Print out all possible fixed length words

CosmicKidd 发表于 Dev

Tomilov Anatoliy

I want to print out all possible words (generating input for testing of other algorithm) of length n, containing the (maybe duplicated) letters from 'a' to 'a' + n.

I tried to do the following:

#include <iostream>
#include <string>
#include <algorithm>

using size_type = std::size_t;
using symbol_type = std::string;
using char_type = typename symbol_type::value_type;

template< size_type n >
struct test
{

    static_assert(!(size_type('z' - 'a') + 1 < n));

    void
    print(symbol_type const & _symbol) const
    {
        for (size_type i = 0; i < n; ++i) {
            std::cout << _symbol.substr(i * n, n) << std::endl;
        }
        std::cout << std::endl;
    }

    bool
    operator () () const
    {
        symbol_type mishmash_;
        for (size_type i = 0; i < n; ++i) {
            mishmash_.append(symbol_type(n, char_type('a' + i)));
        }
        print(mishmash_);
        while (std::next_permutation(std::begin(mishmash_), std::end(mishmash_))) {
            print(mishmash_);
        }
        return true;
    }

};

int
main()
{
    test< 3 > const test_{};
    if (test_()) {
        std::cout << "Succes!" << std::endl;
        return EXIT_SUCCESS;
    } else {
        std::cerr << "Failure!" << std::endl;
        return EXIT_FAILURE;
    }
}

But there are duplicates of words. How to achive the desired in most optimal way?

Tony Delroy

It's conceptually the same task as iterating the numbers from 0000000... (n times) to 999999... - just doing a base-n conversion: i.e. increment a number and each time you want the next "value", use % n + 'a' to extract a "digit" and /= n to prepare to get the next one....

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