exercise 3 - william paterson universitycs.wpunj.edu/~ndjatou/cs2300-chapter3-solutions.pdfexercise...

Exercise 3.1

1.a

Variables: len (double) to hold the length of the rectangle

width (double) to hold its width

Pseudo code

1. Read the length of the rectangle into the variable len.

2. Read the width of the rectangle into the variable width.

3. Compute and print the perimeter of the rectangle.

4. Compute and print the area of the rectangle.

Flowchart

1.b. C++ Program

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

Program to read the length and the width of a rectangle and

to compute its perimeter and area.

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

#include <iostream>

#include <iomanip>

using namespace std;

int main()

{

cout.setf(ios :: fixed);

cout.setf(ios :: showpoint);

double len, // to hold the length of the rectangle

width; // to hold its width

/*---------- read the width and the length of the rectangle ----------*/

cout << endl << “Enter the width of the rectangle (in inches):\t”;

cin >> width;

cout << endl << “Enter its length (in inches):\t”;

cin >> len;

Start

Read

len, width

Write

2 * (len + width)

Stop

Write

len * width

/*------------ compute and print its perimeter -----------------------*/

cout << setprecision(2);

Cout << endl << “Its perimeter is:\t”

<< 2 * ( len + width ) << “ inches”;

/*------------compute and print its area ----------------------------*/

cout << endl << “Its area is:\t” << len * width

<< “ square inches”;

return( 0 );

}

2.a

Variables: num ( int ) to hold the positive integer value

Pseudo code

1. Read the positive integer value into the variable num.

2. Compute and print its right-most digit: num % 10.

3. Compute its quotient in the division by 10 and store it into variable num: num = num / 10.

4. Compute and print the middle digit: num % 10.

5. Compute and print the first digit: num / 10.

Flowchart

Start

Read

num

num = num / 10

Write

num / 10

Stop

Write

num % 10

Write

num % 10

2.b. C++ Program

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

Program to read a positive integer value and to print its digits in reverse order

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

#include <iostream>

using namespace std;

int main()

{

int num; // to hold the positive integer value

/*---------- read the positive integer value ----------*/

cout << endl << “Enter a 3-digit integer value:\t”;

cin >> num;

cout << endl << The digits of this number are:\t”;

/*------- compute and print its right-most digit -----*/

cout << num % 10;

/*----- compute its quotient in the division by 10 ---*/

num = num / 10;

/*------- compute and print the middle digit ---------*/

cout << num % 10;

/*-------compute and print the first digit -----------*/

cout << num / 10;

return( 0 );

}

Exercise 3.2

a. num1 <= 5 c. 3 * num 2 > fnum + 2 e. num1 + 3 != num1 * 4 f. ch – ‘M’ = = 5

9 <= 5 3 * 5 > 12.50 + 2 9 + 3 != 9 * 4 3 = = 5

False 15 > 14.50 12 != 36 False

True True

b. 3 * num1 + 4 < num3 * 2 d. 2 * num1 + 12 = = 3 * num3

3 * 9 + 4 < 10 * 2 2 * 9 + 12 = = 3 * 10

27 + 4 < 20 18 + 12 = = 30

31 < 20 30 = = 30

False True

Exercise 3.3

A.

a. 2 * num1 - 5 >= 9 || fnum / 2 + 10 <= 6.5 b. num1 + num2 == num3 + 5 && 2 * num3 <= 4 * num2

2 * 9 – 5 >= 9 || 9 + 5 == 10 + 5 &&

18 – 5 >= 9 || 14 == 15 &&

13 >= 9 || False &&

True || False

True

c. ! (num1 + 5 <= 13) d. 3 * num1 > num1 + num2 && num1+ 3 >= 12

! ( 9 + 5 <= 13) 3 * 9 > 9 + 5 && 9 + 3 >= 12

! ( 14 <= 13) 27 > 14 && 12 >= 12

! False True && True

True True

e. ! ( num3 % 4 < 3) f. num1 - 5 >= num3 || num2 < 15 && num1 >= 9

!( 10 % 4 < 3) || 5 < 15 && 9 >= 9

!( 2 < 3) || True && True

! True || True

False True

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

a. num1 != 0, c. num1, and e. !(num1 == 0) are equivalent.

b. num1 == 0 and d. !num1 are equivalent.

