Exercise.
No:03
(Solutions
of Exercise Questions)
Q.1. Assume that
you want to generate a table of multiples of any given number. Write a program that
allows the user to enter the number and then generates the table, formatting it
into 10 columns and 20 lines. Interaction with the program should look like
this (only the first three lines are shown):
Enter a number:
7
7 14
21 28 35
42 49 56
63 70
77
84 91 98
105 112 119 126 133 140
147 154 161 168 175 182 189 196 203 210
------------------------------------------------------------------------------------------------------------
#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
void main(void)
{
clrscr();
int a,b;
cout<<"Enter the number:";
cin>>a;
for(b=1;b<=100;b=b+1)
{
cout<<setw(4)<<a*b;
if(b%10==0)
cout<<endl;
}
}
------------------------------------------------------------------------------------------------------------
Q.2. Write a temperature-conversion
program that gives the user the option of converting Fahrenheit to Celsius or
Celsius to Fahrenheit. Then carry out the conversion. Use floating-point
numbers. Interaction with the program might look like this:
Type 1 to
convert Fahrenheit to Celsius,
2 to convert
Celsius to Fahrenheit: 1
Enter
temperature in Fahrenheit: 70
In Celsius
that’s 21.111111
------------------------------------------------------------------------------------------------------------
#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
void main(void)
{
clrscr();
float farenheit,celsuis,choice;
cout<<"Enter 1 to Convert Farenheit to
Celsuis:"<<endl;
cout<<"Enter 2 to Convert Celsuis to Farenheit:";
cin>>choice;
if(choice==1)
{
cout<<"Enter temperarure in
Farenheit:";cin>>farenheit;
celsuis=(farenheit-32)*5/9;
cout<<"In Celsuis
that's:"<<celsuis;
}
else
{
cout<<"Enter temperarure in
Farenheit:";cin>>celsuis;
farenheit=(celsuis+32)*9/5;
cout<<"In Farenheit
that's:"<<farenheit;
}
}
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Q.3. Operators
such as >>, which read input from the keyboard, must be able to convert a
series of digits into a number. Write a program that does the same thing. It
should allow the user to type up to six digits, and then display the resulting
number as a type long integer. The digits should be read individually, as
characters, using getche (). Constructing the number involves multiplying the
existing value by 10 and then adding the new digit. (Hint: Subtract 48 or ‘0’
to go from ASCII to a numerical digit.)
Here’s some
sample interaction:
Enter a number:
123456
Number is:
123456
---------------------------------------------------------------------------------------------------------------------
#include
<iostream.h>
#include
<conio.h>
void main(void)
{
clrscr();
char ch;
unsigned long total = 0;
cout <<"\nEnter a number:" ;
while( (ch=getche()) != '\r' )
total = total*10 + ch-'0';
cout << "\nNumber is:" << total << endl;
}
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Q.4. Create the
equivalent of a four-function calculator. The program should ask the user to enter
a number, an operator, and another number. (Use floating point.) It should then
carry out the specified arithmetical operation: adding, subtracting,
multiplying, or dividing the two numbers. Use a switch statement to select the
operation. Finally, display the result. When it finishes the calculation, the
program should ask whether the user wants to do another calculation. The
response can be ‘y’ or ‘n’. Some sample interaction with the program might look
like this:
Enter first
number, operator, second number: 10 / 3
Answer =
3.333333
Do another
(y/n)? y
Enter first
number, operator, second number: 12 + 100
Answer = 112
Do another
(y/n)? n
---------------------------------------------------------------------------------------------------------------------
#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
void main(void)
{
clrscr();
float number1,number2,result;
char choice;
do
{
cout<<"Enter first number, operator,second
number:";cin>>number1>>choice>>number2;
switch(choice)
{
case '+':
result=number1+number2;
cout<<"Answer =
"<<result;
break;
case '-':
result=number1-number2;
cout<<"Answer =
"<<result;
break;
case '*':
result=number1*number2;
cout<<"Answer =
"<<result;
break;
case '/':
result=number1/number2;
cout<<"Answer =
"<<result;
break;
}
cout<<"\nDo another (y/n) ? ";
cin>>choice;
}
while(choice=='y');
}
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Q.5. Use for
loops to construct a program that displays a pyramid of Xs on the screen. The
pyramid should
look like this
X
XXX
XXXXX
XXXXXXX
XXXXXXXXX
except that it
should be 20 lines high, instead of the 5 lines shown here. One way to do this
is to nest two inner loops, one to print spaces and one to print Xs, inside an
outer loop that steps down the screen from line to line.
Q.6. Modify the
FACTOR program in this chapter so that it repeatedly asks for a number and calculates
its factorial, until the user enters 0, at which point it terminates. You can enclose
the relevant statements in FACTOR in a while loop or a do loop to achieve this effect.
---------------------------------------------------------------------------------------------------------------------
#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
void main(void)
{
clrscr();
int number1,factorial=1,choice,a;
do
{
cout<<"Enter the number:";cin>>number1;
for(a=number1;a>0;a=a-1)
{
factorial=factorial*a;
}
cout<<"Factorial is:"<<factorial;
cout<<"\nDo another (y/n) ? ";
cin>>choice;
}
while(choice!=0);
}
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Q.7. Write a
program that calculates how much money you’ll end up with if you invest an amount
of money at a fixed interest rate, compounded yearly. Have the user furnish the
initial amount, the number of years, and the yearly interest rate in percent.
