Control Structures · Controlled structures may also be specified using the following flowchart symbols: Flowchart Symbols SYMBOL (a) end of an algorithm. operation.(b) (c) Flow Lines(d)

© 2013 Gilbert Ndjatou Page 57

Control Structures

The basic operations of a C/C++ programming language are specified using the following

statements or structures:

the input statement

the output statement

the assignment statements

the single-way selection structure

the two-way selection structure

the multiple-way selection structure

the counter-controlled iteration structure, and

the logically-controlled iteration structure.

The single-way selection, the two-way selection, the multiple-way selection, the counter-controlled

iteration and the logically-controlled iteration are collectively called control structures.

The specification of a controlled structure depends on a logical expression called condition.

Controlled structures may also be specified using the following flowchart symbols:

Flowchart Symbols

SYMBOL

(a)

(b)

(c)

(d)

(e)

(f)

NAME

Terminal

Input/Output

Process

Flow Lines

Decision

Connector

DESCRIPTION

Indicates the beginning or the

end of an algorithm.

Indicates an input or an output

operation.

Indicates a computation or a

data manipulation.

Used to connect the symbols

and to indicate the logic flow.

Indicates a decision point in an

algorithm.

Indicates an entry or an exit

from another part of the

flowchart.


Logical Expressions

A logical expression (or condition) is an expression that evaluates to either true or false.

There are two types of conditions:

simple conditions and

Compound conditions.

Simple Conditions

A simple condition has the following syntax:

<arithmetic expression> <relational operator> <arithmetic expression>

Relational Operators

C/C++ Symbol Meaning

< is less than

> is greater than

== is equal to

<= is less than or equal to

>= is greater than or equal to

!= is not equal to

Evaluations of simple conditions

Assuming that the variables are defined and initialized as follows:

int num1 = 5 , num2 = 7 , num3 = 2;

float fnum = 11.75;

char ch = ‘K’;

Evaluate the following conditions:

a) num1 >= 5 d) num2 + 3 == num3 * 4

b) fnum + 7.2 != fnum / 2 e) 3 * num1 + 4 < num3 * 2

c) fnum * 2 > num1 + 20 f) ‘A’ <= ch


Solutions

a) num1 >= 5 d) num2 + 3 == num3 * 4

5 >= 5 7 + 3 == 2 * 4

True 10 == 8

False

b) fnum + 7.2 != fnum / 2 e) 3 * num1 + 4 < num3 * 2

11.75 + 7.2 != 11.75 / 2 3 * 5 + 4 < 2 * 2

18.95 != 5.875 19 < 4

True False

c) fnum * 2 > num1 +20 f) ‘A’ <= ch

11.75 *2 > 5 + 20 ‘A’ <= ‘K’

13.75 > 25 True

False

Characters are ordered according to their ASCII code representations:

‘0' < ‘1' < ‘2' < . . . < ‘9' < ‘A’ < . . . < ‘Z’ < ‘a’ . . . < ‘z’.

Relational operators have a lower precedence than arithmetic operators:

o In the absence of parentheses, arithmetic operations are evaluated before any relational operation is

evaluated.

However, it is a good programming practice to use parentheses to clarify the meaning of logical

expressions.

For example: (fnum * 2) > (num1 + 20)

Compound Conditions

A compound condition is built from simple conditions using logical operators.

Logical Operators

C++ Symbol Meaning Evaluation

&&

AND

<condition1> && <condition2> is true if and only if

both conditions are true

||

OR

<condition1> || <condition2> is true if and only if at

least one of the two conditions is true

! NOT !<condition> is true if and only if <condition> is false


Evaluations of compound conditions

True || True True || False False || True False || False

True True True False

True && True True && False False && True False && False

True False False False

There is a short-circuit evaluation of a compound condition when you do not have to evaluate the

right-most condition in order to get the true value of the compound condition:

True || True and True || False Therefore True || -

True True True

Also,

False && True and False && False Therefore False && -

False False False

Exercise CS1*


int num1 = 5, num2 = 7, num3 = 2;

float fnum = 11.75;

char ch = ‘K’;

Compute the true value of each of the following compound conditions:

a) num1 >= 5 || num2 + 3 == num3 d) num2 + 3 == num3 * 4 || ch >= ‘Z’

b) ‘A’ <= ch && fnum + 7.2 != 2.5 e) num1 < 3 && num2 > num3

c) !(3 * num1 + 4 < num3 * 2) f) !(fnum * 2 < num1 + 20)


Precedence of C/C++ Operators

In the evaluation of an expression, operators with higher precedence are evaluated before those with

lower precedence.

Operators with the same precedence are evaluated in the order specified in the “order of evaluation”

column.

Exercise CS2


int num1 = 9 , num2 = 5 , num3 = 10;

float fnum = 12.50;

char ch = ‘P’;

Evaluate the following conditions (using short circuit evaluation whenever possible):

a) 2 * num1 - 5 >= 9 || fnum / 2 + 10 <= 6.5

b) num1 + num2 == num3 + 5 && 2 * num3 <= 4 * num2

c) ! (num1 + 5 <= 13)

d) num1 + 5 = = 24 || num1 – 1 > num2 – 5

e) 2 * num1 - 5 >= 9 && fnum / 2 + 10 <= 6.5

f) num1 - num2 <= num1 / 3 || num1 * num2 > 100

g) num1 - 5 >= num3 || num2 < 15 && num1 >= 9

Operator Order of Evaluation Precedence

!

Unary –

right to left

7

*

/

%

left to right

6

+

-

left to right

5

<

<=

>

<=

left to right

4

==

!=

left to right

3

&& left to right 2

|| left to right 1


Accuracy of Floating-Point Values

The fact that floating-point values are approximated inside a computer makes it difficult to test for

the equality of floating-point values.

For example, if the variable fnum is defined as follows:

float fnum = 7.1;

Then, the condition: fnum / 2 == 3.55 may not be true.

The problem may be solved by assuming that two floating-point values are equal if their difference

is relatively very small.

This is done by testing if the absolute value of their difference is less than a certain value chosen by

the programmer (for example 0.000001).

Using the library function fabs() that takes as argument a floating-point value and returns its

absolute value, the condition:

value1 == value2

is replaced with the condition: fabs(value1 - value2) < 0.000001

which tests whether the difference between the two values is small enough so that we can make them

equal.

Evaluation of True and False

In the C/C++ programming language,

false is represented by the integer value 0, and

any value other than 0 represents true.

Therefore, the following two conditions are equivalent (have the same true value):

o Conditions value != 0 and value (value is not zero means that it is true)

o Conditions value == 0 and ! value (value is zero means that !value is true)

bool Data Type

A variable with data type bool can only hold the bool values true and false.

Examples:

bool flag = true;

bool condValue = false, HighTemperature, ExtremeTemperature;

double temperature;

HighTemperature = false

cin >> temperature;

HighTemperature = (temperature >= 120)

ExtremeTemperature = (HighTemperature || (temperature <= -20);


Exercise CS3

1. Which of the following conditions are equivalent (that means have the same true value)?

a) num != 0 d) !num

b) num == 0 e) !(num == 0)

c) num

2. Suppose that a person’s age is stored in the variable age, his number of children in the variable

numChild, his salary in the variable salary, and his height in the variable height. Write the relational

expressions to specify the following conditions:

a. he is 45 year old.

b. he is more than 5.7 feet tall.

c. his salary is between 35,000 and 50,000.

d. he does not have 3 children.

e. he is either 6.0 feet tall, or he has less than 4 children.

f. he is not older than 35 and he has 2 or 3 children


Two-Way Selection using the if-else Structure

The solutions of some problems require the CPU to test a condition and to select the action to be

performed based on whether or not that condition is true.

Example:

Suppose that at a party, you serve alcohol drinks only to guests who are 21 or older, and a juice to

guests who are less than 21.

When a guest comes to get a drink, you first test his age: if he is 21 or older, you serve him an

alcohol drink; otherwise you serve him a juice.

The solutions of these types of problems are written using a two-way selection structure which is

specified using a flowchart as follows:

False True

It says to do the following:

Test condition

a. if it is true, perform the actions (T-Action) specified by T-Statement,

b. otherwise, perform the actions (F-Action) specified by F-Statement.

Next-Statement is the statement that says what to do after any of the actions (F-Action or T-Action)

above has been executed.

A two-way selection structure is specified in C++ using the if - else structure as follows:

if (<condition>)

<T-statement>

else

<F-statement>

<Next-Statement>

Condition

F-Statement T-Statement

Next-Statement


<T-statement> is a single statement that specifies the T-Action

<F-statement> is a single statement that specifies the <F-Action>.

Case Study CS1

Problem Statement

Write a program to read the age of an individual, and to output the message “serve alcohol drink” if he is

21 or older, and the message “serve juice” otherwise. Your program will then output the message

“thank you for using this program.”

Program Logic

output: “Serve alcohol” or “Serve Juice”, depending on the individual’s age.

Input: an individual’s age.

Variables: age (int) to hold an individual’s age.

Algorithm Specification (using a Flowchart)

False True

start

Read

age

age >= 21

Write “Thank you for

using this program”

Stop

Write “serve alcohol drink”

Write

“serve a juice”


Figure CS1 Using the if-else Structure

/*************************************************************************

Program to read the age of an individual and to output the type of drink that he

should be served

*************************************************************************/

#include <iostream>

using namespace std;

#define DRINKAGE 21

int main()

{

int age; // to hold an individual’s age

/*----------------- read the individual’s age------------------------*/

cout << “\n\nEnter the individual’s age please:\t”;

cin >> age;

/*--------determine what type of drink he should be served ---------*/

if (age >= DRINKAGE) // he is over the drinking age

cout << endl << “Serve alcohol drink”;

else // he can not drink alcohol

cout << endl << “Serve juice”;

cout << endl << “Thank you for using this program”;

return (0);

}

Case Study CS2

Problem Statement

Write a program to read a positive integer value and to determine if it is even or odd. If it is even,

print it with the message “EVEN”; otherwise, print it with the message “ODD”.

Program Logic

Output: the input value with the message “EVEN” or “ODD”, depending on whether the input value

is even or odd.

Input: an integer value.

Variable: num (int) to hold the input value.

Note: An integer value is even if the remainder in its division by 2 is 0; otherwise, it is odd.