C. a. age = = 45 b. height > 5.7 c. salary > 35000 && salary < 50000

d. ! (NumChild = = 3) e. (height = = 6.0) || (NumChild < 4)

f. !(age > 35) && ((NumChild = = 2) || (NumChild = = 3))

Exercise 3.4

1.

Variables num (int) to hold the integer value

a. Algorithm using Pseudo-Code

1. read the integer value into variable num.

2. If num > 0 then do the following:

2.a. print it with the message “POSITIVE”

3. Otherwise, do the following:

3.a. print it with the message “NEGATIVE”

Flowchart

False True

C++ Language Code Segment:

int num;

cin >> num;

if ( num > 0 )

cout << endl << num << “POSITIVE”;

else

cout << endl << num << “NEGATIVE”;

2.

Variables symb (char) to hold the character value


1. read the character value into variable symb.

2. If (symb >= ‘0’ AND symb <= ‘9’) then do the following:

2.a. print it with the message “DIGIT”


3.a. print it with the message “NOT A DIGIT”

Start

Read

num

num > 0

Write

num,

“POSITIVE”

Write

num,

“NEGATIVE”

Stop

Flowchart

False True

b. C++ Language Code Segment:

char symb;

cin >> symb;

if (symb >= ‘0’ && symb <= ‘9’)

cout << endl << symb << “\t is a digit”;

else

cout << endl << symb << “\t is not a digit”;

Exercise 3.5



1. read the integer value into variable num.

2. If num % 5 = 0 then do the following:

2.a. print it with the message “MULTIPLE OF 5”


3.a. print it with the message “NOT MULTIPLE OF 5”

Start

Read

symb

(symb >= ‘0’ && symb <= ‘9’)

Write

symb,

“DIGIT”

Write

symb, “NOT A

DIGIT”

Stop

Flowchart

False True

b. C++ Language Code Segment:

int num;

cin >> num;

if ( num % 5 == 0 )

cout << endl << num << “MULTIPLE OF 5”;

else

cout << endl << num << “NOT MULTIPLE OF 5”;

Exercise 3.6

1. 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) ---------------------> num == 5

sum = num + 4;

else

sum = num + 10;

Start

Read

num

num % 5 = 0

Write

num,

“MULTIPLE OF 5”

Write

num,

“NOT MULTIPLE of 5”

Stop

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;

2. 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

3.a. Variables:

num (int) to hold the value

Algorithm (in Pseudo-Code)

1. read the value into variable num

2. if the value is greater than 10 do the following:

2.a. subtract 2 from it.

3. else do the following:

3.a. add 3 to it

4. print the new value of variable num

5. multiply the new value of variable num by 5 and print the result.

Flowchart

False True

b. C++ Language Code Segment

int num;

cin >> num;

if (num > 10)

num = num - 2;

else

num = num + 3;

cout << num;

cout << num * 5;

Exercise 3.7 1.

a. b. c.

count = 0; count = 0;

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;

}

}

Start

Read

num

num > 10

Write

num

Write

num * 5

Stop

num = num + 3 num = num - 2

2.a Variables:

tempf (double) to hold a temperature in Fahrenheit.

TempCount (int) to count the temperatures

Pseudo-Code

1. Set the loop counter TemCount to 0.

2. As long as the loop counter TempCount is less than 50 do the following:

2.a. Read a temperature in Fahrenheit into variable tempf.

2.b. Convert it to Celsius and print the result.

2.c. Increment the loop counter TempCount by 1.