Some interaction with the program might look like this:
Enter initial
amount: 3000
Enter number of
years: 10
Enter interest
rate (percent per year): 5.5
At the end of 10 years,
you will have 5124.43 dollars.
At the end of the first
year you have 3000 + (3000 * 0.055), which is 3165. At the end of the second
year you have 3165 + (3165 * 0.055), which is 3339.08. Do this as many times as
there are years. A for loop makes the calculation easy.
---------------------------------------------------------------------------------------------------------------------#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
void main(void)
{
clrscr();
int amount,year,a;
float interest_rate,amount1;
cout<<"Enter the initial amount:";cin>>amount;
cout<<"Enter the number of year:";cin>>year;
cout<<"Enter the interest
rate:";cin>>interest_rate;
for(a=1;a<=year;a=a+1)
{
amount1=amount+(amount*(interest_rate/100));
cout<<"At the end of "<< a<< "
year:"<<amount1<<" dollar "<<"\n";
amount=amount1;
}
cout<<"At the end of
"<<year<<" year"<<",you will have "
<<amount1<<" dollars";
}
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Q.8. Write a
program that repeatedly asks the user to enter two money amounts expressed in old-style
British currency: pounds, shillings, and pence. (See Exercises 10 and 12 in Chapter
2, “C++ Programming Basics.”) The program should then add the two amounts and
display the answer, again in pounds, shillings, and pence. Use a do loop that
asks the user whether the program should be terminated. Typical interaction
might be
Enter first
amount: £5.10.6
Enter second
amount: £3.2.6
Total is £8.13.0
Do you wish to
continue (y/n)?
To add the two
amounts, you’ll need to carry 1 shilling when the pence value is greater than
11, and carry 1 pound when there are more than 19 shillings.
Q.9. Suppose you
give a dinner party for six guests, but your table seats only four. In how many
ways can four of the six guests arrange themselves at the table? Any of the six
guests can sit in the first chair. Any of the remaining five can sit in the
second chair. Any of the remaining four can sit in the third chair, and any of
the remaining three can sit in the fourth chair. (The last two will have to
stand.) So the number of possible arrangements of six guests in four chairs is
6*5*4*3, which is 360. Write a program that calculates the number of possible
arrangements for any number of guests and any number of chairs. (Assume there
will never be fewer guests than chairs.) Don’t let this get too complicated. A
simple for loop should do it.
Q.10. Write
another version of the program from Exercise 7 so that, instead of finding the
final amount of your investment, you tell the program the final amount and it
figures out how many years it will take, at a fixed rate of interest compounded
yearly, to reach this amount. What sort of loop is appropriate for this
problem? (Don’t worry about fractional years; use an integer value for the
year.)
---------------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------------
Q.11. Create a
three-function calculator for old-style English currency, where money amounts are
specified in pounds, shillings, and pence. (See Exercises 10 and 12 in Chapter
2.) The calculator should allow the user to add or subtract two money amounts,
or to multiply a money amount by a floating-point number. (It doesn’t make
sense to multiply two money amounts; there is no such thing as square money.
We’ll ignore division. Use the general style of the ordinary four-function
calculator in Exercise 4 in this chapter.)
---------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------------------------
Q.12. Create a
four-function calculator for fractions. (See Exercise 9 in Chapter 2, and
Exercise 4 in
this chapter.) Here are the formulas for the four arithmetic operations
applied to
fractions:
Addition: a/b + c/d = (a*d + b*c) / (b*d)
Subtraction: a/b - c/d = (a*d - b*c) / (b*d)
Multiplication:
a/b * c/d = (a*c) / (b*d)
Division: a/b / c/d = (a*d) / (b*c)
The user should type
the first fraction, an operator, and a second fraction. The program
should then display the
result and ask whether the user wants to continue.
---------------------------------------------------------------------------------------------------------------------#include<iostream.h>
#include<conio.h>
#include<iomanip.h>
void
main(void)
{
clrscr();
int a,b,c,d;
float answer;
char choice,ch;
do
{
cout<<"Enter the first fraction:\n";
cin>>a;cout<<"/";cin>>b;cout<<endl;
cout<<"Enter the first fraction:\n";cin>>c;
cout<<"/";cin>>d;
cout<<"Enter your choice :";
cin>>choice;
switch(choice)
{
case '+':
answer=(a*d+b*c)/(b*d);
cout<<"Answer
:"<<answer;
break;
case '-':
answer=(a*d-b*c)/(b*d);
cout<<"Answer
:"<<answer;
break;
case '*':
answer=(a*c)/(b*d);
cout<<"Answer
:"<<answer;
break;
case '/':
answer=(a*d)/(b*d);
cout<<"Answer
:"<<answer;
break;
}
cout<<"\n Do another (y/n) ?";
ch=getche();
}
while(ch!='n');
}
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