False True

Figure CS2 Using the if - else Structure

/************************************************************************

Program to read an integer value and to determine if it is even or odd

************************************************************************/

#include <iostream>


int main()

{

int num; // to hold the value read

/*---------------- read in an integer value--------------------------*/

cout << “\n\nEnter an integer value please:\t”;

cin >> num;

/*----------------determine if it is even or odd --------------------*/

if (num %2 == 0) // it is even

cout << endl << num << “\tEVEN”;

else // it is odd

cout << endl << num << “\tODD”;

return ( 0 );

}

Exercise CS4

Write a C++ code segment to read an integer value and to determine if it is a multiple of 5. If it is a

multiple of 5, print it with the message “MULTIPLE OF 5”, otherwise, print it with the message NOT

MULTIPLE OF 5.”

Start

Read

num

num % 2 = 0

Write

num, “odd”

Write

num, “even”

Stop


Exercise CS5

Write a C++ code segment to read a character and to determine if it is a letter of the alphabet. If it is a

letter of the alphabet, print it with the message, “it is a letter” otherwise, print it with the message, “it is

not a letter.”

Compound Statement

A compound statement has the following syntax:

{

(One or more statements to be executed as a block)

}

Example

{

num2 = num1 + 5;

num1 = num1 +1;

cout << endl << “num1=\t” << num1 << “\n num2=\t” << num2;

}

The statements of a compound statement are executed one after another, starting with the first one in

the sequence.

A compound statement may appear anywhere in a C++ program where a single statement may

appear.

If you have to specify two or more statements as the T-statement or the F-statement of an if-else

structure, these statements must be specified as a compound statement.

Case study CS3

Problem Statement

Write a program to read an integer value and to do the following:

If the value read is zero, print it with the message “INVALID DIVISOR”.

Otherwise read a second value and then compute and print the quotient and the remainder in the

division of the second value by the first.

At the end write the message “Thank you for using this program.”


Program Logic

Output: the quotient and the remainder in the division of the second value by the first or the

message “0 is invalid divisor” depending on the input.

Input: one or two integer values, depending on the value of the first.

Variables:

divisor (int) to hold the first value.

dividend (int) to hold the second value.

Note:

The second value is input and the quotient and the remainder in the division of the second value

by the first are computed only if the first value is not zero.


False True

Start

Read

divisor

divisor == 0

Write

divisor

“invalid divisor”

Read

dividend

Write

dividend / divisor

dividend % divisor

Write

“Thank you”

Stop


Figure CS3 Using a Compound Statement in an if-else Structure

/*************************************************************************

Program to read two integer values and to compute the quotient and the remainder in

the division of the second value by the first.

*************************************************************************/

#include <iostream>


int main()

{

int divisor, // to hold the first value

dividend; // to hold the second value

/*---------------------- read in the divisor-----------------------*/

cout << “\n\nEnter the divisor please:\t”;

cin >> divisor;

/*-----read the dividend and compute the quotient and the remainder--*/

if (divisor == 0) // it is invalid

cout << endl << divisor << “\tIS INVALID DIVISOR”;

else

{

cout << “\n\nEnter the dividend please:\t”;

cin >> dividend;

cout << “\nThe quotient in the division of:\t”

<< dividend << “ by ” << divisor << “\tis:\t”

<< (dividend / divisor);

cout << “\nand the remainder is:\t” << (dividend % divisor);

}

cout << “\n\nThank you for using this program”;

return( 0 );

}

Exercise CS6*

Assuming that all variables are properly defined and initialized, what are the error(s) in each of the

following program segments?

a. cin >> num;

if (num = 5)

sum = num + 4;

else

sum = num + 10;

b. cin >> num;

if (num > 5 && <= 10)

cout << (num + 15);

else

cout << (num - 3);


c. cin >> num;

if (num < 15)

result1 = 2 * num;

result2 = num + 20;

else

result1 = 5 * num;

result = result1 + result2;

cout << “\nThe result is:\t” << result;

Exercise CS7*

Trace the execution of the following program segment and show its output for each of the following

input values:

a. input: 4 b. input: 20

Line # Statements

1 cin >> num;

2 if (num <10 )

{

3 num = num + 6;

4 cout << endl << “num =\t” << num;

}

else

5 cout << endl << “num / 4 =\t” << (num 4);

6 cout << endl << “result =\t” << (2 * num);

Exercise CS8

Write a C++ code segment to read an integer value into variable num1and another integer value into

variable num2 and to do the following: if the value in variable num1 is greater than the value in variable

num2 do the following: add 1 to the value in num1 and 5 to the value in num2 and then print the result

of the product of the new value in num1 and the new value in num2; otherwise, do the following:

subtract 1from the value in num1 and 5 from the value in num2 and then print the result of the product of

the new value in num1 and the new value in num2.

Exercise CS9

A department store offers a rebate of 5% if the total amount of the purchase is at least 100 and a rebate

of 2% otherwise. A sale tax of 8.25 % is also imposed on the price of each purchase (after the rebate).

Write a C++ code segment to read the unit price and the number of items purchased, and to output the

amount of the rebate and the price (including taxes) of the purchase.


Counter-Controlled Iteration using the while Structure

The solutions of some problems require the CPU to repeat the execution of one or more statements a

certain number of times.

Example:

Suppose that a candy bar costs 70 cents in a candy machine that only accepts dimes. To buy a candy

bar from this machine, a user has to deposit a dime in the machine seven times.

The solutions of these types of problems are written using a counter-controlled iteration (or loop)

which is specified using a flowchart as follows:

False True

Body-of- the- Loop consists of the statements whose executions are repeated by the CPU.

counter is called the loop counter

it is a variable that is used to hold the number of repetitions

Next-Statement is the first statement to be executed after the repetitions.

Initial/Final Value

If the initial value is 0 (no execution is done yet), then the final value must be the number of

executions.

If the initial value is 1 (one execution is about to be done), then the final value must be the number

of execution plus 1.

Loop Increment

After each execution of the body-of-the-loop, the loop counter must be updated to reflect the number

of executions.

The statement that you use to update the loop counter is called the loop increment.

The loop increment is a statement in the body-of-the-loop.

counter = initial-value

Counter < final-value

Body-of-the-Loop

Next-Statement


Algorithm of the Candy Bar Machine (using a flowchart)

False True

Specification of a Counter-Controlled Iteration in C++

A counter-controlled iteration may be specified in C++ using the while structure as follows:

counter = initial-value;

while (counter < final-value)

{

<Body-of-of the-loop (including the increment-statement)>

}

<Next-statement>

Start

count = 0

count < 7

Read

coin

count = count + 1 Write

“Thank you”

Stop

total = 0

total = total + coin


Algorithm for the Candy Bar Machine Problem (in C++)

int total, // to hold the total value of the coins

coin, // to hold the value of a coin

count; // to hold the number of repetitions

total = 0;

count = 0;

while ( count < 7 )

{

cin >> coin;

total += coin;

count ++;

}

cout << endl << “Thank You”;

Case Study CS4

Problem Statement

Write a program to read 30 weight measurements in pounds and to convert them into kilograms. Note

that 1 Lb = .454 Kgs.

Program Logic

Input: 30 weight measurements (in Lbs).

Output: 30 weight measurements (in Kgs).

Variables:

count (int) to count the weight measurements: Initial value is 0 - Final value is 30.

pound (double) to hold a weight measurement in Lbs.


Algorithm Specification (using a flowchart)

False True

Figure CS4 Counter-Controlled Iteration using the while Structure

Line Number

1 /*************************************************************

2 Program to read 30 weight measurements in pounds and convert

3 them into kilograms

4 *************************************************************/

5 #include <iostream.h>

6 #include <iomanip.h>

7 #define MAXCOUNT 30

8 #define COEFICIENT .454

9

10 int main()

11 {

12 int count; // to count weight measurements

13 double pound, // the current weight measurement in Lbs

14

Start

count = 0

count < 30

Read

pound

Write

.454 * pound

count = count + 1

Write

“Thank you”

Stop


15 cout << fixed << showpoint << setprecision(2);

16

17 /*read all weight measurements (Lbs) and convert them to Kgs*/

18 count = 0; // no weight measurement is read so far

19 while (count < MAXCOUNT) // repeat thirty times

20 {

21 cout << “\nEnter a weight measurement please:\t”;

22 cin >> pound;

23 cout << “\t = ” << (.454 * pound);

24 count ++;

25 }

26

27 cout << endl << “Thank you”;

28 return (0);

29 }

The body of the loop consists of the statements in line 21 to line 24.

The loop-counter is initialized in line 18; and the loop-increment statement is in line 24.

Exercise CS10*

Assuming that all variables are properly defined, find the mistake(s) in the specification of each of the

following while structures:

a. b. c.

while( count < 10 ) while(count < 10) count = 0;

{ { while(count < 10 )

cin >> num; count = 0; {

cout << 2 * num; cin >> num; cin >> num;

count = count + 1; cout << 2 * num; cout << 2 * num;

} count = count + 1; }

}

Exercise CS11

Write a C++ code segment to read 50 temperature values in Fahrenheit and to convert them to Celsius.

You convert a Fahrenheit temperature to Celsius by using the following formula: Celsius = 5.0 / 9 *

(Fahrenheit - 32).


Using a Running Total

A running total is a variable that is used to compute the sum of the values processed at ach iteration of

a loop: it is first initialized to 0 and each new value processed in the loop is added to the previous total.

Case Study CS5

Problem Statement Write a program to read 20 integer values and to compute their sum.

Program Logic

Input: 20 integer values.

Output: their sum.

Variable:

count (int) to count the values: Initial value is 0 (no value is read so far); Final value is 20.

value (int) to hold the integer value read.

totalValue (int) to hold the sum of the integer values read so far: Initial value is 0.


False True

Start

totalValue = 0

count = 0

count < 20

totalValue = totalValue + value

Read

value

count = count + 1 Write

totalValue

Stop



Line Number

1 /*************************************************************

2 Program to read twenty integer values and to compute their sum.

3 *************************************************************/

4 #include <iostream>

5 using namespace std;


7

8 int main()

9 {

10 int count; // to count values

11 value, // to hold the value read

12 totalValue; // the hold the sum of the values read so far

13

14 /*------read all values and compute their sum-------*/

15 totalValue = 0;

15 count = 0; // no value has been read so far

16 while (count < MAXCOUNT) // repeat twenty times

17 {

18 cout << “\nEnter an integer value please:\t”;

19 cin >> value;

20 totalValue += value;

21 count++;

22 }

23

24 /*-----print the sum of all the values read----------*/

25 cout << “\n\nThe sum of all the values read is:\t”

26 << totalValue;

27 return (0);

28 }

Exercise CS12*

Each of the following code segments is supposed to read 10 integer values and to compute their sum.