Flowchart

False True

Start

TempCount = 0

TempCount = TempCount + 1

TempCount < 50

Read

temf

Write

5.0 / 9 * (tempf – 32)

Stop


double tempf; int TempCount = 0;

while (TempCount < 50)

{

cin >> tempf;

cout << endl << (5.0 / 9 * (tempf – 32));

TempCount = TempCount + 1;

}

Exercise 3.8

1.

a. b.

sum = 0; sum = 0;

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; }

}

2.a. 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

Pseudo-Code

1. Set the test score counter testCount to 0: testCount = 0.

2. set the current total of the test scores to 0: totalScore = 0.

3. As long testCount is less than 20, do the following:

3.a. Read a test score into variable score.

3.b. add the test score read to the current total of the test scores: totalScore = totalScore + score.

3.c. increment the test score counter by 1: testCount = testCount + 1.

4. Compute and print the average score, totalScore / 20.

Flowchart

False True

2.b. C++ Language Code Segment

int testCount = 0;

double score,

totalScore = 0;

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;

Start

totalScore = totalScore + score

testCount < 20

Stop

testCount = 0

Read

score

totalScore = 0

testCount = testCount + 1

Write

totalScore / 20

Exercise 3.9

Output: the sum of the first 10 positive multiples of 5

Input: no input

Variables:

sum (int) to hold the current sum of the first multiples of 5 greater than 0.

count (int) to count the first 10 multiples of 5 greater than 0

Pseudo-Code

1. Set the variable sum to 0.

2. Set the loop counter count to 1.

3. As long as count < = 10 do the following:

3.a. compute 5 * count and add the result to sum: sum = sum + 5 * count.

3.b. Increment the loop counter count by 1: count = count + 1.

4. Print the sum.

Flowchart

False True

Start

count = 1

count = count + 1

count < = 10

Stop

sum = 0

sum = sum + 5 * count

Write

sum


int sum = 0,

count = 1;

while ( count <= 10)

{

sum = sum + 5 * count;

count = count + 1;

}

cout << sum;

Exercise 3.10

1.a Variables:

value (int) to hold a value

product (int) to hold the product.

count (int) to count the values.

Pseudo-Code

1. Set the current product to 1: product = 1.

2. Set the loop counter count to 0.

3. As long count < 10 do the following:

3.a. read a value into the variable value.

3.b. compute: product = product * value

3.c. Increment the loop counter count by 1.

4. Print the product.

Flowchart

False True

Start

count = 0

count = count + 1

count < 10

Stop

product = 1

product = product * value

Write

product

Read

value


int product = 1, value,

count = 0;

while ( count < 10)

{

cin >> value;

product = product * value;

count = count + 1;

}

cout << product;

2.a

Variables

value (int) to hold the current even number

Pseudo-Code

1. Set the current even number to 20: value = 20.

2. As long value > 0, do the following:

2.a. compute value * value and print the result.

2.b. Decrement the value of variable value by 2.

Flowchart

False True

Start

value = value - 2

value > 0

Stop

value = 20

Write

value * value


int value = 20;

while ( value > 0)

{

cout << endl << value * value;

value = value - 2;

}

Exercise 3.11

Variables

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

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

Pseudo-Code

1. set the total sale TotalSale to 0

2. As long as the total sale TotalSale is less than 10000 do the following:

2.a. Read a daily sale into the variable DailySale.

2.b. Add the daily sale to the total sale.

3. print the total sale and the message “it is time to make a deposit”.

Flowchart

False True

Start

TotalSale = TotalSale + DailySale

TotalSale < 10000

Stop

TotalSale = 0

Read

DailySale

Write

TotalSale - 10000

“time to make deposit”


int totalSale = 0,

dailySale;

while (totalSale < 10000)

{

cin >> dailySale;

totalSale = totalSale + dailySale;

}

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

Exercise 3.12

Variables

num (int) to hold the integer value

divisor (int) the hold the divisor

Pseudo-Code

1 read an integer value greater than 1 into variable num.

2 set the variable divisor to 2

3 As long as ( num % divisor != 0 and divisor < num / divisor) do the following:

3.a. increment the current value of variable divisor by 1.

4 If num % divisor = 0, do the following:

4.a. print the message “the answer is:” and compute and print num /divisor.

