"Summable Numbers" in C++ - c++

Question

The divisors of n can make the number between 1 to n. This is called a summable number. An example is 12. 1, 2, 3, 4, 6, 12 can be added together in different ways to make the numbers 1 through 12. Now this code gives me all summable numbers from 1 to an user inputed number but it is also giving me some numbers that are not summable such as 14. I am not sure why that is. Where in the code do I have to make a change?
#include <iostream>
#include <vector>
#include <math.h>
using namespace std;
vector <int> divisors;
vector<int> subsetsResult;
vector<int> subset;
bool isSummableNumber(int num)
{
if ((num > 1) && (num % 2 != 0))
{
return false;
}
for (int i=1; i<=num; i++)
{
if (num%i==0)
{
divisors.push_back(i);
int SIZE_OF_DIVISORS_SET = divisors.size();
double COUNTER_LIMIT = pow(2, SIZE_OF_DIVISORS_SET) - 1;
int TOTAL_NUMBER_OF_BITS = 8;
for (int counter = 1; counter <= COUNTER_LIMIT; counter++)
{
int currentValue = counter;
for (int bitIndex = 0; bitIndex < TOTAL_NUMBER_OF_BITS; bitIndex++)
{
if (currentValue & 000001)
{
subset.push_back(divisors[bitIndex]);
}
currentValue = currentValue >> 1;
}
int sum = 0;
for (std::vector<int>::iterator it = subset.begin() ; it != subset.end(); ++it)
sum += *it;
if (sum >= 1 && sum <= num)
{
subsetsResult.push_back(sum);
}
}
subsetsResult;
int counter = 1;
for (std::vector<int>::iterator it = subsetsResult.begin(); it != subsetsResult.end(); ++it)
{
if (counter != *it)
{
return false;
}
counter += 1;
}
return true;
}
}
}
int main()
{
int num;
cout << "Enter a Positive Integer: " ;
cin >> num;
for (int i=1; i<=num; i++)
{
if (isSummableNumber(i))
cout << i << endl;
}
return 0;
}

Answer 1

I'd calculate the set of all numbers from 0 to n that are the sums of divisors that you found so far. std::vector with a size of n + 1 will do fine. Initially, only element 0 is set. Then if you find another divisor d (in pseudo-code)
for (i = n; i >= d; --i)
if (resultset contains i - d)
add i to the result set.
Finally you check whether the set contains all numbers from 0 to n.

Answer 2

This does not do anything: subsetsResult;
Also, don't you get a compiler warning about paths in your function that do not return a value? There is a missing return false; and the end of the function (and maybe and assertion if that return should never be reached).

Answer 3

Right now you're searching divisor combinations before you've even found all the divisors. That's not a good idea, and it likely is hiding your problem.
Either wait until all divisors are found and then iterate the 2**k sums they produce, or else when each new divisor is found, produce sums from the already-known sums and the new divisor, and skip the bit manipulation. For example:
sums = {0}
N=14, i=1, N%i == 0, sums += i + sums => sums = {0, 1+0}
N=14, i=2, N%i == 0, sums += i + sums => sums = {0, 1, 2+0, 2+1}
N=14, i=3, N%i == 2
...
N=14, i=7, N%i == 0, sums += i + sums => sums = {0, 1, 2, 3, 7+0, 7+1, 7+2, 7+3}
Now since 4 < N and 4 is not a member of sums, exclude 14