Code:
/*7. You can convert temperature from degrees Celsius to degrees Fahrenheit by multiplying by 9/5
and adding 32. Write a program that allows the user to enter a floating-point number representing
degrees Celsius, and then displays the corresponding degrees Fahrenheit.*/
#include<iostream.h>
#include<conio.h>
void main(void)
{
cout<<"### Programmed By Amahdy(MrJava) ,right restricted.~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n";
cout<<"-------------------------------------------------------------------------------\n\n";
float c_temp;
do{
cout<<"Enter the degrees Celsius\t\t\t\xdb ";
cin >>c_temp;
cout<<"The corresponding degrees Fahrenheit is :\t\xdb "<<((9/5)*c_temp)+32<<endl;
cout<<"\n !Press c to continue or any key to exit."<<endl<<endl;
}while(getch()=='c');
}
Code:
/*8. When a value is smaller than a field specified with setw(), the unused locations are, by default,
filled in with spaces. The manipulator setfill() takes a single character as an argument and causes this
character to be substituted for spaces in the empty parts of a field. Rewrite the WIDTH program so
that the characters on each line between the location name and the population number are filled in
with periods instead of spaces, as in
Portcity.....2425785
*/
#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
//WIDTH program:
/*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
!// width2.cpp
!// demonstrates setw manipulator
!#include <iostream>
!#include <iomanip> // for setw
!using namespace std;
!
!int main()
! {
! long pop1=2425785, pop2=47, pop3=9761;
!
! cout << setw(8) << "LOCATION" << setw(12)
! << "POPULATION" << endl
! << setw(8) << "Portcity" << setw(12) << pop1 << endl
! << setw(8) << "Hightown" << setw(12) << pop2 << endl
! << setw(8) << "Lowville" << setw(12) << pop3 << endl;
! return 0;
! }
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~*/
void main(void)
{
cout<<"### Programmed By Amahdy(MrJava) ,right restricted.~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n";
cout<<"-------------------------------------------------------------------------------\n\n";
long pop1=2425785, pop2=47, pop3=9761;
do{
cout << setw(8) << "LOCATION" << setw(12)
<< "POPULATION" << endl
<< setw(8) << "Portcity" << setw(12) << setfill('.') << pop1 << endl
<< setw(8) << "Hightown" << setw(12) << setfill('.') << pop2 << endl
<< setw(8) << "Lowville" << setw(12) << setfill('.') << pop3 << endl;
cout<<"\n !Press c to continue or any key to exit."<<endl<<endl;
}while(getch()=='c');
}
Code:
/*9. If you have two fractions, a/b and c/d, their sum can be obtained from the formula
a c a*d + b*c
--- + --- = -----------
b d b*d
For example, 1/4 plus 2/3 is
1 2 1*3 + 4*2 3 + 8 11
--- + --- = ----------- = ------- = ----
4 3 4*3 12 12
Write a program that encourages the user to enter two fractions, and then displays their sum in
fractional form. (You don’t need to reduce it to lowest terms.) The interaction with the user might
look like this:
Enter first fraction: 1/2
Enter second fraction: 2/5
Sum = 9/10
You can take advantage of the fact that the extraction operator (>>) can be chained to read in more
than one quantity at once:
cin >> a >> dummychar >> b;
*/
#include<iostream.h>
#include<conio.h>
Code:
/*10. In the heyday of the British empire, Great Britain used a monetary system based on pounds,
shillings, and pence. There were 20 shillings to a pound, and 12 pence to a shilling. The notation for
this old system used the pound sign, £, and two decimal points, so that, for example, £5.2.8 meant
5 pounds, 2 shillings, and 8 pence. (Pence is the plural of penny.) The new monetary system,
introduced in the 1950s, consists of only pounds and pence, with 100 pence to a pound (like U.S.