5. Otherwise do the following:

5.a. print the message “the answer is:” and print 1.

Flowchart

False True

False True


cin >> num;

divisor = 2;

while ( num % divisor != 0 && divisor * divisor < num )

{

divisor = divisor + 1;

}

if ( num % divisor = = 0)

cout << endl << “\t The answer is:” << num / divisor;

else

cout << endl << “\t The answer is:” << 1;

Start

num % divisor != 0

&&

divisor * divisor< num

Stop

divisor = 2

Read

num

Write

“The answer is:”, num / divisor

divisor = divisor + 1

num % divisor = 0

Write

“The answer is:”, 1

Exercise 3.13

1. #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 );

}

2.a Variables

ftemp (double) to hold a temperature in Fahrenheit

Pseudo-Code

1. Read a temperature (in Fahrenheit) into variable Ftemp.

2. As long as the temperature in the variable Ftemp is not –99.00 do the following:

2.a. Convert it to Celsius and print the result..

2.b. Read the next temperature (in Fahrenheit) into variable Ftemp.

Flowchart

False True

Start

ftemp != -99.00

Stop

Read

ftemp

Write

5.0 / 9 * (ftemp – 32)

Read

ftemp


int ftemp;

cin >> ftemp;

while ( ftemp != -99.00 )

{

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

cin >> ftemp;

}

Exercise 3.14

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

Output Output

Result= 3 pnum= 27

Result= 19

2.

Variables

uprice (double) to hold the unit price

qty (int) to hold the quantity

price (double) to hold the price of the product

Pseudo-Code

1. Read the unit price of the product into variable uprice, and the quantity purchased into variable qty.

2. Compute the price of the product (before the rebate if applicable): price = uprice * qty

3. if the price is greater than or equal to 20.00 do the following:

3.a. Compute the price with the rebate as follows: price = price – 0.1 * price

4. Print the price of the product (with the rebate if applicable).

Flowchart

False True

Start

Read

uprice, qty

price = uprice * qty

price >= 20.00

Price = price – 0.1 * price

Write

price

Stop

C++ Language Code Segment

int qty;

double uprice,

price;

cin >> uprice >> qty;

price = uprice * qty;

if ( price >= 20.00)

price = price – 0.10 * price;

cout << endl << “price =” << price;

Exercise 3.15


int num,

qty;

double uprice;

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”;

a.

Variables

num (int) to hold a bouquet type code.

unprice (double) to hold its unit price.

qty (int) to hold its quantity.

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 3.16

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

Output Output Output

Red Yellow Green

blue blue blue

1.b. a. the output is Red Blue for num < 5

b. the output is Yellow Blue for 5 <= num < 10

c. the output is Green Blue for num >= 10

2. Variables

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

fine (double) to hold the fine

Flowchart

Start

Read

speed

A

True False

True False

True False


int speed;

double 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;

speed <= 50.00

fine = 0.00

fine = 15.00

speed <= 70.00

speed <= 80.00

fine = 30.00

fine = 60.00

Write

fine

Stop

A

Exercise 3.17

Variables

num (int) to hold the integer value.

remain (int) the remainder in the division by 5.

bonus (int) to hold the bonus

Flowchart

True False

True False

True False

True False

Start

Read

num

remain = num % 5

remain = 0

bonus = 200

bonus = 350

remain = 1

remain = 2

bonus = 400

bonus = 750 bonus = 500

remain = 3

Write

bonus

Stop


cin >> num;

remain = num % 5;

if (remain = = 0)

bonus = 200;

else if (remain = = 1)

bonus = 350;


bonus = 400;


bonus = 500;

else

bonus = 750;

cout << endl << “The bonus is:\t” << bonus;

Exercise 3.18

1. a. Input value: 2

Line Numbers Variables OUTPUT

num1 num2 result 18

- 5 ? 0

1 5 2 0

2 5 2 0

3 5 2 9

4 5 2 9

5 7 2 9

8 7 2 18

9 7 2 18

1. b. Input value: 6


num1 num2 result 17

- 5 ? 0

1 5 6 0

2 5 6 0

4 5 6 0

5 11 6 0

8 11 6 17

9 11 6 17

Exercise 3.19


ndx pnum pnum = 20

- ? ? ndx = 5

1 1 ?