What is wrong in each of these code segments?

a. b.

count = 0; count = 0;

while (count < 10 ) while (count < 10 )

{ {

sum = 0; cin>>num;

cin >> num; sum = sum + num;

sum = sum + num; count = count + 1;

count = count + 1; }

} cout << endl << sum;

cout << endl << sum;


Exercise CS13*

Write a C++ code segment to read the test scores of twenty students and to compute and print their

average.

Exercise CS14

A client has purchased 20 products in a store. Write a C++ code segment to read the unit price and the

number of items of each product and to compute and print the total price of all these products.

Case Study CS6

Problem Statement

Write a program to compute the sum of the first 20 positive even integer values (2, 4, . . .).

Program Logic

Input: none.

Output: the sum of the first 20 positive integer values.

Variable:

count (int) to count the values: Initial value is 1; Final value is 21.

totalValue (int) to hold the sum of the integer values processed so far: Initial value is 0.


False True

Start

totalValue = 0

count = 1

count < 21

totalValue = totalValue + 2 * count

count = count + 1

Write

totalValue

Stop



Line Number

1 /*************************************************************

2 Program to read twenty integer values and to compute their sum.

3 *************************************************************/




7

8 int main()

9 {

10 int count; // to count the values

11 totalValue; // the hold the sum of the values so far

12

13 totalValue = 0;

14 count = 1;

15 while (count < MAXCOUNT) // repeat twenty times

16 {

17 totalValue += 2 * count;

18 count ++;

19 }

20

22 /*-----print the sum of the first 20 positive even values ------*/

23 cout << “\n\nThe sum of the first 20 positive even values is:\t”

24 << totalValue;

25 return (0);

26 }

Exercise CS15

Write a C++ code segment to add the first 20 positive multiples of 5 ( 5 + 10 + 15 + 20 + . . . ) and to

output the result.

Using a Running Product and a Counter with the initial value Greater

than the final Value

A running product is a variable that is used to compute the product of values processed at each

iteration of a loop: it is first initialized to 1 and each new value processed in the loop is multiplied to it.

Case Study CS7

Problem Statement

For a positive integer value n, n factorial (written n!) is the product of all the integer values from 1

to n (1 * 2 * 3 * . . . * n).

Write a program to read a positive integer value n, and to compute and print n!


Program Logic

Input: a positive integer value.

Output: the product of all positive integer values less than or equal to the value read.

Variables:

number (int) to hold the value read

factorial (int) to hold the product of the integer values: initial value = 1

After

Iteration Value of number value of factorial

- n 1

1 n - 1 1 x n

2 n - 2 1 x n x (n - 1)

.

.

.

n - 1 1 1 x n x (n - 1) x . . . x 2


False True

Start

Read

number

factorial = 1

number > 1

Factorial = factorial * number

number = number - 1

Write

factorial

Stop



Line Number

1 /*************************************************************

2 Program to read a positive integer value and to compute and print its

factorial.

3 *************************************************************/



6 int main()

7 {

8 int number, //integer value/loop counter

9 factorial = 1; // to hold the product

10

11 /*------------read the integer value --------------------*/

12 cout << endl << “enter a positive integer value:\t”;

13 cin >> number;

14

15 /*-------compute the factorial of the number read -------*/

16 while (number > 1)

17 {

18 factorial *= number;

19 number --;

20 }

21

22 /*------------------print the factorial------------------*/

23 cout << endl << “n factorial is:\t”

24 << factorial;

25 return (0);

26 }

The body of the loop consists of the statements in lines 18 and 19.

The loop-counter is initialized in line 13; and the loop-increment statement is in line 19.

Exercise CS16

Write a C++ code segment to generate the squares (n*n) of the integer values 20, 18, 16, 14, . . . , 2,

starting with 20.

Exercise CS17

Write a C++ code segment to read ten integer values and to compute and print their product.


Logically-Controlled Iteration using the while Structure

The solutions of a class of problems require the CPU to repeat the execution of one or more statements

as long as a certain condition is true (that means, until a certain condition becomes false).

Case Study CS8

Problem Statement

A can of soda costs $ 1.00 in a soda machine, and a user buys a can of soda by depositing coins (nickels,

dimes or quarters) in the machine until the total value of the coins deposited in the machine is greater

than or equal to $ 1.00.

Write a program to read the value of each coin deposited in the soda machine (5, 10, or 25) until the

total value of the coins deposited in the machine is greater than or equal to 100. The program will then

output the message “Thank you for using the soda machine” and the amount of the change.

Program Logic

input: the value of a coin (5, 10 or 25).

Output: the message “Thank you for using this soda machine” and the amount of change.

Note:

To read the value of each coin deposited in the machine and to compute their sum, we need the

following:

A variable coinValue to hold the input value of a coin.

A variable totalValue to hold the sum of the values of the coins read so far.

It is initialized to 0 before the body-of-the-loop.

The loop condition is: totalValue < 100.

Variables:

coinValue ( int ) to hold the value of a coin.

totalValue ( int ) to hold the total value of the coins deposited so far.



False True

Start

totalValue = 0

totalValue < 100

Read

coinValue

totalValue = totalValue + coinValue

Write

“Thank you”

Write

totalValue - 100

Stop


Figure CS8 Logically-Controlled Iteration using the while Structure

Line Number

1 /*****************************************************************

2 Program to monitor the purchase of a can of soda in a soda machine

3 *****************************************************************/



6 int main()

7 {

8 const int SODAPRICE = 100; // the price of a can of soda

9 int coinValue, // the value of a coin

10 totalValue = 0; // the total value of the coins

11 while( totalValue < SODAPRICE )

12 {

13 /*------------read the value of the next coin------------*/

14 cout << endl << “enter the value of the next coin:\t”;

15 cin >> coinValue;

16

17 /* compute the total value of the coins deposited so far */

18 totalValue = totalValue + coinValue;

19 }

20

21 /*--- say “thank you” and give the change back to the user */

22 cout << “\n\nThank you for using this soda machine”

23 << “\n\nYour change is:\t” << (totalValue - SODAPRICE);

24 return ( 0 );

25 }

Exercise CS18*

The manager of a store is required to make a deposit of $ 10,000 whenever the sum of the daily sales in

that store is $ 10,000 or more. Write a C++ code segment to read the daily sales in that store, and to

compute their sum until it becomes greater than or equal to $ 10,000. The sum of the daily sales and the

message “it is time to make a deposit” are then printed.

Exercise CS19

A store has 120 bags of coffee and every day, the amount of bags sold is deducted from the current stock

until it becomes less than or equal to 30. The manager of the store must then place a new order.

Write a C++ code segment to read the number of bags sold daily and to update the current stock until it

becomes less than or equal to 30. The message “place a new order” and the amount of bags to purchase

in order to have 120 bags are then printed.


while Structure with one Statement in the Body-of- the-Loop

When the body-of-the-loop consists of just one statement, the braces may be omitted from the while

structure.

Case Study CS9

Problem Statement Write a program to compute the smallest power of 5 greater than 10000 (1, 5,

25, 125, . . . ).

Program Logic

Input: none.

Output: the smallest power of 5 greater than 10000.

Note:

To compute the smallest power of 5 greater than 10000, we need:

A variable power to hold the powers of 5. It is initialized to 1 before the body-of-the-loop.

The current value of variable power is multiplied by 5 in the body-of-the-loop until it becomes

greater than 10000.

The loop condition is therefore: power <= 10000.

After Value of

Iteration variable power

- 1

1 1 x 5 = 5

2 5 x 5 = 25

3 25 x 5 = 125

. . .

Variable: power (int) to hold the powers of 5.

©2013 Gilbert Ndjatou Page 87


False True

Figure CS9 while Structure with a single statement in the body-of-the-Loop

Line Number

1 /***************************************************************

2 Program to compute the smallest power of 5 greater than 10000

3 ****************************************************************/



6 int main()

7 {

8 int power; // to hold the current power of 5

9 power = 1; // first power of 5

10 while ( power <= 10000)

11 Power = power * 5;

12

13 /*-----------------print the result-------------------------*/

14 cout << endl << power

15 << “ is the smallest power of 5 greater than 10000";

16 return (0);

17 }

Although the body-of-the-loop consists of just one statement, it could also be specified as a

compound statement as follows:

while (power <= 10000)

{

power = power * 5;

}

Start

power = 1

power <= 10000

Power = power * 5

Stop


Sentinel-Controlled Iteration

When you want to process a list of values and you do not know how many values are in the list, you

can use a sentinel value (dummy value, flag value, or signal value) to mark the end of the list.

A sentinel value is in general chosen in such a way that they can not possibly be one of the data

items in the list.

Example:

For a program to read and process a list of test scores received by students in an exam, a sentinel

value might be -1 or -99. With -99, the list could be:

87 69 95 65 73 86 93 68 -99

The data items in the list are accessed and processed one at a time until the sentinel value is

accessed.

However, the sentinel value must not be processed.

The algorithm to perform this type of repetition is specified using a flowchart as follows:

False True

Read/Access the first data item

data item != sentinel

Process a data item

Read/Access the next data item

Next-Statement


It is specified using the while structure as follows:

<statement-to read/access-the first-value>

While(<value-read/access != sentinel-value>)

{

<processing-statements>

<statement-to-read/access-the-next-value>

}

<processing-statements> consists of one or more statements used to process a data item in the

list.

Case Study CS10

Problem Statement

Write a program to read one or more weight measurements in pounds and to convert them into

kilograms. The dummy value -99.0 is entered to end the input of the weight measurements. Note that 1

Lb = .454 Kgs.

Program Logic

Input: one or more weight measurement (in Lbs) followed by the sentinel value -99.00.

Output: weight measurements in (Kgs).

Variables: pound (double) to hold a weight measurement in Lbs.



False True

Figure CS10 Sentinel-Controlled Iteration using the while Structure

Line Number

1 /*************************************************************

2 Program to read one or more weight measurements in pounds and

3 to convert them into kilograms.

4 *************************************************************/


6 #include <iomanip>


8 #define COEFICIENT .454

9 #define DUMMYWEIGHT -99.0

10

11 int main()

12 {

13 double pound, // the current weight measurement in Lbs

14

15 cout << fixed << showpoint) << setprecision(2);

16

Start

Read

pound

pound != -99

Read

pound

Write

“Thank you”

Stop

Write

.454 * pound


17 /*read the first weight measurement or the dummy value -99.0*/

18 cout << endl

<< “Enter a weight measurement or -99.0 to terminate:\t”;

19 cin >> pound;

20 while (pound != DUMMYWEIGHT)

21 {

22 cout << “\t = ” << (.454 * pound);

23 cout << endl

<< “Enter a weight measurement or -99.0 to terminate:\t”;

24 cin >> pound;

25 }

26 cout << endl << “Thank you”;

27 return (0);

28 }

Exercise CS20*

The following program that should read a list of integer values and compute their sum until the sentinel

value -99 is read, contains some error(s). Rewrite it with these errors corrected.

