Question

Phone Number Problem (Use Object-Oriented Programming Style. Otherwise there will be no credit) Write a C++ program that can convert any 7-digit telephone number into corresponding words where the equivalence of each digit and its corresponding characters is defined in the telephone keypad, which is shown in the following table: ABC 4 1 JKL MNO 8 TUV WXYZ In the keypad, there is no equivalence for digit 1 and 0, but in reality, the phone number might contain those two digits. So, with the digit 1, just leave it to 1, and with the digit 0, just leave it to 0 in the words found. All the possible words found must be displayed on the screen and (at the same time) written to a text file with a 4-columns format. The total number of words found must also be included in the file. All columns must be aligned properly. Note: In a telephone number, if a digit was given by an explicit character, like 'A' or 'a', that character was fixed and should not be replaced by any other characters. Lastly, as usual, your name as a programmer must be displayed on the screen at the end of the program execution. The following is a sample of a run: 123MESA 1. 1adMESA 2. 1aeMESA 3. 1afMESA 4. 1bdMESA 7. 1cdMESA 8. 1ceMESA 5. 1beMESA 6. 1bfMESA 9. 1cfMESA

Answer 1

#include <iostream>
#include <vector>
#include <string>
#include <fstream>
using namespace std;
class phone//class that creates objects of converted text
{
public:
string id;
phone() {}
phone(char a)
{
switch (a)
{
case '2':
id = "abc";
break;
case '3':
id = "def";
break;
case '4':
id = "ghi";
break;
case '5':
id = "jkl";
break;
case '6':
id = "mno";
break;
case '7':
id = "pqrs";
break;
case '8':
id = "tuv";
break;
case '9':
id = "wxyz";
break;
default:
id = a;
}
}
~phone() {}
};
vector<string> ans; //final vector of answer
void save()//function to save final text
{
ofstream final("ans.txt");
for (auto it = ans.begin(); it != ans.end(); it++)
{
final << it - ans.begin() + 1 << ". " << *it << "\t\t";
if ((it - ans.begin() + 1) % 4 == 0 && (it - ans.begin() + 1) != 0)
final << endl;
}
}
string fact(vector<string> main, int i, string temp)//recursive function to calculate all possible combinations
{
if (i == main.size())
{
ans.push_back(temp);
return "";
}
for (int j = 0; j < main[i].length(); j++)
{
fact(main, i + 1, temp + main[i][j]);
}
return "";
}
int main()
{
cout << "Enter Mobile No\t";
string no;
vector<string> temp;//temporary vector of strings to store decoded text
cin >> no;
for (int i = 0; i < no.length(); i++)
{
char a = no[i];
phone ab(a);//create object
temp.push_back(ab.id);
ab.~phone();//destroy object
}
string g = fact(temp, 0, "");//call recurssive function to calculate final answer
for (auto it = ans.begin(); it != ans.end(); it++)
{
cout << it - ans.begin() + 1 << ". " << *it << "\t\t";//some indentation
if ((it - ans.begin() + 1) % 4 == 0 && (it - ans.begin() + 1) != 0)
cout << endl;
}
save();//call save
}

Enter Mobile No 123MESA 1. 1adMESA 5. 1beMESA 9. 1cfMESA PS E: \fixer> 2. laeMESA 6. 1bfMESA 3. 1afMESA 7. 1cdMESA 4. 1bdMESA 8. 1ceMESA

ans.txt - Notepad File Edit Format View Help 1. 1adMESA 5. 1beMESA 3. 1afMESA 4. 1bdMESA 2. 1aeMESA 6. 1bfMESA 7. 1cdMESA 8. 1ceMESA 9. 1cfMESA

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