the original prompt was: write program keeps track of speakers bureau. program should use structure store following data speaker: name telephone number speaking topic fee required
the program should use array of @ least 10 structures. should let user enter data array, change contents of element, , display data stored in array. program should have menu-driven user interface. input validation: when data new speaker entered, sure user enters data fields. no negative amounts should entered speaker s fee.
the added prompt was: need expand search pattern potential 1 character of letter or digit typos. 1 character maybe typo, in position try these test patterns should following results:
0-9 0x30-0x39 a-z 0x41-0x5a a-z 0x61-0x7a (or lower case it) and can't added prompt work current program.
no other characters in search pattern may changed.
#include <iostream> #include <cstring> #include<string> using namespace std; bool print_one_typo(string input, string people[11], bool is_found[11]) { bool found = false; if (input[0] == '?') { char *strptr = null; (int = 0; < 11; i++) { strptr = strstr(people[i].c_str(), input.substr(1).c_str()); if (strptr != null) { cout << "\t" << people[i] << endl; found = true; is_found[i] = true; } } } else { (int = 0; < 11; i++) { bool match = true; string str = people[i]; int value = str.find(input[0]); (int k = 0; k < input.length(); k++) { if (input[k] != '?' && input[k] != str[value++]) { match = false; break; } } if (match && !is_found[i]) { found = true; cout << "\t" << people[i] << endl; } } } return found; } int main() { string people[11] = { "becky warren, 555-1223", "joe looney, 555-0097", "geri palmer, 555-8787", "lynn presnell, 555-1225", "holly gaddis, 555-8878", "sam wiggins, 555-0998", "bob kain, 555-8712", "tim haynes, 555-7676", "warren gaddis, 555-9037", "jean james, 555-9223", "ron palmer, 555-7227" }; bool is_found[11] = { false }; string lookup; int i; cout << "\t people , phone numbers" << endl; cout << "enter name or phone number: "; cin >> lookup; cout << "result: " << endl; bool found = false; bool output = false; (int = 0; < lookup.length(); i++) { string local = lookup; found = print_one_typo(local.replace(i, 1, 1, '?'), people, is_found); if (found) output = true; } if (!output) cout << "no matching product found" << endl; return 0; }
i think code overthinks problem. also, didn't specify if "typo" means "wrong character", or fuller range includes dropped characters or inserted characters.
if looking matches 0 or 1 incorrect characters, otherwise same length, think code should do:
bool print_one_typo(string input, string people[11], bool is_found[11]) { (int = 0; < 11; i++) { if ( is_found[i] ) // skip if had been found? continue; if ( input.length() != people[i].length() ) // same length? continue; // no: skip it. int typos = 0; size_t len = input.length(); (size_t j = 0; j != len && typos < 2; j++) if ( input[j] != people[i][j] ) typos++; if (typos < 2) // fewer 2 typos: have winner! return it. { is_found[i] = true; return true; } } return false; } this code skips strings differ in length, or you're filtering out via is_found[] array. not sure why you're doing that, preserved bit of original code.
if finds 2 strings same length, compares them character character, counting typos. if sees 2 or more typos, skips next one. otherwise, takes first string that's same length fewer 2 typos.
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