#include <iostream>


#defineDUMMYVALUE -99

int main( )

{

int total, value;

cout << “\nEnter all values to add followed by”

<< DUMMYVALUE <<“ to stop\n”;

while ( value != DUMMYVALUE )

{

cin >> value;

total = total + value;

}

cout << endl << “The total of these value is:\t” <<total;

return( 0 );

}

Exercise CS21*

Write a C++ code segment to read one or more temperature values in Fahrenheit and to convert them to

Celsius. The sentinel value -99 marks the end of the input temperatures. You convert a Fahrenheit

temperature to Celsius by using the following formula: Celsius = 5.0 / 9 *(Fahrenheit - 32).

Exercise CS22

Write a C++ code segment to read the pay rate and the number of hours of work of one or more

employees and to compute each employee’s gross pay: pay rate * number of hours. The dummy pay

rate -99.00 marks the end of the input values.


Exercise CS23

Write a C++ code segment to read the test scores of one or more students (until the sentinel value -99 is

read) and to compute their average.

One-Way Selection using the if Structure

A one-way selection structure is used to specify that the execution of one or more actions be

performed only if a test performed on a condition is true.

It is specified using a flowchart as follows:

False True

C-Statement is one or more statements that specify the operations to be performed when

condition evaluates to true.

It is specified in C++ language using the if structure as follows:

if (<condition>)

<Statement>

<Next-Statement>

where <Statement> is a single statement or a compound statement.

Note that when <condition> evaluates to false, no action is performed. Instead, the execution of

the program continues with the execution of <Next-Statement>.

condition

C-Statement

Next-Statement


Case Study CS11

Problem Statement

A rebate of 5% is given in a store only if the price of the merchandise is at least $100.00. Write a

program to read the unit price and the number of quantity of the merchandise, and to compute the

amount of the purchase after the rebate if applicable.

Program Logic

Input: the unit price and the quantity of the merchandise.

Output: the total amount of purchase.

Variables:

unitPrice ( double ) to hold the unit price.

quantity ( int ) to hold the quantity.

purchase ( double ) to hold the amount of purchase.


False True

Start

Read

unitPrice, quantity

purchase = unitPrice * quantity

purchase >= 100.00

purchase = purchase - .05 * purchase

Write

purchase

Stop


Figure CS11 One-Way Selection using the if Structure

Line Number

1 /****************************************************************

2 Program to compute the total amount of purchase in a store

3 ****************************************************************/



6 int main()

7 {

8 const double MINPURCHASE = 100.00; //minimum purchase

9 const double RRATE = 0.05; // rebate rate

10 double unitPrice, // unit price

11 Purchase // amount of purchase

12 int quantity; // number of quantity

13

14 /*--------read in the unit price and the quantity -------*/

15 cout << “\nEnter the merchandise unit price:\t”;

16 cin >> unitPrice;

17 cout << ‘\nEnter its quantity:\t”;

18 cin >> quantity;

19

20 /*-----------compute the amount of purchase --------------*/

21 purchase = unitPrice * quantity;

22 if(purchase >= MINPURCHASE)

23 purchase = purchase - RRATE * purchase;

24

25 /*-------------- print the amount of purchase --------------*/

26 cout << “\nYour total amount of purchase is:\t”

27 << purchase;

28 return( 0 );

29 }

The Null Statement

The null statement is a statement that consists of just the semicolon.

It says to do nothing, but to continue with the next operation.

You may also use the if-else structure and the null statement to specify a one-way selection as

follows:

if (<condition>)

<Statement>

else;

<Next-Statement>


Example

The if structure in lines 22 and 23 of the program in Figure 3.19 could be written as follows:

if (purchase >= 100.00 )

purchase = purchase - RRATE * purchase;

else;

/*-------------- print the amount of purchase --------------*/

cout << “\nYour total amount of purchase is:\t”

<< purchase;

Exercise CS24*

What is the output of the following program segment for each of the following input values?

a. input value: 4 b. input value: 20

cin >> num;

if( num >= 5)

{

pnum = num + 7;

cout << endl << “pnum=\t” << pnum;

}

cout << “result=\t” << (num - 1);

Exercise CS25*

greatt, smallt, and somet are integer variables with initial values. Write a C++ code segment to read an

integer value and to do the following: add 1 to smallt if the value read is less than 5; then add 2 to greatt

if the value read is greater than 20, otherwise subtract 3 from smallt. Then, regardless of the value read

add 7 to somet.

Exercise CS26

In a department store, a 10% rebate is given for every purchase of at least $ 20.00. Write a C++ code

segment to read the unit price of a product, the quantity purchased, and to compute the amount of the

purchase after the rebate if applicable.

Exercise CS27

sum, large_ct, middle_ct, and small_ct are integer variables with initial values. Write a C++ code

segment to carry out the following tasks: read a value into the integer variable num. if the value read is

greater than 100, add 1 to variable large_ct; if it is less than 50, add 2 to variable small_ct; otherwise,

add 3 to variable middle_ct; regardless of the value read, add it to variable sum.


Multiple-Way Selection using if-else Structures

A multiple-way selection structure is used to specify that the action to be performed be chosen from

a list of three or more actions based on the true values of two or more conditions.

A multiple-way selection with four actions and three conditions is specified as follows:

True False

True False

True False

Condition1 is first tested, and if it is true, Action-1 is executed, followed by Next-Action;

Otherwise, condition2 is tested, and if it is true, Action-2 is executed, followed by Next-Action;

Otherwise, condition3 is tested, and if its true, Action-3 is executed, followed by Next-Action;

Otherwise, Action-4 is executed, followed by Next-Action.

Note that only one of the actions, Action-1, Action-2, Action-3, or Action-4 is executed.

In some situations, Action-4 may also be the null action.

condition1

condition2

condition3

Action-1

Action-2

Action-3 Action-4

Next-Action


This behavior is specified in C/C++ using if-else structures as follows:

if ( <condition1> )

<Statement-1>

else if ( <condition2> )

<Statement-2>


<Statement-3>

else

<Statement-4>

<Next-Statement>

<Statement-1>, <Statement-2>, <Statement-3>, and <Statement-4> are single or compound

statements that specify respectively, Action-1 , Action-2 , Action-3 , and Action-4 .

<Next-Statement> is a statement that specifies the Next-Action.

In the case that <Statement-4> is the null statement, this behavior may also be specified as follows:

if ( <condition1> )

<Statement-1>


<Statement-2>


<Statement-3>

<Next-Statement>

Case Study CS12

Problem Statement

Four letters M, S, D, and W are used to encode the marital status of an individual as follows: M (married), S

(single), D (divorced), and W (widowed). Any other character is considered invalid. Write a program to read a

marital code and to output the corresponding marital status.

Program Logic

Input: a marital code (capital M, S, D, or W).

Output: a marital status. Any other character is considered invalid.

Variables:

mcode (char ) to hold the marital code.



True False

True False

True False

True False

Figure CS12 Multiple-Way Selection using if-else Structures

Line Number

1 /****************************************************************

2 Program to read a marital code and to output the corresponding

3 marital status.

4 ****************************************************************/



mcode = ‘M’

mcode = ‘S’

mcode = ‘D’

Start

Read

mcode

Write

“married”

Write

“single”

Mcode = ‘W’ Write

“divorced”

Stop

Write

“invalid

”

Write

“Widowed”


7 int main()

8 {

9 char mcode; // to hold a marital code

10

11 /*------------------read in a marital code------------------*/

12 cout << “\nEnter a marital code:\t”;

13 cin >> mcode;

14

15 /*---determine and print the corresponding marital status --*/

16 if (mcode == ‘M’)

17 cout << “\nThis individual is married”;

18 else if (mcode == ‘S’)

19 cout << “\nThis individual is single”;

20 else if (mcode == ‘D’)

21 cout << “\nThis individual is divorced”;

22 else if (mcode == ‘W’)

23 cout << “\nThis individual is a widow”;

24 else

25 cout << “\nThe marital code is invalid”;

26 return( 0 );

27 }

Exercise CS28*

A florist who sells four different types of bouquets of flowers has numbered them 101, 202, 303, and

404 with their unit prices specified as follows:

ITEM NUMBER UNIT PRICE

101 $ 5.25

202 $ 3.10

303 $ 9.75

404 $ 6.50

Write a C++ code segment to read an item number and the number of bouquets of that type purchased,

and to compute and print the price of those bouquets.

Exercise CS29

A game consists of selecting a positive integer value and rewarding bonuses to the player according to

the remainder in the division of that number by 5 as follows:

REMAINDER BONUS

0 200

1 350

2 400

3 500

4 750

Write a C++ code segment to read a positive integer value and to output the bonus received by the user.


Case Study CS13

Problem Statement

Write a program to read a test score and to determine and print the corresponding letter grade as follows:

A ( score >= 90.00) B (80.00 <= score < 90.00) C (70.00 <= score <80.00)

D (60.00 <= score < 70.00), F (score < 60.00).

It is assumed the input value is a valid test score.

Program Logic

Input: a test score (floating-point value in the range 0.00 to 100.00).

Output: the corresponding letter grade.

Variables: score (double ) to hold a test score.

Algorithm Specification Using a Flowchart

True False

True False

True False

True False

score >= 90.00

score >= 80.00

Score >= 70.00

Start

Read

score

Write

‘A’

Write

‘B’

score >= 60.00 Write

‘C’

Stop

Write

‘F’ Write

‘D’


Figure CS13 Multiple-Way Selection using if-else structures

Line Number

1 /****************************************************************

2 Program to read a test score and to determine and output the

3 corresponding letter grade.

