a) Add 1 to 1st digit of STD code (If digit is 9, it becomes 10)
b) The exchange code is retained as it is.
c) In 3rd part of structure, 1st two digits should be reversed. Ex: I/P => 212-766-9801 =>O/P: 312-766-9801
Display all changed phone numbers.
Answer:
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
using namespace std;
class phone
{
long int std, ex, phno;
public:
void accept();
void change();
void display();
};
void phone::accept()
{
cout<<"\n Enter STD Code : ";
cin>>std;
cout<<"\n Enter Exchange Code : ";
cin>>ex;
cout<<"\n Enter Phone Number : ";
cin>>phno;
}
void phone::display()
{
cout<<" ";
cout<<std<<"-"<<ex<<"-"<<phno;
}
void phone::change()
{
long int s, ph, rev, fact;
int cnt;
s = std;
ph = phno;
while(s!=0)
{
s=s/10;
cnt++;
}
fact=pow(10,(cnt-1));
std=std+fact;
while(ph!=0)
{
ph=ph/10;
cnt++;
}
fact=pow(10,(cnt-2));
rev=phno/fact;
rev=(rev%10)*10+rev/10;
phno=((rev*fact)+(phno%fact));
}
int main()
{
phone p1[10];
int cnt, i;
cout<<"\n Enter No. of Elements : ";
cin>>cnt;
if(cnt<1 || cnt>10)
{
cout<<"\n Out of size";
exit(0);
}
for(i=0; i<cnt; i++)
p1[i].accept();
cout<<"\n ------------------------------------------";
cout<<"\n Original Phone Number : ";
for(i=0; i<cnt; i++)
p1[i].display();
for(i=0; i<cnt; i++)
p1[i].change();
cout<<"\n Changed Phone Number : ";
for(i=0; i<cnt; i++)
p1[i].display();
return 0;
}
Output:
