12 to 19 Experiements

8/2/2019 12 to 19 Experiements

1/22

12 to 19EXPERIMENT

CONDITIONAL STATEMENTS

Ex No: 12 ROOTS OF A QUADRATIC EQUATION

AIM:

To write a program to find the roots of the Quadratic Equation.

Formula:

D=B2-4AC

Condition 1: D= =0

ROOT1=ROOT2= (-B+ D)/ (2*A)

Condition 2: D>0

ROOT1= (-B+D)/ (2*A)

ROOT2= (-B-D)/ (2*A)Condition 3: D


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Step 3: If D is equal to zero then

3.1 Print Roots are real and equal

3:2 Find the two roots as ROOT1 = (-B + SQRT ((D)) / (2*A)

3.3 Print the ROOT1 two times.

Step 4: If D is greater than zero then

4.1 Print Roots are real and unequal

4.2 Find the two roots as

ROOT1 = (-B + SQRT ((D)) / (2 * A)

ROOT2 = (-BSQRT ((D)) / (2 * A)

4.3 Print ROOT1 and ROOT2

Step 5: If D is less than zero then

5.1 Print the Roots are imaginary

5.2 Find the two roots as

X = (-B) / (2 * A)

Y=-D/ (2*A)

5.3 Print them in the format of (X + iY), (XiY)


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/* Root of a Quadratic Equation */

#include

#include

#include

void main()

{

float a,b,c,d,r,r1,r2;clrscr();

printf("\n\n Enter the a,b and c values : ");

scanf("%f %f %f",&a,&b,&c);

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

r=sqrt(d);

if(d==0)

{

printf("\n\n Roots are Equal");

r1=(-b+r)/(2*a);

printf("\n\n Both Roots Values : %f",r1);

}

else if(d>0)

{

printf("\n\n Roots are Unequal");

r1=(-b+r)/(2*a);

r2=(-b-r)/(2*a);printf("\n\n Root1 and Root2 Values are : %f , %f",r1,r2);

}

else

{

printf("\n\n Roots are Imaginary");

r1=-b/(2*a);


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r2=r/(2*a);

printf("\n\n Root1 and Roots are : (%f + i%f) , (%f - i%f)",r1,r2,r1,r2);

}

getch();

OUTPUT FOR ROOTS OF QUADRATIC EQUATION

Enter the a,b and c values : 10 5 6

Roots are Imaginary

Root1 and Roots are: (0.250000 + i0.000000), (0.250000i0.000000)

Enter the a, b and c values: 5 20 10

Roots are Unequal

Root1 and Root2 Values are: 0.585786,3.414214

Enter the a, b and c values: 10 20 10

Roots are Equal

Both Roots Values: 1.000000


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Ex No: 13 ELECTRICITY BILL PREPARATION

AIM

To write a program to prepare the electricity bill.

An electric power distribution company charges its domestic consumers as

follows:

CONSUMPTION UNITS RATE OF CHARGE

0-200 Rs.0.50 per unit

201-400 Rs 100+ Rs. 0.65 per unit excess of 200

401-600 Rs 230+ Rs. 0.80 per unit excess of 400

601 and above Rs 390+ Rs. 1 per unit excess of 600

ALGORITHM:

Step 1: Enter the value units

Step 2: If units are in the range 0-200

Charges=0.50*units


Charges=100 + 0.65*units



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Charges=230 + 0.80*units

Step 5: If units are in the range 601 and above

Charges=390 + 1*units

Step 6: Print the charges

/* Electricity Bill Preparation */#include

#include

void main()

{

float units,charges;

clrscr();

printf("\n\n Enter the Units Values : ");

scanf("%f",&units);

if(units>=0 && units=201 && units=401 && units


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Enter the Units Values: 553

Total Amount Is Rs. 672.40

Enter the Units Values: 1129

Total Amount Is Rs. 1519.40

Ex No: 14 REVERSING THE GIVEN NUMBER

AIM:

To write a program to reverse the given integer using while loop.

ALGORITHM:

Step 1: Initialize SUM as 0

Step 2: Enter the number, N to be reversed

Step 3: While the number, N is not less than zero do

3.1 R = Find the remainder of N divided by 10

3.2 SUM = Multiply SUM by 10 and add to R

3.3 N = Find the Quotient of N divided by 10

Step 4: Print the value of SUM, which is the reversal of a given number.


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/* Reverse Order Number */#include

#include

void main()

{

int num,mod,rev=0;

clrscr();

printf("\n\n Enter a Number: ");

scanf("%d", &num);while(num>0)

{

mod=num%10;

rev=(rev*10)+mod;

num=num/10;

}

printf("\n\n Reverse of the given Number: %d", rev);

getch();

}

OUTPUT FOR REVERSING THE GIVEN NUMBER

Enter a Number: 5867

Reverse of the given Number: 7685


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Ex No: 15 ARMSTRONG NUMBER GENERATION

AIM:

To write a program to generate Armstrong numbers using do while loop.

ALGORITHM:

Step 1: Get limit of the generation N

Step 2: Do the following for N times

Step 2.1: SUM = sum of cubes of each digit of original number

Step 2.2: If original number and SUM is same, then print the original

number.


