C++ TUTORIAL 3

SJEM2231 STRUCTURED PROGRAMMING

NAME: WAN MOHAMAD FARHAN BIN AB RAHMAN

TUTORIAL 3/ LAB 3 (13rd OCTOBER 2014)

(Use different loop structures for problems 2, 4 and 5, and understand how it works)

QUESTION 1

Write a program to find the given number is odd or even. First test whether the given

number is non zero positive number.

#include <iostream>

using namespace std;

int main()

{

int number;

ULANG:

cout << "Please enter any positive integer: ";

cin >> number;

if(number > 0)

{

cout << "The given number is non zero positive number"<<endl;

}

else

{

cout << "The given number is not a positive number. Try again" <<endl;

goto ULANG;

}

if(number % 2 ==0)

cout <<"Thus, " << number << " is an even number" << endl;

else

cout <<"Thus, " << number << " is an odd number" <<endl;

return 0;

}

// Output :

Please enter any positive integer: -1

The given number is not a positive number. Try again

Please enter any positive integer: 0

The given number is not a positive number. Try again

Please enter any positive integer: 5

The given number is non zero positive number

Thus, 5 is an odd number

//Alternative method for question 1

#include<iostream>


int main()

{

int n;

cout << "Enter a number: ";

cin >> n;

if (n>0 && n%2==0)

{

cout<<"\nThe number is non zero positive number"<<endl;

cout << "The number is even number"<<endl;

}

else if (n>0 && n%2!=0)

{

cout<<"\nThe number is non zero positive number"<<endl;

cout << "The number is odd number"<<endl;

}

else

cout<<"\nThe number is zero or negative number"<<endl;

return 0;

}

//Output:

Enter a number: 0

The number is zero or negative number

Enter a number: 8

The number is non zero positive number

The number is even number

Enter a number: 11

The number is non zero positive number

The number is odd number

QUESTION 2

Write a program to print the multiplication table as shown below(lower triangular matrix).

1

2 4

3 6 9

4 8 12 16

5 10 15 20 25

…

10 20 30 40 50 60 70 80 90 100

//Using for loop :

#include <iostream>


int main()

{

for (int i=1 ; i<=10; i++)

{

for(int j=1; j<=10; j++)

{

if (j>i) break;

cout << i*j << "\t";

}

cout << endl;

}

return 0;

}

// Or we can write this way, still using for loop but slightly different:

#include <iostream>


int main()

{

for (int i=1 ; i<=10; i++)

{

for(int j=1; j<=i; j++)

{

cout << i*j << "\t";

}

cout << endl;

}

return 0;

}

//Using while loop

#include<iostream>


int main()

{

int i=1;

while(i<=10)

{

int j=1;

while(j<=i)

{

cout<<i*j <<"\t";

j++;

}

cout <<endl;

i++;

}

return 0;

}

//Using do while loop

#include<iostream>


int main()

{

int i=1, j=1;

do

{

do

{

/*We may replace int j=1 here instead of declaring j=1 above */

cout << i*j <<"\t";

j++;

} while(j<=i);

j=1; /*we can eliminate this j=1 if we declare int j=1 above */

cout<<endl;

i++;

} while(i<=10)

return 0;

}

//The output we will get is just the same:

1

2 4

3 6 9

4 8 12 16

5 10 15 20 25

6 12 18 24 30 36

7 14 21 28 35 42 49

8 16 24 32 40 48 56 64

9 18 27 36 45 54 63 72 81

10 20 30 40 50 60 70 80 90 100

QUESTION 3

Write a program that solves quadratic equations of the form ax2 + bx + c = 0, where a, b,

and c are given real number and x is the unknown

(This will NOT apply if a is zero, so that condition must be checked separately. The

formula also fails to work (for real numbers) if the expression under the square root is

negative)

#include <iostream>

#include <cmath>


int main()

{

// Variable Declarations

double a, b, c;

//Variable Inputs

cout << "Enter the value of a: ";

cin >> a;

cout << "Enter the value of b: ";

cin >> b;

cout << "Enter the value of c: ";

cin >> c;

//Computations

double discriminant = (pow(b,2) - 4*a*c);

double positive_root = (((-b) + sqrt(discriminant))/(2*a));

double negative_root = (((-b) - sqrt(discriminant))/(2*a));

if (discriminant == 0)

{

cout << "\nThe discriminant is "<< discriminant << endl;

cout << "The equation has a single root.\n";

}

else if (discriminant < 0)

{

cout << "\nThe discriminant is "<< discriminant << endl;

cout << "The equation has two complex roots.\n";

}

else

{

cout << "\n\nThe discriminant is " << discriminant << endl;

cout << "The equation has two real roots.\n";

}

//Final Root Values

cout << "The roots of the quadratic equation are x = ";

cout << negative_root<< "," << positive_root << endl;

return 0;

}

//Output:

Enter the value of a: 1

Enter the value of b: 6

Enter the value of c: 8

The discriminant is 4

The equation has two real roots.