4 ****************************************************************/



7 int main()

8 {

9 double score; // to hold a test score

10

11 /*------------------read in a test score--------------------*/

12 cout << “\nEnter a test score:\t”;

13 cin >> score;

14

15 /*----determine and print the corresponding letter grade ---*/

16 if ( score >= 90.00)

17 cout << “\nThe letter grade is:\t” << ‘A’;

18 else if (score >= 80.00)

19 cout << “\nThe letter grade is:\t” << ‘B’;


21 cout << “\nThe letter grade is:\t” << ‘C’;


23 cout << “\nThe letter grade is:\t” << ‘D’;

24 else

25 cout << “\nThe letter grade is:\t” << ‘F’;

26 return ( 0 );

26 }

The multiple-way selection specified in line 16 to line 25 could also be written as follows:

if ( score < 60.00)

cout << “\nThe corresponding letter grade is:\t” << ‘F’;

else if (score < 70.00)

cout << “\nThe corresponding letter grade is:\t” << ‘D’;


cout << “\nThe corresponding letter grade is:\t” << ‘C’;


cout << “\nThe corresponding letter grade is:\t” << ‘B’;

else

cout << “\nThe corresponding letter grade is:\t” << ‘A’;


Exercise CS30*

What is the output of the following program segment for each of the following input values?

a. input value: 3 b. input value: 7 c. input value: 12

cin >> num;

if (num < 5)

cout << "Red\n";

else if (num < 10)

cout << "Yellow\n";

else

cout << "Green\n";

cout << “blue”

Exercise CS31*

The decision table below shows fines imposed for speeding violations. Write a C++ code segment to

read the speed of a car and to output the corresponding fine.

Speed <= 50 mph Fine = $ 0.00

50 < Speed <= 70 Fine = $ 15.00

70 < Speed <= 80 Fine = $ 30.00

Speed > 80 Fine = $ 60.00

Exercise CS32

A department store offers rebates to customers according to the price of the items purchased as follows:

Price >= $ 100.00 Rebate = $ 10.00

$ 50 <= Price < $ 100.00 Rebate = $ 5.00

$ 20 <= price < $ 50.00 Rebate = $ 2.00

price < $ 20.00 No rebate

A sale tax of 8.25% is also imposed on the price of each purchase (after the rebate, if any).

Write a C++ code segment to read the price of a product and to output the rebate received by the

customer, the price (including the rebate), and the amount due (including taxes).


Iterations using the for Structure

The for structure has the following general syntax:

for(<init-expression-list> ; <loop-condition> ; <increment-expression-list>)

{

<body-of-the-loop>

}

It is executed in the same way that the following while structure is executed:

<init-expression-Statements>

while ( <loop-condition> )

{

<body-of-the-loop>

<increment-expression-Statements>

}

Example

1. The following counter-controlled iteration (specified using the while structure) may be specified

using the for structure in any of the following ways in (a), (b), and ( c ):

totalscore = 0;

testcount = 0; // no test score has been read so far

while (testcount < 20) // repeat twenty times

{

cout << “\nEnter a test score please:\t”;

cin >> score;

totalscore + = score;

testcount ++;

}


( a )

The <init-expression-list> consists of the expression, testcount = 0,

The <loop-condition> is testcount < 20, and

The <increment-expression-list> consists of the expression, testcount ++ .

totalscore = 0;

for (tescount = 0 ; testcount < 20 ; testcount ++) // repeat twenty times

{


cin >> score;


}

( b )

Any expression that is evaluated before the <loop condition> is tested for the first time may also be

moved into the <init-expression-list>.

So, we may also specify the expression totalscore = 0 in the <init-expression-list> as follows:

for (totalscore = 0, tescount = 0 ; testcount < 20 ; testcount ++)

{ // repeat twenty times


cin >> score;


}

( c )

Expressions that are evaluated in the <body-of-the-loop> may be moved into the <increment-

expression-list>.

However, keep in mind that expressions in an expression list are evaluated from left to right.

So, we may move the expression totalscore + = score from the <body-of-the-loop> to the

<increment-expression-list> before the loop increment expression as follows:


for (totalscore = 0, tescount = 0 ; testcount < 20 ; totalscore += score, testcount++)

{ // repeat twenty times


cin >> score;

}

2. The following logically-controlled iteration (specified using the while structure) may be specified

using the for structure in any of the following ways in (a) and (b):

cin >> number;

cout << “\n\nIts digits in reverse order are as follows:\t”;

while (number != 0)

{

cout << (number % 10);

number / = 10;

}

( a )

The <init-expression-list> consists of the expressions,

cin >> number and cout << A\n\nIts digits in reverse order are as follows:\t@.

The <loop-condition> is number != 0, and

The <increment-expression-list> consists of the expression number / = 10.

Each of these components of the structure is written on a separate line for readability.

for(cin >> number , cout << “\n\nIts digits in reverse order are as follows:\t”;

number != 0 ;

number /= 10)

{


}


( b )

The cin and the cout expressions specified in the <init-expression-list> of the for structure in (a)

make it hard to read.

One could improve the readability of the for structure by just leaving the <init-expression-list>

empty as in the following for structure.

Notice that although the <init-expression-list> is empty, the semicolon is still required before the

<loop-condition>.

cin >> number ;

cout << A\n\nIts digits in reverse order are as follows:\t@;

for( ; number != 0 ; number /= 10)

{


}

for Structures with one or no Statement in the Body- of-the-Loop

The braces may be omitted in a for structure when the <body-of-the-loop> consists of just one

statement.

When the <body-of-the-loop> is empty (that means without any statement), the right parenthesis

must be followed by a semicolon ( ; ).

Example

The following logically-controlled iteration specified using the while structure may be specified

using the for structure in any of the following ways in (a) and (b):

power = 1; // 0th power of 5

while ( power <= 10000) // repeat until a power of 5 is greater than 10000

power * = 5;


( a )

The <init-expression-list> consists of the expression, power = 1,

The <loop-condition> is power <= 10000, and

The <increment-expression-list> consists of the expression, power * = 5.

The <body-of-the-loop> is empty.

Notice the semicolon after the right parenthesis. It specifies the null statement. That means to do nothing.

/* Repeat until a power of 5 is greater than 10000 */

for (power = 1 ; power <= 10000 ; power *= 5) ;

( b )

In the above for structure, the expression, power * = 5 could be moved into the <body-of-the-loop>, making the

<increment-expression-list> empty.

Notice that the braces are not required to specify to <body-of-the-loop>, and

Although the <increment-expression-list> is empty, the semicolon is still required after the <loop-condition>.

for (power = 1 ; power <= 10000 ; )

power *= 5; // Repeat until a power of 5 is greater than 1000

Defining Variables in a for Structure

Variables such as the loop iteration counter that are used exclusively in a for structure may be

defined in the <init-expression-list> when they are initialized.

Example

In the following for structure, variable tescount is referenced exclusively in the structure, and could

be defined when it is initialized.

However, the same thing cannot be done for variable totalscore, because it is referenced outside of

the for structure.


int totalscore;

for (totalscore = 0, int tescount = 0 ; testcount < 20 ; testcount++)

{

int score; // repeat twenty times


cin >> score;

totalscore += score;

}

cout << endl << “The sum of the scores is:\t” << totalscore;

Exercise CS33

Rewrite the following program segment as an equivalent segment that uses a for structure:

product = 1;

next = 1;

while (next <= m)

{

product * = next;

next ++ ;

}

Exercise CS34

Rewrite the following program segment as an equivalent segment that uses a while structure:

for(result = 0, count = 1; count < 20; count++)

{

cin >> num;

result += num + 2 * count;

}


Iterations using the do/while Structure

Another structure provided in the C/C++ programming language to specify iterations is the do/while

structure.

It has the following syntax:

do

<body-of-the-loop>

while (<condition>);

where <body-of-the-loop> is a single or a compound statement.

Note that the <body-of-the-loop> must contain one or more statements that will eventually make the

loop-condition false.

Also notice the semicolon after the right parenthesis.

Unlike the for structure, the execution of the do/while structure is different from that of the while

structure:

the <body-of-the-loop> is first executed,

then the <condition> is tested;

and the <body-of-the-loop> will continue to be executed as long as the <condition> is true.

It follows that with the do/while structure, the <body-of-the-loop> is executed at least once

Whereas with the while or the for structure, this may not be the case.

Example

Problem 1 write a program to compute the sum of the first ten positive integer values (1, 2, 3, . . . 10).


Solution:

#include <iostream>


#define MAXVALUE 10 /* the maximum value */

int main()

{

int sum = 0 , count;

count = 1;

do

sum += count;

while ( ++ count <= MAXVALUE);

cout << “\nThe sum of the first ten positive integer values is:\t” << sum;

return 0;

}

Problem 2 Write a program to prompt the user to enter Y (for yes) and N (for no). The program will

continuously prompt the user until one of these characters is entered.

Solution:

#include <iostream>


int main()

{

char answer; // to hold the answer from the user

do

{

cout << “\nEnter Y (for yes) or N (for no):\t”;

cin >> answer;

}

while (answer != ‘Y’ && answer != ‘N’);

cout << “\nThank you”

<< “\nYou did a good job”;

return 0;

}


Which Iteration Structure to Use?

Most programmers have a tendency to do the following:

Use the for structure to specify counter-controlled iterations.

Use the while structure to specify logically-controlled iterations.

The do/while structure is rarely used, except for the solution of problems similar to that of Problem

2 in which the <body-of-the-loop> must be executed at least once.

Extra Credit Exercises 1

1. Write a C++ code segment to add the first 20 positive integer values ( 1 + 2 + 3 + . . . + 20) and to

output the result.

2. Write a C++ code segment to read a double precision floating-point value and to calculate and

output its 6th power ( n * n * n * n * n * n).

3. Write a C++ code segment to read 20 non zero integer values and to count the number of positive

values and the number of negative values, and to output the results.

4. Write a C++ code segment to read 20 non zero integer values and to compute the average of positive

values and the average of negative values, and to output the results.

5. The Fibonacci sequence is the sequence: 1, 1, 2, 3, 5, 8, . . . , such that the first and the second values

are 1, and the value in any position n >= 3 is the sum of the values in the two previous positions: the

value in position 3 is the sum of the values in positions 2 and 1; the value in position 4 is the sum of

the values in positions 3 and 2, . . . etc. Using two integer variables previous1 and previous2 to hold

the values in the two previous positions from the current position, and a variable new to hold the

value in the current position, Write a C++ code segment to read an integer value greater than or

equal to 3, and to output the Fibonacci sequence from position 3 to the position that corresponds to

the value read. For example, if the input value is 8, the output will be: 2, 3, 5, 8, 13, 21.

6. A positive integer value is a prime number if it is divisible only by 1 and by itself. Write a C++ code

segment to read a positive integer value greater than 2 and to output a message indicating whether it

is a prime number or not. To find out whether a positive integer value n greater than 2 is a prime

number, you may repeatedly divide it by i= 2, then 3, then 4, . . . , until the remainder is 0 or

i*i>=n. The given value is a prime number if the remainder is not 0; otherwise, it is not a prime

number.

