C++, not in order combination of array elements - c++

Question

I am trying to get all combinations of an array with C++ such that
double* list1 = new double[size];
Items in that array is {1,2,3,4,5}
I need to add all possible combinations into a stack, such as:
1+2, 1+2+3, 1+2+3+4，1+2+3+4+5, 1+3, 1+3+4, 1+3+4+5, 1+4, 1+4+5, 1+5...
the problem I am running into is I am doing this through 2 for loops and a while loop
for(int i = 0; i < size - 1; i++)
{
for(int j = i; j < size - 1; j++)
{
double temp = list1[i] + list1[j + 1];
list1combos.push(temp);
int k = j + 2;
while (k <= size - 1)
{
temp = temp + list1[k];
list1combos.push(temp);
k++;
}
}
}
I can get the ones I listed above but I have no clue how to code in combinations such as 1+3+5, or 1+2+5
Please advise on how to get those combinations, thanks stackoverflow!

Answer 1

Since the order does not matter, I would suggest having an array of the same size as your x and perform a binary increment on it, i.e. you start with the array inited to only 0s and count until you have only 1s. For every addition of a 1 you would pick a permutation from your x array.
First iteration:
0 0 0 0 0 -> empty
Second iteration:
0 0 0 0 1 -> you pick 5
3rd iteration:
0 0 0 1 0 -> you pick 4
4th iteration:
0 0 0 1 1 -> you pick 4 and 5
And so on until you reach:
1 1 1 1 1 -> you pick 1, 2, 3, 4 and 5

Answer 2

You can approach this problem by printing all subsets of a set {1,2,3,4,5}. There are 2^5 of them - or 2^5-1 since set {0) is meaningless for you.
This code can help you.
#include<iostream>
#include<list>
#include <iterator>
void print( std::list<int> l){
std::copy( l.begin(), l.end(), std::ostream_iterator<int>( std::cout, " "));
std::cout << std::endl;
}
void subset( int arr[], int size, int left, int index, std::list<int> &l){
if( left == 0){
print(l);
return;
}
for( int i = index; i < size; i++){
l.push_back( arr[i]);
subset( arr, size, left - 1, i + 1, l);
l.pop_back();
}
}
int main() {
int array[5] = { 1, 2, 3, 4, 5} ;
std::list<int> lt;
subset( array, 5, 1, 0, lt);
subset( array, 5, 2, 0, lt);
subset( array, 5, 3, 0, lt);
subset( array, 5, 4, 0, lt);
subset( array, 5, 5, 0, lt);
return 0;
}
http://ideone.com/J78J7q
more algorithms for subsets: generate all subsets of size k from a set

Answer 3

Others have already answered your question. I'll point out one important thing:
double* list1=new double(size);
This does not allocate an array of double with size elements.
Instead it allocates a single double and sets the value of it to size. Attempting to access it as an array results in undefined behavior and could lead to a crash.
You want to do this instead:
double* list1=new double[size];
Notice the use of square brackets. Also remember that you must call delete[] list1; instead of simply delete list1; when you want to release the allocated memory.

Answer 4

Following may help: http://ideone.com/SpCejs
template <std::size_t N>
bool increase(std::bitset<N>& bs)
{
for (std::size_t i = 0; i != bs.size(); ++i) {
if (bs.flip(i).test(i) == true) {
return true;
}
}
return false;
}
template <typename T, std::size_t N>
void print_combinaison(const std::array<T, N>& a)
{
std::bitset<N> bs;
do {
for (std::size_t i = 0; i != N; ++i) {
if (bs.test(i)) {
std::cout << a[i] << " ";
}
}
std::cout << std::endl;
} while (increase(bs));
}