The roots of the quadratic equation are x = -4,-2

//Alternative method for question 3

#include<iostream>

#include<cmath>


int main()

{

double a, b, c, discriminant, root_1, root_2;

cout<<"Please enter value of a: ";

cin>>a;

cout<<"\nPlease enter value of b: ";

cin>>b;

cout<<"\nPlease enter value of c: ";

cin>>c;

discriminant= (b*b)-(4*a*c);

if (discriminant>0)

{

cout<<"\nThe equation has 2 distinct root" <<endl;

}

else if(discriminant==0)

{

cout<<"\nThe equation has equal root"<<endl;

}

else if(discriminant<0)

{

cout<<"\nThe equation has complex root" <<endl;

}

root_1= ((-b + (sqrt(d)))/(2*a));

root_2= ((-b - (sqrt(d)))/(2*a));

cout<< "The first root is "<<root_1 <<"\n"<<"while the second root is " << root_2 <<endl;

return 0;

}

//Output:

Please enter value of a: 1

Please enter value of b: 6

Please enter value of c: 8

The equation has 2 distinct root

The first root is -2

while the second root is -4

QUESTION 4

Write a program to find the mean (average) of N numbers.

//Using while loop

#include<iostream>


int main()

{

int N, i, b, sum=0;

float average;

cout<<"Please enter the value of N: ";

cin>> N;

i =1;

while(i<=N)

{

cout << "Value " << i << " is:";

cin >> b;

sum= sum + b;

i++;

}

cout<<"The sum is: "<< sum <<endl;

average= float (sum)/ N;

cout<<"Thus, the average is: "<<average<< endl;

return 0;

}

//Output for while loop:

Please enter the value of N: 5

Value 1 is:10

Value 2 is:7

Value 3 is:3

Value 4 is:2

Value 5 is:114

The sum is: 136

Thus, the average is: 27.2


#include<iostream>


int main()

{

int i=1, b,sum=0,N;

float average;

cout<<"Please enter a number N: ";

cin>> N;

do

{

cout << "Value " << i << " is:";

cin >> b;

sum=sum+b;

i++;

} while(i<=N);

cout<<"\nThe summation of N number is: " << sum <<endl;

average= float (sum)/N;

cout<<"The average is " << average <<endl;

return 0;

}

//Output for do while loop


Value 1 is:10

Value 2 is:7

Value 3 is:3

Value 4 is:2

Value 5 is:114

The sum is: 136


//Using for loop

#include<iostream>


int main()

{

int i, b, sum=0, N;

float average;


cin>> N;

for (i=1; i<=N ; i++)

{

cout << "Value " << i << " is: ";

cin >> b;

sum= sum +b;

average= float(sum)/N;

}

cout<<"\nThe summation of N number is: "<< sum <<endl;

cout<<"The average is " << average <<endl;

return 0;

}

//Output for for loop:


Value 1 is:10

Value 2 is:7

Value 3 is:3

Value 4 is:2

Value 5 is:114

The sum is: 136


QUESTION 5

Write a program to find the sum of EVEN numbers (2+4+6+…+N) and ODD numbers

(1+3+ ... +N) up to N numbers.

//Using for loop

#include<iostream>


int main()

{

int i, N, sum_odd=0, sum_even=0;


cin>>N;

for(i=1; i<=N; i++)

{

if(i%2==0)

sum_even=sum_even +i;

}

cout<<"\nThe summation of even number of N is: " <<sum_even <<endl;

for(i=1; i<=N; i++)

{

if(i%2!=0)

sum_odd=sum_odd +i;

}

cout<<"The summation of odd number of N is: " <<sum_odd <<endl;

return 0;

}

//Output for for loop

Please enter a number N: 10

The summation of even number of N is: 30

The summation of odd number of N is: 25


#include <iostream>


int main()

{

int i=1, N, sum_even=0, sum_odd=0;

cout << "Enter any positive integer, N: ";

cin >> N;

while (i<= N)

{

if (i%2 ==0)

sum_even = sum_even + i;

else

sum_odd = sum_odd + I;

i++;

}

cout << "\nThe summation of even numbers is " << sum_even << endl;

cout << "The summation of odd numbers is " << sum_odd << endl;

return 0;

}

//Output for while loop

Enter any positive integer, N: 10

The summation of even numbers is 30

The summation of odd numbers is 25

//Using do while loop:

#include <iostream>


int main()

{

int i=1, N, sum_even=0, sum_odd=0;

cout << "Enter any positive integer, N: ";

cin >> N;

do

{

if (i%2 ==0)

sum_even = sum_even + i;

else

sum_odd = sum_odd + i;

i++;

} while (i<= N);

cout << "\nThe summation of even numbers is " << sum_even << endl;

cout << "The summation of odd numbers is " << sum_odd << endl;

return 0;

}

//Output for do while loop

Enter any positive integer, N: 10

The summation of even numbers is 30

The summation of odd numbers is 25

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C++ TUTORIAL 3