7. Write a C++ code segment to read an integer value greater than 1 and to compute its greatest divisor

that is less than the value itself. For example, if the value is 15, the answer is 5, and if the value is

13, the answer is 1. You compute the greatest divisor of an integer value n greater than 1 that is less

than n, by repeatedly dividing that value n by i= 2, then 3, then 4, . . . etc., until you get a remainder

of 0 or i *i>= n. If the remainder is 0, then the greatest divisor is n/i, otherwise, it is 1.

8. Write a C++ code segment to read the prices of one or more books, and to compute their average

price. Book prices are input one after another, and terminated with the sentinel value -99.00.


9. Write a C++ code segment to read one or more positive integer values and to compute their product.

The sentinel value 0 marks the end of the input values.

10. Write a C++ code segment to read 50 integer values and to determine the number of even values

(among these integer values) and print it.

11. Write a C++ code segment to read one or more integer values and to compute the average of

positive values and print it. If there are no positive values, print an appropriate message.

12. Write a C++ code segment to read an integer value, followed by an operator ( +, -, *, %, or / ), and

then another integer value (for example, 23 + 5), and to compute the expression consisting of the

first value, followed by the operator, which is followed by the second value, and to output the result.

Note that in the division of the first value by the second, you must make sure that the second value is

not zero.

13. Write a program segment to read a character from the keyboard and to determine if it is a letter (‘a’ -

‘z’ or ‘A’ - ‘Z’), a digit (‘0' - ‘9'), or a special character (any other character). If it is a letter, print

the message “letter”; if it is a digit, print the message “digit”; and if it is a special character, print the

message “special character”.

Tracing the Execution of a Program

Example 1:

Given the following definitions of variables: int num1 = 5, num2, result = 0;

Trace the execution of the following program segment for each of the following input values:


Line Number Statement

1 cin >> num2;

2 if(num1 > num2)

3 result = num1 + 4;

4 if (num2 < 7)

5 num1 = num2 + num1;

6 else if (num2 > 10)

7 num1 = 2 * num1;

8 result = result + num1 + num2;

9 cout << result;


a. Trace for input value: 8

Line Number Variables

num1 num2 result Output:

- 5 ? 0 13

1 5 8 0

2 5 8 0

4 5 8 0

6 5 8 0

8 5 8 13

9 5 8 13

b. Trace for input value: 3

Line Number Variables

num1 num2 result Output:

- 5 ? 0 20

1 5 3 0

2 5 3 0

3 5 3 9

4 5 3 9

5 8 3 9

8 8 3 20

9 8 3 20

Exercise CS35

Trace the execution of the code segment in the example above for each of the following input values:


Example 2: Given the following definitions of variables: int ndx = 10, num, result = 0;

Trace the execution of the following program segment:

Line Number Statement

1 num = 0;

2 while( ndx > 7 )

3 {

4 num = num + 3;

5 result = result + num + ndx;

6 ndx = ndx - 2;

7 }

8 cout << “\nndx =\t” << ndx;

9 cout << “\nresult=\t” << result;


Trace:

Line Number Variables Output:

ndx num result ndx = 6

- 10 ? 0 result = 27

1 10 0 0

2 10 0 0

4 10 3 0

5 10 3 13

6 8 3 13

2 8 3 13

4 8 6 13

5 8 6 27

6 6 6 27

2 6 6 27

8 6 6 27

9 6 6 27

Exercise CS36

Trace the execution of each of the following program segments:

a. 1 num = 1; b. 1 ndx = 1;

2 while (num <= 20) 2 pnum = 60;

3 num = num * 2; 3 while(ndx < 5)

4 cout << “\nnum=” << num; {

4 pnum = pnum / ndx;

5 ndx = ndx + 2;

}

6 cout << “\npnum=” << pnum

<< “\nndx=” << ndx;

Exercise CS37

Trace the execution of the following program segment:

Line # 1 int num = 20, result, count;

2 for( result = num, count = 0; count < 3; count++)

3 {

4 num /= 2;

5 result += num + count;

6 }

7 cout << “\nresult=\t” << result << “\nnum=\t” << num;

8 cout << “\ncount=\t” << count;


Multiple-Way Selection using the switch Structure

A multiple-way selection structures in which the value of the same expression is tested at each

decision step against an integer or a character value as follows:

if(<expression> = = value-1)

<Statement-1>

else if (<expression> = = value-2)

<Statement-2>

.

.

.

else if (<expression> = = value-n)

<Statement-n>

else

<Default-Statement>

may be specified in C/C++ using the switch structure as follows:

switch(<expression>)

{

case value-1:

<Statement-1>

break;

case value-2:

<Statement-2>

break;

.

.

.

case value-n:

<Statement-n>

break;

default:

<Default-Statement>

}


Example 1

Given the following multiple-way selection using if-else Structures

/*------------determine and print the corresponding marital status ------------------------*/

if (mcode == ‘M’)

cout << “\nThis individual is married”;

else if (mcode == ‘S’)

cout << “\nThis individual is single”;

else if (mcode == ‘D’)

cout << “\nThis individual is divorced”;

else if (mcode == ‘W’)

cout << “\nThis individual is a widow”;

else

cout << “\nThe marital code that you have typed is invalid”;

It is specified using the switch structure as follows:

/*------------determine and print the corresponding marital status ------------------------*/

switch(mcode)

{

case ‘M’:

cout << “\nThis individual is married”;

break;

case ‘S’:

cout << “\nThis individual is single”;

break;

case ‘D’:

cout << “\nThis individual is divorced”;

break;

case ‘W’:

cout << “\nThis individual is a widow”;

break;

default:

cout << “\nThe marital code that you have typed is invalid”;

}


Example 2 Using the switch Structure without the Default-Statement

Given the following multiple-way selection using if-else structures:

/*------ if the car has a rebate, print it and compute its price with the rebate------------*/

if ( CarType == 2)

{

cout << endl << “This car has a rebate of:\t$” << MIDREBATE;

price = price - MIDREBATE;

}

else if ( CarType == 3)

{

cout << endl << “This car has a rebate of:\t$” << SEDANREBATE;

price = price - SEDANREBATE;

}

else if ( CarType == 5)

{

cout << endl << “This car has a rebate of:\t$” << TRUCKREBATE;

price = price - TRUCKREBATE;

}

It is specified using the switch structure as follows:

/*------ if the car has a rebate, print it and compute its price with the rebate------------*/

switch (CarType)

{

case 2:

cout << endl << “This car has a rebate of:\t$” << MIDREBATE;

price - = MIDREBATE;

break;

case 3:

cout << endl << “This car has a rebate of:\t$” << SEDANREBATE;

price - = SEDANREBATE;

break;

case 5:

cout << endl << “This car has a rebate of:\t$” << TRUCKREBATE;

price - = TRUCKREBATE;

}

Note that the statements specified for each case do not have to be enclosed between the braces.


The break Statement and the switch Structure

Example: Role of the break Statement in a switch Structure

Given the following multiple-way selection using the switch structure (specified from line 14 to line 33):

Line Number



3 int main()

4 {

5 int num, // to hold an integer value

6 bonus; // to hold the bonus earned

7

8 /*----------------read an integer value---------------------*/

9 cout << “\nEnter an integer value:\t”;

10 cin >> num;

11

12 /*--------determine and print the corresponding bonus-------*/

13 bonus = 5;

14 switch(num)

15 {

16 case 2:

17 cout << “\nYou have entered my lucky number”;

18 bonus += 2;

19 break;

20 case 3:

21 case 7:

22 cout << “\nI do not like this number”;

23 - - bonus;

24 break;

25 case 5:

26 cout << “\nIt is OK!”;

27 case 9:

28 cout << “\nYou could do better”;

29 ++ bonus;

30 break;

31 default:

32 cout << “\nGood try”;

33 }

34 cout << “\nYour bonus is:\t” << bonus;

35 return 0;

36 }


It is rewritten using if-else structures as follows:

if (num == 2)

{

cout << “\nYou have entered my lucky number”;

bonus += 2;

}

else if (num == 3 || num == 7)

{

cout << “\nI do not like this number”;

- - bonus;

}

else if (num == 5)

{

cout << “\nIt is OK!”;

cout << “\nYou could do better”;

++ bonus; }

else if (num == 9)

{

cout << “\nYou could do better”;

++ bonus; }

else

cout << “\nGood try”;

Exercise CS38

1. Rewrite each of the following code segments using the switch structure.

a. char letter; b. int num, value;

int num = 10; cin >> num;

cin >> letter; if (num + 5 == 0 || num + 5 == -1)

if (letter == ‘A’) value = 10;

{ else if (num + 5 == 2)

cout << endl << “Apple”; value = 7;

num += 5; else if (num + 5 == 5)

} value = 4;

else if (letter == ‘Y’) cout << endl << “value =” << value;

{

cout << endl << “Yellow”;

num ++ ;

}

else if (letter == ‘G’)

cout << endl << “Green”;

else

cout << endl << “Invalid”;

cout << endl << num:


2. Rewrite each of the following code segments using if-else structures.

a. int ctl; b. int ctl;

cin >> ctl; cin >> ctl;

switch (ctl) switch (ctl)

{ {

case 0: case 1:

case 1: cout << “red”;

cout << "red "; break;

break; case 2:

case 2: cout << “blue”;

cout << "blue "; case 3:

break; cout << “green”;

case 3: break;

cout << "green "; case 4:

break; cout << “yellow”;

default: }

cout << "yellow";

}

cout << endl << ctl;

3. Provide the output of the program segments in question 2a. for each of the following input values:

i. 1 ii. 2 iii. 3 iv. 4

4. Provide the output of the program segments in question 2b. for each of the following input values:

i. 1 ii. 2 iii. 3 iv. 4

Exercise CS39

A florist sells four different types of bouquets of flowers that he has numbered 101, 202, 303, and 404

with the unit prices specified as follows:

ITEM NUMBER UNIT PRICE

101 $ 5.25

202 $ 3.10

303 $ 9.75

404 $ 6.50

Write a program segment to read an item number and the number of bouquets of that type purchased,

and to compute and print the total price of those bouquets (using the switch structure).


Nesting if - else Structures

The following flowchart is specified in C++ as follows:

True False

True False True False

Specification in C++:

if( anum < 1000)

if( bnum < 5 )

(a) result = 1;

else

(c) result = 2;

else

if(bnum > 10 )

(b) result = 3;

else

result = 4;

cout << endl << “result=\t” << result;

if-else structures (a) and (b) are nested inside if-else structure ( c ).