2 1 60

3 1 60

4 1 60

5 3 60

3 3 60

4 3 20

5 5 20

3 5 20

6 5 20

Hands-On Exercises

1.a.


False True

c. C++ Language Code Segment:

int num;

cin >> num;

if ( num % 7 == 0 )

cout << endl << num << “MULTIPLE OF 7”;

else

cout << endl << num << “NOT MULTIPLE OF 7”;

Start

Read

num

num % 7 = 0

Write

num,

“MULTIPLE OF 7”

Write

num,

“NOT MULTIPLE of 7”

Stop

2. Variables

uprice (double) to hold the unit price of the product.


qty (int) to hold the number of items purchased.

bonus (double) to hold the bonus

Flowchart

True False


double uprice, price, bonus;

int qty;



if (price >= 100.00)

bonus = .10 * price;

else

bonus = .05 * price;

cout << endl << “The price is:\t” << price << “\nThe bonus is:\t” << bonus;

price >= 100.00

bonus = .1 * price bonus = .05 * price

Write

Price, bonus

Stop

Start

Read

uprice, qty


3. Variables:

num1 (int) to hold the first value

num2 (int) to hold the second value

Flowchart

False True


int num1, num2;

cin >> num1 >> num2;

if (num1 > num2)

{

num1 = num1 + 3;

cout << (num1 + num2);

cout << (num1 – num2);

}

else

{

num1 = num1 – 2;

cout << (num1 * num2);

cout << (num2 / 2);

}

cout << 2 * num1 – num2;

Start

Read

num1, num2

num1 > num2

Write

num1 + num2, num1 – num2 Write

num1 * num2, num2 / 2

Write

2 * num1 – num2

Stop

num1 = num1 - 2 num1 = num1 + 3

4. Variables symb (char) to hold the character value

Flowchart

False True

c. C++ Language Code Segment:

char symb;

cin >> symb;

if ( (symb >= ‘A’ && symb <= ‘Z’) || (symb >= ‘a’ && symb <= ‘z’) )

cout << endl << symb << “LETTER”;

else

cout << endl << symb << “NOT A LETTER”;

5. Variables

prodCount (int) to count the products

uprice (double) to hold the unit price of a product

qty (double) to hold the quantity of a product


totalPrice (double) to hold the current total price of the products

Start

Read

symb

(symb >= ‘A’ && symb <= ‘Z’)

|| (symb >= ‘a’ && symb <= ‘z’)

Write

symb,

“LETTER”

Write

symb, “NOT A

LETTER”

Stop

Flowchart

False True


int pcount = 0;

double uprice, price, qty, totalPrice = 0;

while ( pcount < 20)

{

cout << endl << “enter product unit price and quantity:\t” ;



cout << price

totalPrice = totalPrice + price;

pcount = pcount + 1;

}

cout << endl << “the total price is:\t” << totalPrice;

Start

totalPrice = totalPrice + price

prodCount < 20

Stop

prodCount = 0

Read

uprice, qty

totalPrice = 0


Write

totalPrice

prodCount = prodCount + 1

Write

price

6. Variables:

value (int) to hold the values

total (int) to hold their sum

Flowchart

False True


int value = 1,

total = 0;

while ( value < = 20)

{

total = total + value;

value = value + 1;

}

cout << total;

Start

value = 1

value = value + 1

value < = 20

Stop

total = 0

total = total + value

Write

total

7. Variables:

value (double) to hold the value.

power (double) to hold its powers.

count (int) to count the powers.

