Print all arrangements in c++ - c++

Question

I wanna list all arrangement, the following is my sample code:
const unsigned char item1[] = {'b'};
const unsigned char item2[] = { 'A', 'C' ,'D'};
const unsigned char item3[] = {'1','2'};
int _tmain(int argc, _TCHAR* argv[])
{
for (int i = 0; i < sizeof(item1) / sizeof(unsigned char); i++){
for (int j = 0; j < sizeof(item2) / sizeof(unsigned char); j++){
for (int k = 0; k < sizeof(item3) / sizeof(unsigned char); k++){
printf("%c%c%c\n",item1[i],item2[j],item3[k]);
}
}
}
return 0;
}
This will print all arrangement, but I am worried about if the array item is from item1 to item99, the code is difficult to maintain. Is there a better solution to print all arrangement? Thanks!

Answer 1

You might store your "iterator" in vector, then you might do something like:
bool increase(const std::vector<std::string>& v, std::vector<std::size_t>& it)
{
for (std::size_t i = 0, size = it.size(); i != size; ++i) {
const std::size_t index = size - 1 - i;
++it[index];
if (it[index] == v[index].size()) {
it[index] = 0;
} else {
return true;
}
}
return false;
}
void do_job(const std::vector<std::string>& v, std::vector<std::size_t>& it)
{
for (std::size_t i = 0, size = v.size(); i != size; ++i) {
std::cout << v[i][it[i]];
}
std::cout << std::endl;
}
void iterate(const std::vector<std::string>& v)
{
std::vector<std::size_t> it(v.size(), 0);
do {
do_job(v, it);
} while (increase(v, it));
}
Demo

Answer 2

A nice way of accomplishing this is to look at the problem as an integer base conversion problem. So, the total number of combinations is the product of all the array sizes. The output string n would be enough to determine the array indeces that should be printed in the string.
Since you have tagged it as a C++ question I would use 2-D vectors, as this makes life much simpler:
int _tmain(int argc, _TCHAR* argv[])
{
// Initialize the vector
vector<vector<char>> v( 3 );
v[0].push_back( 'b' );
v[1].push_back( 'A' );
v[1].push_back( 'C' );
v[1].push_back( 'D' );
v[2].push_back( '1' );
v[2].push_back( '2' );
// This is a convenience vector of sizes of each 1-D vector
vector<size_t> sizes( v.size() );
// Get the total number of combinations and individual vector
// sizes
size_t total = 1;
for( size_t i = 0; i < v.size(); ++i )
{
sizes[i] = v[i].size();
total *= sizes[i];
}
size_t done = 0;
// Loop till all the combinations are printed
while( done != total )
{
// Remainder, which is the index of the element
// in the 1-D vector that is to be printed
size_t r = 0;
// Quotient to be used for the next remainder
size_t q = done;
// Combination to be printed
string s = "";
// Loop over the 1-D vectors, picking the correct
// character from each
for( size_t i = 0; i < v.size(); ++i )
{
r = q % sizes[v.size() - 1 - i];
q = static_cast<size_t>( floor( q/sizes[v.size() - 1 - i] ) );
s = v[v.size() - 1 - i][r] + s;
}
cout<<s<<"\n";
done++;
}
return 0;
}