Some examples of the output of this code segment are given as follows:

anum = 500 and bnum = 2 result = 1




anum < 1000

bnum < 5 bnum > 10

result = 1 result = 2 result = 3 result = 4

Write result


Interpreting Nested if-else Structures

In a nested if-else structure, an else balances the nearest preceding unbalanced if in the same

program block.

Example Interpreting Nested if-else Structures

1.

And its output is specified for each range of values as follows:

Range of Values Output

num > 40 Z W

25 < num <= 40 Y W

15 < num <= 25 X W

num <= 15 W

The above program segment can be rewritten using multiple-way selection as follows:

if (num > 40)

cout << ‘Z’;

else if (num > 25)

cout << ‘Y’;

else if (num > 15)

cout << ‘X’;

cout << “\t W”;

Nested if-else Structures

are Interpreted as Follows

if (num > 15)

if (num > 25)

if (num > 40)

cout << ‘Z’;

else cout << ‘Y’;

else cout << ‘X’;

cout << “\tW”;

if (num > 15)

if (num > 25)

if (num > 40)

cout << ‘Z’;

else

cout << ‘Y’;

else

cout << ‘X’;

else;

cout << “\t W”;


2. The following nested if-else structures have different interpretations:

cin >> status >>

value;

if (status == 1)

if (value == 500)

cout << ‘A’;

else

cout << ‘B’;

cout << ‘C’;

cin >> status >> value;

if (status == 1)

{

if (value == 500)

cout << ‘A’;

}

else

cout << ‘B’;

cout << ‘C’;

In the first nested structure, the else balances the second if, whereas in the second, it balances the

first if. The input/output relation is specified as follows:

First Nested if-else Structure Second Nested if-else Structure

Input: status = 1 value = 500 Output: A C Output: A C

Input: status = 1 value != 500 Output: B C Output: C

Input: status != 1 value = 500 Output: C Output: B C

Input: status != 1 value != 500 Output: C Output: B C

Exercise CS40*

1. Provide the output of the program segment specified by the flowchart above for the following

values:

a. anum = 200, bnum = 4; c. anum = 500, bnum = 35

b. anum = 1500, bnum = 17 d. anum = 2000, bnum = 8

2. For each of the following program segments (a) and (b), provide the output for the following input

values:

a. status is 0, and value is 50 c. status is 0, and value is 200

b. status is 1, and value is 50 d. status is 1, and value is 200


( a )


if (status == 0)

if (value < 100)

cout << ‘A’;

else

cout << ‘B’;

cout << ‘C’;

( b )


if (status == 0)

{

if (value < 100)

cout << ‘A’;

}

else

cout << ‘B’;

cout << ‘C’;

Exercise CS41

1. Use proper indentation to specify the interpretation of the following nested if-else structures:

a. cin >> code >> num1 >> num2; b. if ( num <= 200)

if (code == ‘T’) if ( num < 100)

if (num1 > 100) if ( num <= 0)

{ cout << "A\n";

num++ ; else

num2 * = 5; cout << "B\n";

} else

else cout << "C\n";

num - - ; else

else cout << "D\n";

if ( num2 < 50 )

num1 /= 2;

else

{

num2 += num1;

num1 * = 10;

}

2. For what range of values of the variable num will the output of the program segment 1.b be:

i. A ii. B iii. C iv. D.

3. Rewrite the program segment 1.b using multiple-way selection with if-else structures.


Case Study CS13

Problem Statement

The relative risk for individuals to have a certain disease depends on their age and whether they smoke

or not as specified in the following table.

Smoker

Yes No

Age >= 40 6 3

Age < 40 4 1

Write a program to read the age of an individual (integer value) and its smoking status (1 for smoker and

0 for non smoker) and to print its relative risk.

Program Logic

input: the age of an individual and its smoking status (0 or 1).

output: its relative risk to have a certain disease.

variables: age (int) to hold the age of an individual.

status (int) to hold its smoking status.


True False


Start

Read

age, status

age >= 40

status = 1 status = 1

Write

6

Write

3

Write

4

Write

1

Stop


Figure CS13 Using Nested if-else Structures

Line Number

1 /*************************************************************

2 Program to compute the relative health risk of an individual

3 *************************************************************/



5 #define RISK1 6 /* highest risk */

6 #define RISK2 4 /* second highest risk */

7 #define RISK3 3 /* third highest risk */

8 #define RISK4 1 /* lowest risk */

9 #define AGELIMIT 40 /* age limit */

10 #define SMOKER 1 /*smoking status*/

11

12 int main()

13 {

14 int age, //the individual=s age

15 status; //smoking status: 1(smoker) 0 (non smoker)

16

17 /*------read the individual=s age and smoking status ------*/

18 cout << A\nEnter age and smoking status in this order:\t@;

19 cin >> age >> status;

20

21 /*---- determine this individual=s relative health risk ---*/

22 if ( age >= AGELIMIT ) //age is equal or above the age limit

23 if ( status == SMOKER ) // he is a smoker

24 cout << endl << “relative risk is:\t”

25 << RISK1;

26 else // he is a non-smoker


28 << RISK3;

29 else // he is under the age limit

30 if ( status == SMOKER ) // he is a smoker


32 << RISK2;

33 else // he is a non-smoker


35 << RISK4;

36

37 return 0;

38 }


Case Study CS14

Problem Statement

The simple interest rate on loans at a local bank depends on the client’s account balance and the amount

of the loan as follows:

Amount Borrowed

< $ 750.00 >= $ 750.00

Balance < $ 5000.00 9 % 8 %

--------------------------------------------------------------------------------------------------

< $ 1000.00 >= $ 1000.00

Balance >= $ 5000.00 5 % 4 %

Write a program to read a client’s account balance and the amount of the loan, and to compute the

simple interest and print it.

Program Logic

Input: a client’s balance, and the amount of the loan.

Output: the simple interest on the loan.

Variables:

balance (double) to hold the account balance.

amount(double) to hold the loan amount.

rate (double) to hold the simple interest rate.


True False


Start

Read

balance, amount

status

balance < 5000

amount < 1000 amount < 750

Stop

rate = .09 rate = .08 rate = .05 rate = .04

Write

rate * amount


Figure CS14 Using Nested if-else Structures

Line Number

1 /**************************************************************

2 Program to compute the simple interest on a loan

3 **************************************************************/



6 int main( )

7 {

8 double balance, // the account balance

9 amount, // the amount borrowed

10 rate; // the interest rate

11

12 /*----------read the balance and the amount of loan --------*/

13 cout << endl << AEnter the balance and amount of loan :\t@;

14 cin >> balance >> amount;

15

16 /*---------------- get the interest rate ------------ ------*/

17 if (balance < 5000.00)

18 if (amount < 750.00)

19 rate = .09;

20 else

21 rate = .08;

22 else

23 if (amount < 1000.00)

24 rate = .05;

25 else

26 rate = .04;

27

28 /*---------- compute and print the simple interest ---------*/

29 cout << endl << “The interest is:\t” << (rate * amount);

30

31 return 0;

32 }

Exercise CS42

The contribution fees of the members of a club depend on whether they are students or not, and whether

it is a family membership or a single membership as follows:

Famil membership

Yes No

Student $ 30.00 $ 25.00

Non-Student $ 150.00 $ 100.00

Assuming that the status of a member is encoded by an integer value (1 for student, and 0 for non-

student) and that the type of membership is also encoded by an integer value (1 for family, and 0 for

non-family membership), write a program segment to read the status of a member and its type of

membership, and to output its membership fees.


Nested Iterations

Case Study CS15

Problem Statement 15 salesmen have reported their total sales (in dollars) in 10 different cities as follows:

Cities #

salesman # 1 2 3 . . . 10

1 7000.00 4000.00 6000.00 . . . 7000.00

2 4000.00 5000.00 1000.00 . . . 500.00

. .

. .

. .

15 2000.00 7000.00 4000.00 . . . 2000.00

Write a program segment to read the total sales of each salesman in all 10 cities, and to compute and

print their sum. Your program must also compute the sum of the total sales for the company. That

means, the sum of the total sales of the 15 salesmen in the 10 cities.

Program Logic

Input: the total sales in 10 cities for 15 salesmen.

Output: each salesman’s sum of total sales, and the company’s total sales.

The algorithm consists of repeating the following steps 15 times:

1. Read a salesman’s total sales in the 10 cities and compute their sum.

2. Add this salesman’s sum of the cities total sales to the company’s total sale.

3. Print this salesman’s sum of the cities total sales.

We therefore have here a counter-controlled iteration: we need a counter (let’s call it scount) to

count the salesmen: its initial value is 0, and its final value is 15.

We need a variable to hold the sum of the total sales of a salesman in all ten cities (let’s call it

salesmanTotal).

To compute the sum of the total sales for the company, we need a variable (running total) that is first

initialized to 0 (outside of the body-of-the-loop) and then incremented by the sum of the total sales

in all ten cities for each salesman (salesmanTotal). Let’s call this variable companyTotal.

Step 1 says to read the total sales for a salesman in the 10 cities and compute their sum. This is done

by repeating the following steps 10 times:

1.1 Read a salesman’s total sales in a city and add it to the variable salesmanTotal.

We therefore have here a counter-controlled iteration: we need a counter (let’s call it cityCount) to

count the cities (initial value 0, and final value 10)


We also need a variable to hold a salesman’s total sales in a city (let’s call it citySale).

The algorithm follows:

Variables:

companyTotal (double) to hold the total sales in the company.

scount (int) to count the salesmen.

salesmanTotal (double) to hold a salesman’s total sales (in all the cities).

citySale (double) to hold a salesman’s total sales in a city.

cityCount (int) to count the cities.