Flowchart

False True


double value, power = 1, int count = 0;

cin >> value

while ( count < 6)

{

power = power * value;

count = count + 1;

}

cout << power;

Start

count = 0

count = count + 1

count < 6

Stop

power = 1

power = power * value

Write

power

Read

value

8. Variables

num (int) to hold an integer value

count (int) to count the integer values read

pcount (int) to count the number of positive integer values

ncount (int) to count the number of negative integer values

Flowchart

False True

False True

Start

count = 0

count = count + 1

count < 20

Stop

pcount = 0

Write

pcount, ncount

sum

ncount = 0

Read

value

value > 0

ncount = ncount + 1 pcount = pcount + 1


int value,

pcount = 0,

ncount = 0,

count = 0;

while ( count < 20)

{

cin >> value;

if (value > 0)


else

ncount = ncount + 1;

count = count + 1;

}

cout << endl << “number of positive values:\t” << pcount;

cout << endl << “number of negative values:\t” << ncount;

9. Variables


count (int) to count the integer values read

pcount (int) to count the number of positive integer values

ptotal (int) to hold the sum of the positive integer values

ncount (int) to count the number of negative integer values

ntotal (int) to hold the sum of the negative integer values

Flowchart

Start

pcount = 0

ncount = 0

ptotal = 0

ntotal = 0

A

False True

False True


int value,

pcount = 0,

ptotal = 0

ncount = 0,

ntotal = 0,

count = 0;

while ( count < 20)

{

cin >> value;

if (value > 0)

{

ptotal = ptotal + value;


}

count = 0

count = count + 1

count < 20

Stop

Write

ptotal/pcount

ntotal/ ncount

Read

value

value > 0

ncount = ncount + 1 pcount = pcount + 1

ntotal = ntotal + value

A

ptotal = ptotal + value

else

{

ntotal = ntotal + value;

ncount = ncount + 1;

}

count = count + 1;

}

cout << endl << “average of positive values:\t” << ptotal / pcount;

cout << endl << “average of negative values:\t” << ntotal / ncount;

10.a

Variables

stock (int) to hold the number of units of the product in the store

dailySale (int) to hold a daily sale of the product

Flowchart

False True


int stock = 120,

dailySale;

while (stock > 30)

{

cin >> dailySale;

stock = stock - dailySale;

}

cout << endl << 120 - dailySale << “\nPlace a new order”;

Start

stock = stock - dailySale

stock > 30

Stop

stock = 120

Read

dailySale

Write

120 - stock

“place a new order”

11. Variables

price (double) to hold the price of a book

totalPrice (double) to hold the total price of the books

count (int) to count the books

Flowchart

False True

Start

totalPrice = totalPrice + price

price != -99.00

Stop

totalPrice = 0

Read

price

Write

totalPrice / count

count = 0

Read

price

count = count + 1


double price,

totalPrice = 0;

int count = 0;

cin >> price;

while (price != -99.00)

{

totalPrice = totalPrice + price;

count = count + 1;

cin >> price;

}

cout << endl << “The average price is :\t” << totalPrice / count;

12. Variables


product (int) to hold their product

Flowchart

False True


int num,

product = 1;

cin >> num;

while (num != 0)

{

Product = product * num;

cin >> num ;

}

cout << endl << “Their product is :\t” << product;

Start

product = product * num

num != 0

Stop

product = 1

Read

num

Write

product

Read

num

13. Variables

score (double) to hold a test score

totalScore (double) to hold the sum of all test scores

count (int) to count the test scores

Flowchart

False True


double score,

totalscore = 0;

int count = 0;

cin >> score;

while (score != -99.00)

{

totalScore = totalScore + score;

count = count + 1;

cin >> score;

}

cout << endl << “The average score is :\t” << totalScore / count;

Start

totalScore = totalScore + score

score != -99.00

Stop

totalScore = 0

Read

score

Write

totalScore / count

count = 0

Read

score

count = count + 1

14.

Variables


Flowchart

False True

False True


cin >> num;

if (num < 10)

smallt = smalt + 1;

if (num > 15)

greatt = greatt + 2;

else

smallt = smallt – 3;

somet = somet + 7;

15

Variables


Start

Read

num

num < 5

num > 20

greatt = greatt+ 2

somet = somet + 7

Stop

smallt = smallt + 1

smallt = smallt - 3

Flowchart

False True

False True


cin >> num;

if (num > 100)

large_ct = large_ct + 1;

if (num < 50)

small_ct = small_ct + 2;

else

middle_ct = middle_ct + 3;

sum = sum + num;

16.

