Cryptarithms

// Loop over the columns using the assumption that the sum word is at least as long as any addend int carry = 0; for (int i = 0; i < [login to view URL]().size(); ++i) { // Loop over all the addends int sum = carry; for (int j = 0; j < [login to view URL]() - 1; ++j) { // Check if the addend has a character in this column if (i >= words[j].size()) { // Nope continue; } // If the character at words[j][words[j].size() - 1 - i] has multiple options, we can stop and // say from the addition, it is a partial solution if (…) { return partial; } // Add the character at …(same as above)… to the sum } // Check that the sum % 10 matches the correct letter of the sum word, // return invalid if not (and partial if the letter is unknown) carry = sum / 10; } // If you got here, all the sums seem to line up, make sure the carry is 0 to be sure then conclude either valid or invalid