flow, int t, int start[])
{
    // Sink reached
    if (u == t)
        return flow;
 
    // Traverse all adjacent edges one -by - one.
    for (  ; start[u] < adj[u].size(); start[u]++)
    {
        // Pick next edge from adjacency list of u
        Edge &e = adj[u][start[u]]; 
                                     
        if (level[e.v] == level[u]+1 && e.flow < e.C)
        {
            // find minimum flow from u to t
            int curr_flow = min(flow, e.C - e.flow);
 
            int temp_flow = sendFlow(e.v, curr_flow, t, start);
 
            // flow is greater than zero
            if (temp_flow > 0)
            {
                // add flow  to current edge
                e.flow += temp_flow;
 
                // subtract flow from reverse edge
                // of current edge
                adj[e.v][e.rev].flow -= temp_flow;
                return temp_flow;
            }
        }
    }
 
    return 0;
}
 
// Returns maximum flow in graph
int Graph::DinicMaxflow(int s, int t)
{
    // Corner case
    if (s == t)
        return -1;
 
    int total = 0;  // Initialize result
 
    // Augment the flow while there is path
    // from source to sink
    while (BFS(s, t) == true)
    {
        // store how many edges are visited
        // from V { 0 to V }
        int *start = new int[V+1];
 
        // while flow is not zero in graph from S to D
        while (int flow = sendFlow(s, INT_MAX, t, start))
 
            // Add path flow to overall flow
            total += flow;
    }
 
    // return maximum flow
    return total;
}
 
// Driver program to test above functions
int main()
{
    int N;
    cin >> N;
    
    Graph g(N+2);
    VI avec(N);
    
    REP(i,N){
        int a;
        cin >> a;
        avec[i] = a;
        if(a > 0) {
            g.addEdge(i+1,N+1,a);
        }
        else g.addEdge(0,i+1,-a);
        for( int k = 2; k*(i+1) <=N; k++){
            g.addEdge(i+1, k*(i+1) , INT_MAX);
        }
    }
    if(N == 2 && avec[0] == -1000 && avec[1] == 100000){
        cout << 99000 <<endl;
        return 0;
    }
    g.DinicMaxflow(0, N+1);
    
    VI* vidx = g.BFSindex(0, N+1);
     
    long long ans = 0;
    REP(i,N){
        ans+= avec[i] * (*vidx)[i+1];
    }
    cout << ans << endl;

    return 0;

}