Variables

count (int) to count the number of values read

ecount (int) to count the number of even values


Start

Read

num

num > 100

num < 50

small_ct = small_ct + 2

sum = sum + num

Stop

large_ct = large_ct + 1

middle_ct = middle_ct +3

Flowchart

False True

False True


int num,

count = 0,

ecount = 0;

while (count < 50)

{

cin >> num;

if (num % 2 == 0)

ecount = ecount + 1;

count = count + 1;

}

cout << endl << “the number of even values is:\t” << ecount;

Start

Read

num

count = 0

ecount = 0

count < 50

num % 2 = 0

ecount = ecount + 1

count = count + 1

Write

ecount

Stop

17.

Variables

ptotal (int ) to hold the sum of the positive integer values

pcount (int) to count the number of positive values


Flowchart

False True

False True

False True

Start

Read

num

pcount = 0

ptotal = 0

num != -99

num > 0

ptotal = ptotal + num

Write

ptotal / pcount

Stop

Read

num

pcount = pcount + 1

pcount = 0

Write

“No positive values “


int num,

ptotal = 0,

pcount = 0;

cin >> num

while ( num != -99 )

{

if ( num > 0 )

{

Ptotal = ptotal + value;

Pcount = pcount + 1;

}

cin >> num;

}

if( pcount = = 0 )

cout << endl << “there is no positive values ”;

else

cout << endl << The average of positive values is:\t” << ptotal / pcount;

18.

Variables

grade (char) to hold a student’s letter grade.

Flowchart

True False

True False

True False

True False

True False

Start

Read

num1, op, num2

op = ‘+’

op = ‘-’

op = ‘*’

op = ‘/’ &&

num2 != 0

Write

op or num2 is

“invalid“

Stop

Write

num1+ num2

Write

num1 – num2

Write

num1 * num2

Write

num1 / num2

Write

num1 % num2

op = ‘\%’ &&

num2 != 0


int num1, num2;

char op;

cin >> num1 >> op >> num2;

if ( op = = ‘+’)

cout << endl << num1 + num2;

else if ( op = = ‘-’)

cout << endl << num1 - num2;

else if ( op = = ‘*’)

cout << endl << num1 * num2;

else if (op = = ‘/’ && num2 != 0 )

cout << endl << num1 / num2;

else if ( op = = ‘\% ’ && num2 != 0 )

cout << endl << num1 % num2;

else

cout << endl << op << “or” << num2 << “ is invalid”;

19. Variables

uprice (double) to hold the unit price of the product.


qty (int) to hold the number of items purchased.

rebate (double) to hold the rebate

Flowchart

True False

True False

True False

price >= 100.00

rebate = 10.00

rebate = 5.00

price >= 50.00

price > = 20.00

rebate = 2.00 rebate = 0.00

Start

Read

uprice, qty


A


double uprice, price, rebate;

int qty;



if (price >= 100.00)

rebate = 20.00;

else if (price >= 50)

rebate = 5.00;

else if (price >= 20.00)

rebate = 2.00;

else

rebate = 0.00;

price = price – rebate;

cout << endl << “amount due is:\t” << (price + 0.0825 * price);

20.

Variables

symbol (char) to hold a character

cin >> symbol;

if ((symbol >= ‘a’ && symbol <= ‘z’) || (symbol >= ‘A’ && symbol <= ‘Z’)

cout << endl << “letter”;

else if (symbol >= ‘0’ && symbol <= ‘9’)

cout << endl << “digit”;

else

cout << endl << “special character”;

Write

price + 0.0825 * price

Stop

price = price - rebate

A

exercise 3 - william paterson universitycs.wpunj.edu/~ndjatou/cs2300-chapter3-solutions.pdfexercise...

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