Figure CS15 Using Nested Iterations

Line Number

1 /****************************************************************

2 Program to process salesmen’s total sales in a company

3 ****************************************************************/


5 #define CITYMAX 10 /* maximum number of cities */

6 #define SALESMENMAX 15 /* maximum number of salesmen */

7

8 int main()

9 {

10

11 /*---compute the total sales of the company and print it----*/

12 double companyTotal = 0; //the total sales of the company

13 for ( int scount = 0; scount < SALESMENMAX ; scount ++ )

14 {

15 /*-- compute a salesman’s sum of total sales and print it -*/

16 double salesmanTotal = 0; // the sum of a salesman’s total sale

17 cout << “\nEnter the total sales of salesman #” << (scount + 1)

18 << “ in all cities, followed by the ENTER key” << endl;

19 for ( int cityCount = 0 ; cityCount < CITYMAX ; cityCount ++ )

20 {

21 /*-- read the total sales in this city for this salesman and add it to its total*/

22 double citySale; //a salesman’s total sale in a city

23 cin >> citySale;

24 salesmanTotal += citySale;

25 }

26 cout << “\nThe sum of total sales for salesman # “

27 << (scount + 1) << “is:\t” << salesmanTotal;

28

29 /*-add salesman’s sum of total sales to the company total*/

30 companyTotal += salesmanTotal;

31 }

32

33 /*---------- output the company total sales ----------------*/

34 cout << “\nThe company total sales is:\t” << companyTotal;

35 return 0;

36 }


Exercise CS43

Each student in a class of 20 students has completed 6 quizzes and has received a score on each of these

quizzes. The quizzes are numbered from 1 to 6, and students are numbered from 1 to 20 as follows:

Quiz #

1 2 3 4 5 6

Student #

1 95.0 80.0 90.0 86.0 97.0 98.0

2 87.0 78.0 79.0 80.0 80.0 83.0

. .

. .

. .

20 78.0 89.0 76.0 88.0 78.0 92.0

Write a program segment to read each student’s quiz scores, and to compute his/her average quiz

score and print it. The program must also compute the class average of the quizzes. The class

average is the sum of all students’ average scores divided by 20.

Exercise CS44

The 9x9 multiplication table of the digits 1 to 9 is produced by multiplying the row number by the

column number as follows:

Column #

1 2 3 4 5 6 7 8 9

Row #

1 1 x 1 1 x 2 1 x 3 1 x 4 . . . 1 x 9

2 2 x 1 2 x 2 2 x 3 2 x 4 . . . 2 x 9

. .

. .

. .

9 9 x 1 9 x 2 9 x 3 9 x 4 . . . 9 x 9

Write a program segment to generate the above multiplication table.


Extra Credit Exercises 2

1. Write a program segment to read an integer value, an operator ( +, -, *, %, or /), and another integer

value in this order (for example, 23 + 5), to compute the expression consisting of the first value,

followed by the operator, which is followed by the second value, and to output the result (using the

switch structure). Note that in the division of the first value by the second, you must make sure that

the second value is not zero.

2. A game consists of reading a positive integer value and rewarding bonuses based on the remainder in

the division of that number by 5 as follows: 0: 200; 1: 350; 2: 400; 3: 500; 4: 750.

Write a program segment to read a positive integer value, and to output the bonus awarded to the

user (using the switch structure).

3. A travel agency rewards bonuses to its clients based on the price of the vacation package and the

number of people in the vacation group as follows:

Number of people

< 3 3 > 3

Price <= $ 2500.00 $ 20.00 $ 10.00 $ 5.00

Price > $ 2500.00 $ 50.00 $ 25.00 $ 15.00

Write a program segment to read the price of a vacation package and the number of people in the

vacation group, and to output the bonus awarded.

4. Rewrite the code segment in Exercise CS43 by assuming that each student has completed one or

more quizzes. For each student, the dummy score of -99 is entered to end his/her list of scores. A

student’s average score is the sum of the scores on the quizzes divided by the number of quizzes

completed.

5. Trace the execution of the following program segment and show its output:

Line Number

1 int row, column;

2 row = 10;

3 while( row > 5)

4 {

5 cout << endl << row;

6 column = 1;

7 while (column < 5)

8 {

9 cout << ‘\t’ << row + column;

10 column ++ ;

11 }

12 row - = 2;

13 }


Solutions of the Exercises

Exercise CS1*

a) num1 >= 5 || num2 + 3 == num3 d) num2 + 3 == num3 * 4 || ch >= ‘Z’

5 >= 5 || 7 + 3 == 2 * 4 || ‘K’ >= ‘Z’

True 10 == 8 || False

True False || False

False False

b) ‘A’ <= ch && fnum + 7.2 != 2.5 e) num1 < 3 && num2 > num3

‘A’ <= ‘K’ && 11.75 + 7.2 != 2.5 5 < 3 &&

True && 18.95 != 2.5 False

True && True False

True

c) !(3 * num1 + 4 < num3 * 2) f) !(fnum * 2 < num1 + 20)

!( 3 * 5 + 4 < 2 * 2) !(11.75 * 2 < 5 + 20)

!( 19 < 4) !( 23.50 < 25)

! False !True

True False

Exercise CS6*

a) cin >> num;

if (num = 5) ---------------------> num == 5

sum = num + 4;

else

sum = num + 10;

b) cin >> num;

if (num > 5 && <= 10) ---------------> num > 5 && num <= 10

cout << num + 15;

else

cout << num - 3;

c) cin >> num; cin >> num;

if (num < 15) if (num < 15)

result1 = 2 * num; ----------------> {

result2 = num + 20; result = 2 * num;

else ---------------- > result2 = num + 20;

result1 = 5 * num; }

result = result1 + result2; else

cout << “\nThe result is:\t” << result; reult1 = 5 * num;

result = result1 + result2;

cout << “\nThe result is:\t” << result;


Exercise CS7*

Program trace

Input: 4 input: 20

Line # num Line # num

1 4 1 20

2 4 2 20

3 10 5 20

4 10 6 20

6 10

Output: num = 10 Output: num / 4 = 5

result = 20 result = 40

Exercise CS10*

a. b. c.


while( count < 10 ) while(count < 10) count = 0;

{ { while(count < 10 )

cin >> num; count = 0; {

cout << 2 * num; cin >> num; cin >> num;

count = count + 1; cout << 2 * num; cout << 2 * num;

} count = count + 1; count = count + 1;

} }

Exercise CS12*

a. b.

sum = 0; sum = 0;


while (count < 10 ) while (count < 10 )

{ {

sum = 0; cin>>num;

cin >> num; sum = sum + num;

sum = sum + num; count = count + 1;

count = count + 1; }

}

Exercise CS13*

Variables

testCount (int) to count the test scores

score (double) to hold a test score

totalScore (double) to hold the current total of the test scores


C++ Language Code Segment

int testCount = 0; // to count the test scores

double score, // to hold a test score

totalScore = 0; // to hold the current total of the test scores

while ( testCount < 20 )

{

cout << endl << “enter a test score:\t” ;

cin >> score;

totalScore = totalScore + score;

testCount = testCount + 1;

}

cout << endl << “the average test score is:\t” << totalScore / 20;

Flowchart

False True

Start

totalScore = totalScore + score

testCount < 20

Stop

testCount = 0

Read

score

totalScore = 0

testCount = testCount + 1

Write

totalScore / 20


Exercise CS18*

Variables

totalSale (double) to hold the total sale in the store

dailySale (double) to hold a daily sale in the store

Flowchart

False True


int totalSale = 0, // to hold the total sale in the store

dailySale; // to hold a daily sale in the store

while (totalSale < 10000)

{

cin >> dailySale;

totalSale + = dailySale;

}

cout << endl << totalSale - 10000 << “\tIt is time to make a deposit”;

Start

totalSale = totalSale + dailySale

totalSale < 10000

Stop

totalSale = 0

Read

dailySale

Write

totalSale - 10000

“time to make deposit”


Exercise CS20*

#define DUMMYVALUE -99

int main ( )

{

int total = 0, value;

cout << “\nEnter all values to add followed by”

<< DUMMYVALUE << “to stop\n”;

cin >> value; // add this statement here

while ( value != DUMMYVALUE )

{

cin >> value; // remove this statement

total = total + value;

cin >> value; // add this statement here

}

cout << endl << “The total of these values is:\t” << total;

return ( 0 );

}

Exercise CS21*

Variables Ftemp (double) to hold a temperature in Fahrenheit

Flowchart

False True


double Ftemp; // to hold a temperature in Fahrenheit

cin >> Ftemp;

while (Ftemp != -99.00)

{

cout << endl << Ftemp << “\t” << 5.0 / 9 * (Ftemp – 32);

cin >> Ftemp;

}

Start

Ftemp != -99.00

Stop Read

Ftemp

Write

5.0 / 9 * (Ftemp – 32)

Read

Ftemp


Exercise CS24*

a. Input value: 4 b. Input Value: 20

Output Output

Result= 3 pnum= 27

Result= 19

Exercise CS25*

Variable num (int) to hold the integer value

Flowchart

False True

False True

Start

Read

num

num < 5

num > 20

greatt = greatt+ 2

somet = somet + 7

Stop

smallt = smallt + 1

smallt = smallt - 3



int num; // to hold the integer value

cin >> num;

if (num < 5)

smallt = smalt + 1;

if (num > 20)

greatt = greatt + 2;

else

smallt = smallt – 3;

somet = somet + 7;

Exercise CS28*

Variables

num (int) to hold a bouquet type code.

unprice (double) to hold its unit price.

qty (int) to hold its quantity.


int num, // to hold a bouquet type code

qty; // to hold its quantity

double uprice; // to hold its unit price

cin >> num >> qty;

if (num = = 101)

uprice = 5.25;

else if (num = = 202)

uprice = 3.10;


uprice = 9.75;


uprice = 6.50;

else

uprice = 0.0;

cout << endl << “The price is:\t” << (uprice * qty);

if (uprice = = 0)

cout << endl << “invalid bouquet number”;


Flowchart

True False

True False

True False

True false

True False

Start

Read

num, qty

num = 101

uprice = 5.25

uprice = 3.10

num = 202

num = 303

uprice = 9.75

uprice = 0.0 uprice = 6.50

num = 404

Write

uprice * qty

Stop

uprice = 0

Write

“invalid bouquet number”


Exercise CS30*

a. Input value: 3 b. Input value: 7 c. Input value: 12

Output Output Output

Red Yellow Green

blue blue blue

Exercise CS31*

Variables

speed (int) to hold the speed of the car.

fine (double) to hold the fine

Flowchart

True False

True False

True False

speed <= 50.00

fine = 0.00

fine = 15.00

speed <= 70.00

speed <= 80.00

fine = 30.00 fine = 60.00

Write

fine

Stop

Start

Read

speed



int speed; // to hold the speed of the car

double fine; // to hold the fine

cin >> speed;

if (speed <= 50)

fine = 0.00;

else if (speed <= 70)

fine = 15.00;

else if (speed <= 80)

fine = 30.00;

else

fine = 60.00;

cout << endl << “the fine imposed is:\t” << fine;

Documents

Control Structures · Controlled structures may also be specified using the following flowchart symbols: Flowchart Symbols SYMBOL (a) end of an algorithm. operation.(b) (c) Flow Lines(d)