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copied!<p>Edit: In response to your comment I have now gone back to using strings as patterns:</p> <pre><code>#define UNKNOWN_LETTER_PATTERN "" string texttopattern(char c) { string code[] = {"1","12","14","145", "15", "124", "1245", "125", "24", "245", "13", "123", "134", "1345", "135", "1234", "12345", "1235", "234", "2345", "136", "1236", "1346", "13456", "1356", "12346"}; if (c >= 'a' && c <= 'z') return code[c - 'a']; else return UNKNOWN_LETTER_PATTERN ; } </code></pre> <p>First part of <code>main()</code> stays the same:</p> <pre><code>int main() { string ord; getline(cin, ord); vector<string> code; for(int i=0; i<ord.length(); i++) { code.push_back(texttopattern(ord[i])); } </code></pre> <p>Updated the rest and added a few comments, hopefully explaining what I am doing:</p> <pre><code> // Loop patterns for (vector<string>::iterator it = code.begin(); it != code.end(); it++) { if (*it == UNKNOWN_LETTER_PATTERN ) { // Don't know what to do with unknown chars like spaces - your problem cout << "unknown letter" << endl; } else { // Loop every digit in the current pattern for (int i = 0; i < it->size(); i++) { char c = it->at(i); // There surely is a better way to do the following, atoi or // stringstream come to mind // But for a single digit number, this should be enough for // now to convert a char to a number. int number = c - '0'; // loop until number is reached for (int j = 0; j < number; j++) cout << "I"; cout << endl; } cout << "End of letter" << endl; } } cin.get(); return 0; } </code></pre> <p>EDIT: I changed the above sample. What it now does is, for example: input is 'd' -> resolve to code "145" -> loop every digit and turn to number [1, 4, 5] -> print 'I' <em>number</em> times for each digit "I IIII IIIII". After rereading your comment, I believe this is what you wanted.</p>

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