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/* Armstrong Number */

#include

#include

void main()

{

int a,n,b=0,t;

clrscr();printf("\n\n Enter the Number : hmm");

scanf("%d",&n);

t=n;

while(n>0)

{

a=n%10;

b=b+(a*a*a);

n=n/10;

}

if(b==t)

{

printf("\n\n This is Armstrong Number");

}

else

printf("\n\n This is not an Armstrong Number");

getch();

}

OUTPUT FOR ARMSTRONG NUMBER

Enter the Number: 547

This is Not an Armstrong Number.


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Enter the Number: 407

This is Armstrong Number.

Ex No: 16 SINE SERIES CALCULATION

AIM:

To write a program to calculate Sine value using its series using for loop.

Formula:

Sin(x) =x - x3/3! +X

5/5!-x

7/7! +-----

Hint: Similarly do for the cosine series

Cos (x) =1-x2/2! +X

4/4!-x

6/6!

ALGORITHM:

Step 1: Get a degree X and No of terms in the series to be generated N

Step 2: Assign X to Val, T, Sum for future purpose & the radian value of X

in X.

Step 3: Do the following for N+ 1 time

Step 3.1: Find out each term in the series T

Step 3.2: Find out sum of the terms T in SUM

Step 4: Print SUM as Sin(X).


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/* Conditional Statements */

/* Sine Series */

#include

#include

void main()

{

int i = 2, n, s = 1, x, pwr = 1, dr;float nr = 1, x1, sum;

clrscr();

printf("\n\n **** Conditional Statements ****");

printf("\n\n **** 8.Sine Series ****");

printf("\n\n Enter the Angle : ");

scanf("%d", &x);

x1 = 3.142 * (x / 180.0);

sum = x1;

printf("\n\n Enter the number of Terms : ");

scanf("%d", &n);

while(i


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**** 8.Sine Series ****

Enter the Angle : 65

Enter the number of Terms : 11

The Sum of the Sine Series is : 1.135

Ex No: 17 SORTING THE GIVEN ARRAY OF NUMBERS

AIM:

To write a program to sort the given array of numbers in ascending /

descending.

ALGORITHM:

Step 1: Enter the size of the array N.

Step 2: Enter the elements of the array A.

Step 3: Set an outer loop up to N-1

Step 3.1: Set an inner loop up to N-1

Step 3.2: If first number is greater than next number, then swap them

Step 4: Print the ascending order of the given array (from 0th

element to nth

element).

Step 5: Print the descending order of the given array (from nth

element to 0th

element).


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/* Arrays */

/* 1. Array Sorting */

#include

#include

void main()

{

int a[100],n,i,j,t;clrscr();

printf("\n\n Enter Integer value for total number of elements to be sorted: ");

scanf("%d",&n);

printf("Enter the values for %d elements",n);

for(i=0;i


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}

OUTPUT:

Enter Integer value for total number of elements to be sorted: 5

Enter the values for 5 elements

1211

10

9

8

Sorted Array

8 9 10 11 12


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Ex No: 18 MATRIX ADDITION

AIM:

To write a program to ADD two matrices.

ALORITHM:

Step1: Enter the row I and column J of the matrix.

Step2: Enter the elements of the matrix A.

Step 3: Enter the elements of the matrix B.

Step 4: Print the matrix A in the matrix form.

Step 5: Print the matrix B in the matrix form.

Step 6: Set a loop up for row values

Step 6.1: Set an inner loop up for column values.

Step 6.2: Add the elements of A and B in column wise and store the

result in matrix C.

Step 7: Print matrix C.


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/* Matrix Addition */

#include

#include

Void main ()

{

int a[3][3],b[3][3],c[3][3],i,j,n,m;

clrscr();printf("\n\n Enter the Rows and Columns Value : ");

scanf("%d %d",&n,&m);

printf("\n enter First Matrix Value :\n");

for(i=0;i


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printf("%d\t",c[i][j]);

}

}

getch();

}

OUTPUT:

Enter the Rows and Columns Value:

2

2

Enter First Matrix Value:

3 4

5 6

Enter the Second Matrix Value:

1 2

7 8

The Addition of two Matrixes

4 6


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Ex No: 19 STUDENTS MARKS CALCULATION USING

STRUCTURE

AIM:

To write a program to prepare students mark list using structure.

ALGORITHM:

Step1: Define the STUDENTS structure with details of Roll no, Name,

Marks

Step2: Get the number of students as N.

Step3: Declare a structure variable with size N.

Step4: Do the following for N times

Step4.1: Get the details of the student.

Step4.2: Calculate total, average, and Grade

Step4.3: Print the students individual Mark list with total, average

and Grade.


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PROGRAM:

/*STUDENT MARKLIST USING STRUCTURES*/

#include

#include

struct student {

char name[25],regno[10],grade;

int mark1,mark2,mark3,total;

float percentage;

}stud[50];

void main()

{

int no,i;

clrscr();

printf("\n\n Enter the Number of Students : ");

scanf("%d",&no);

for(i=0;i


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scanf("%d",&stud[i].mark2);

printf("\n Enter The Student Mark3 : ");

scanf("%d",&stud[i].mark3);

stud[i].total=stud[i].mark1+stud[i].mark2+stud[i].mark3;

stud[i].percentage= stud[i].total/3;

if(stud[i].percentage


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Enter the Student Mark1: 56



Details for student 0

Anand 118071001 90 78 79 247 82.00 B

Details for student 1

Sathiya 118071007 56 67 57 180 60.00 C

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