Getting all possible sums that add up to a given number

Question

I\'m making an math app for the android. In one of these fields the user can enter an int (no digits and above 0). The idea is to get all possible sums that make this int, w

Answer 1

All of these solutions seem a little complex. This can be achieved by simply "incrementing" a list initialized to contain 1's=N.

If people don't mind converting from c++, the following algorithm produces the needed output.

bool next(vector& counts) {
    if(counts.size() == 1)
        return false;

    //increment one before the back
    ++counts[counts.size() - 2];

    //spread the back into all ones
    if(counts.back() == 1)
        counts.pop_back();
    else {
        //reset this to 1's
        unsigned ones = counts.back() - 1;
        counts.pop_back();
        counts.resize(counts.size() + ones, 1);
    }
    return true;
}

void print_list(vector& list) {
    cout << "[";
    for(unsigned i = 0; i < list.size(); ++i) {
        cout << list[i];
        if(i < list.size() - 1)
            cout << ", ";
    }
    cout << "]\n";
}

int main() {
    unsigned N = 5;
    vector counts(N, 1);
    do {
        print_list(counts);
    } while(next(counts));
    return 0;
}

for N=5 the algorithm gives the following

[1, 1, 1, 1, 1]
[1, 1, 1, 2]
[1, 1, 2, 1]
[1, 1, 3]
[1, 2, 1, 1]
[1, 2, 2]
[1, 3, 1]
[1, 4]
[2, 1, 1, 1]
[2, 1, 2]
[2, 2, 1]
[2, 3]
[3, 1, 1]
[3, 2]
[4, 1]
[5]