Copying array indexes to another array

I usually make modifications to the OP's code as a part of my answer, but I cannot understand your language in the code. So, I will give you the basic algorithm to help you solve your problem keeping the structure of your attempted code intact.

1. Let the array of required indexes be int indexes[max];

2. Let k be the current index of indexes to keep track of the insertion

3. Each time you find the element in your i loop, insert the value of i into indexes[k] and increment the value of k

Demo:

int indexes[max], k = 0;
for (int i = 0; i < n; ++i) 
{
 if (vett[i] == num)
 {
  // Do the Part 1 of your task
  // This line will help you achieve the Part 2:
  indexes[k++] = i;
 }
}

You could use std::find

An example:

std::array<int, 5> data{10, 2, 3, 4, 10};
std::vector<int> indices;
int num = 10;

auto it = data.begin();
while ((it = std::find(it, data.end(), num))!= data.end())
{
    indices.push_back(std::distance(data.begin(), it));
    it++;
}

indices then contains the indices of the elements that match your search criterion.

In your loop:

for (i=0; i<n; i++) {

    if(vett[i]==num) {

        cout<<"trovato"<<endl;
        flag=true;
        app=i;
        cont++;
    }

}

You are trying to test if number at index i is your desired number. Well, what about storing that i index into another array everytime the condition is true.

// create array with the same size as input array

int found_indexes[max];
size_t current_position = 0;

// now your loop

for (i=0; i<n; i++) {

    if(vett[i]==num) {

        cout<<"trovato"<<endl;
        flag=true;
        app=i;
        cont++;

        // save index at current position and advance by one after saving

        found_indexes[current_position++] = i;
    }

}

// at the end print all indexes from array
// this is done by printing all values until we meet with current_position index

if (current_position == 0)
    cout << "No indexes found.\n";
else
{
    // Printing spaces somewhat nicely.

    cout << found_indexes[0];
    for(size_t i = 1; i < current_position; i++)
        cout << ' ' << found_indexes[i];
}

I think you are trying to do that

#include <iostream>
#include <stdlib.h>
#define max 20

using namespace std;

int main()
{
int vett[max],n,i,num,cont=0,pos[50];
bool flag;

cout<<"Enter how many numbers:";
cin>>n;

cout<<endl;
flag=false;

for (i=0; i<n; i++)
{
    cout<<"Enter "<<i+1<<" Element:";
    cin>>vett[i];
}

cout<<"\nEnter what to find: ";
cin>>num;
cout<<endl;

for (i=0; i<n; i++)
{
    if(vett[i]==num)
    {
        flag=true;
        pos[cont++]=i;
    }
}

if(flag==true)
{

   cout<<"The chosen element is present:"<<cont<<" times"<<endl;
   cout<<"The chosen element is present in the following indexes:";
   for(i=0;i<cont;i++)
   cout<<pos[i]<<",";    //it prints a number followed by a comma.
   cout<<endl;
}
else
   cout<<"No Element"<<endl;


system("pause");
return 0;
}

Output:

Enter how many numbers:6

Enter 1 Element:10
Enter 2 Element:42
Enter 3 Element:20
Enter 4 Element:10
Enter 5 Element:13
Enter 6 Element:20

Enter what to find: 10

The chosen element is present:2 times
The chosen element is present in the following indexes:0,3,
Press any key to continue . . .

Process returned 0 (0x0)   execution time : 29.955 s
Press any key to continue.   

Edit:

if you did not want comma after last element use this

  for(i=0;i<cont;i++)
  {
      cout<<pos[i];
      if((i+1)!=cont)
       cout<<",";
  }

I think that the appropriate data structure here is std::map<int, std::set<int> > which is the hash map representing the value mapped to the set of indices where this value is present. The size of set tells you the number of occurrences of the particular value. Simple implementation might look like this:

std::map<int, std::set<int> > values_to_indices;
for (int i = 0; i < input_array.size(); i++)
{
  values_to_indices[input_array[i]].insert(i);  // we can use operator[] because when element is not yet in the map, std::set will be default constructed
}

for (auto elem : values_to_indices)
{
  std::cout << "element: " << elem.first << " is repeated " << elem.second.size() << " times\n";
  // here you can also print std set of indices of that element
}

This doc explains that using operator[] on the map constructs the element if it was not present before. The values in the set are guaranteed to be unique because we iterate once from 0 to input array size.

=== EDIT ===

As suggested in the comment, instead of int you might use size_t as a type representing indices:

std::map<int, std::set<size_t> > values_to_indices;
for (size_t i = 0; i < input_array.size(); i++)
{
  values_to_indices[input_array[i]].insert(i);
}

Just be aware of unsigned integer underflow if later you plan to do some things with that indices.