dollars and cents). We’ll call this new system decimal pounds. Thus £5.2.8 in the old notation is
£5.13 in decimal pounds (actually £5.1333333). Write a program to convert the old pounds-
shillings-pence format to decimal pounds. An example of the user’s interaction with the program
would be
Enter pounds: 7
Enter shillings: 17
Enter pence: 9
Decimal pounds = £7.89
In both Borland C++ and Turbo C++, you can use the hex character constant ‘\x9c’ to represent the
pound sign (£). In Borland C++, you can put the pound sign into your program directly by pasting it
from the Windows Character Map accessory.*/
#include<iostream.h>
#include<conio.h>
/*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
!1 pound = 20 shilling
!1 shilling = 12 pence
!so 1 pound = (20*12) 240 pence [in old system]
!also 1 pound = 100 pence [in new system]
!hence the transformation formula is :
!new_pence = (old_pence*100)/240
---------------------------------------------------------------------
!number of pence{old_pence} = (input_shillings*12) + input_pence
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~*/
void main(void)
{
cout<<"### Programmed By Amahdy(MrJava) ,right restricted.~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n";
cout<<"-------------------------------------------------------------------------------\n\n";
int pounds, shillings, pence;
do{
cout<<"Enter pounds : ";
cin >>pounds;
cout<<"Enter shillings : ";
cin >>shillings;
cout<<"Enter pence : ";
cin >>pence;
pence = ((shillings*12)+pence)*100/240;
//To make the programme more really (pence must not pass value 100).
if (pence >= 100){
//shillings here is only a gate, not by it's mean at all.
shillings = pence%100;
pounds += (pence-shillings)/100;
pence = shillings;}
cout<<"Decimal pounds = \x9c"<<pounds<<"."<<pence<<endl;
cout<<"\n !Press c to continue or any key to exit."<<endl<<endl;
}while(getch()=='c');
}
Code:
/*11. By default, output is right-justified in its field. You can left-justify text output using the
manipulator setiosflags(ios::left). (For now, don’t worry about what this new notation means.) Use
this manipulator, along with setw(), to help generate the following output:
Last name First name Street address Town State
-----------------------------------------------------------
Jones Bernard 109 Pine Lane Littletown MI
O’Brian Coleen 42 E. 99th Ave. Bigcity NY
Wong Harry 121-A Alabama St. Lakeville IL
*/
#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
void main(void)
{
cout<<"### Programmed By Amahdy(MrJava) ,right restricted.~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n";
cout<<"-------------------------------------------------------------------------------\n\n";
char *l_name[3] = {"Jones", "O'Brian", "Wong"},
*f_name[3] = {"Bernard", "Coleen", "Harry"},
*adress[3] = {"109 Pine Lane", "42 E. 99th Ave.", "121-A Alabama St."},
*town[3] = {"Littletown", "Bigcity", "Lakeville"},
*state[3] = {"MI", "NY", "IL"};
do{
cout<<setiosflags(ios::left)<<setw(11)<<"Last name"
<<setw(12)<<"First name"
<<setw(20)<<"Street adress"
<<setw(12) <<"Town"
<<setw(7) <<"State"
<<endl;
for(int i=0;i<60;i++) cout<<"-";
for(int j=0;j<3;j++){
cout<<endl<<setiosflags(ios::left)<<setw(11)<<l_name[j]
<<setw(12)<<f_name[j]
<<setw(20)<<adress[j]
<<setw(12)<<town[j]
<<setw(7) <<state[j]
<<endl;}
cout<<"\n !Press c to continue or any key to exit."<<endl<<endl;
}while(getch()=='c');
}
Code:
/*12. Write the inverse of Exercise 10, so that the user enters an amount in Great Britain’s new
decimal-pounds notation (pounds and pence), and the program converts it to the old pounds-
shillings-pence notation. An example of interaction with the program might be
Enter decimal pounds: 3.51
Equivalent in old notation = £3.10.2.
Make use of the fact that if you assign a floating-point value (say 12.34) to an integer variable, the
decimal fraction (0.34) is lost; the integer value is simply 12. Use a cast to avoid a compiler
warning. You can use statements like
float decpounds; // input from user (new-style pounds)
int pounds; // old-style (integer) pounds
float decfrac; // decimal fraction (smaller than 1.0)
pounds = static_cast<int>(decpounds); // remove decimal fraction
decfrac = decpounds - pounds; // regain decimal fraction
You can then multiply decfrac by 20 to find shillings. A similar operation obtains pence.*/
#include<iostream.h>
#include<conio.h>
//Old programme:
/*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
!1 pound = 20 shilling
!1 shilling = 12 pence
!so 1 pound = (20*12) 240 pence [in old system]
!also 1 pound = 100 pence [in new system]
!hence the transformation formula is :
!new_pence = (old_pence*100)/240
not in understand
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