programming language codes

Python program to check prime numbers

2025-08-01T08:23:00.003+05:30

# Check if a number is prime

num = int(input("Enter a number: "))

if num > 1:

for i in range(2, int(num ** 0.5) + 1):

if num % i == 0:

print(f"{num} is not a prime number")

break

else:

print(f"{num} is a prime number")

else:

print(f"{num} is not a prime number")

Python program to check armstrong number

2025-08-01T07:27:00.001+05:30

# Check if a number is an Armstrong number

num = int(input("Enter a number: "))

order = len(str(num))

sum = 0

temp = num

while temp > 0:

digit = temp % 10

sum += digit ** order

temp //= 10

if num == sum:

print(f"{num} is an Armstrong number")

else:

print(f"{num} is not an Armstrong number")

Python program to calculate compund interest

2025-08-01T07:26:00.002+05:30

# Calculate simple interest

principal = float(input("Enter principal amount: "))

rate = float(input("Enter rate of interest: "))

time = float(input("Enter time (in years): "))

si = (principal * rate * time) / 100

print(f"Simple Interest: {si}")

Python program to calculate simple interest

2025-08-01T07:25:00.002+05:30

# Calculate simple interest

principal = float(input("Enter principal amount: "))

rate = float(input("Enter rate of interest: "))

time = float(input("Enter time (in years): "))

si = (principal * rate * time) / 100

print(f"Simple Interest: {si}")

Python program for finding factorial

2025-08-01T07:24:00.004+05:30

# Find factorial of a number

num = int(input("Enter a number: "))

factorial = 1

for i in range(1, num + 1):

factorial *= i

print(f"Factorial of {num}: {factorial}")

Python program to add to numbers

2025-08-01T07:22:00.000+05:30

# Program to add two numbers entered by the user

# Take input from the user

num1 = input("Enter the first number: ")

num2 = input("Enter the second number: ")

# Convert inputs to float (to support decimals)

sum = float(num1) + float(num2)

# Display the result

print("The sum of {0} and {1} is {2}".format(num1, num2, sum))

c program of gauss ELIMINITION method

2012-05-22T11:13:00.000+05:30

/*Gauss Elimination */
# include <stdio.h>
# include <conio.h>
# include <math.h>
# define MAX 10
void main()
{
int i,j,n,k;
float mat[MAX][MAX],x[MAX],temp,pivot,sum=0;
clrscr();
printf("\t\t\t GAUSS ELIMINITION METHOD\n");
printf("-------------------------------------------------------------------\n");
printf("Enter No of Equtions : ");
scanf("%d",&n);
printf("Enter Coefficients of Eqution \n");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%f",&mat[i][j]);
printf("Enter Constant value\n");
for(i=1;i<=n;i++)
{
scanf("%f",&mat[i][n+1]);
x[i]=mat[i][n+1];
}
for(i=2;i<=n;i++)
{
for(j=i;j<=n;j++)
{
pivot=mat[j][i-1]/mat[i-1][i-1];
for(k=i-1;k<=n+1;k++)
mat[j][k]=mat[j][k]-pivot*mat[i-1][k];
}
}
printf("Eliminated matrix as :- \n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n+1;j++)
printf("\t%.2f",mat[i][j]);
printf("\n");
}
for(i=1;i<=n;i++)
{
if(mat[i][i]==0)
{
printf("Since diagonal element become zero\n Hence solution is not possible\n");
exit(1);
}
}
printf("Solution : \n");
for(i=0;i {
sum=0;
for(j=n;j>n-i;j--)
sum=sum+mat[n-i][j];

x[n-i]=(mat[n-i][n+1]-sum*x[n])/mat[n-i][n-i];
printf("X%d = %4.2f\n",n-i,x[n-i]);
}

getch();
}

OUTPUT :-

GAUSS ELIMINITION METHOD
------------------------------------------------------------------------------
Enter No of Equtions : 3
Enter Coefficients of Eqution
4 3 -2
1 1 1
3 -2 1
Enter Constant value
5 3 2
Eliminated matrix as :-
4.00 3.00 -2.00 5.00
0.00 0.25 1.50 1.75
0.00 0.00 28.00 28.00
Solution :
X3 = 1.00
X2 = 1.00
X1 = 1.00

facebook page of the blog

2012-05-22T11:09:00.000+05:30

click here to see the facebook page of this blog.
click on LIKE................. :)

C Program Of NA Using Trapezoidal Interpolation

2012-05-22T10:50:00.003+05:30

/*Trapezoidal rule */
# include <stdio.h>
# include <conio.h>
# define MAX 20
void main()
{
int i,j,n;
float x[MAX],y[MAX],sum=0,h,res;
clrscr();
printf("Solution using Trapezoidal rule \n");
printf("Enter Number of data points : ");
scanf("%d",&n);
printf("Enter value of table value set by set\n");
for(i=1;i<=n;i++)
{
printf("Enter value of x[%d] and y[%d] ",i,i);
scanf("%f%f",&x[i],&y[i]);
}
h=x[2]-x[1];
printf("\nInterval :=%f\n",h);
sum=y[1]+y[n];
for(i=2;i<n;i++)
sum=sum+y[i]*2;

res=sum*(h/2);
printf ("Integral from %3.2f to %3.2f is : = %f",x[1],x[n],res);
getch();
}
OUTPUT :-
Solution using Trapezoidal rule
Enter Number of data points : 6
Enter value of table value set by set
Enter value of x[1] and y[1] 7.47 1.93
Enter value of x[2] and y[2] 7.48 1.95
Enter value of x[3] and y[3] 7.49 1.98
Enter value of x[4] and y[4] 7.50 2.01
Enter value of x[5] and y[5] 7.51 2.03
Enter value of x[6] and y[6] 7.52 2.06

Interval :=0.010000
Integral from 7.47 to 7.52 is : = 0.099652

C Program Of NA Using Simpson 1/3 Rule

2012-05-22T10:48:00.000+05:30

/*Simpson 1/3 Rule */
# include <stdio.h>
# include <conio.h>
# define MAX 20
void main()
{
int i,j,n;
float x[MAX],y[MAX],sum=0,h,res;
clrscr();
printf("Solution of 1/(1+x)dx using Simpson's 1/3 rule \n");
printf("Enter Number of data points : ");
scanf("%d",&n);
printf("Enter value of table value set by set\n");
for(i=1;i<=n;i++)
{
scanf("%f",&x[i]);
y[i]=1/(x[i]+1);
}
printf("X \t\t Y\n");
for(i=1;i<=n;i++)
printf("%f\t%f\n",x[i],y[i]);
h=x[2]-x[1];
printf("\nSegment Interval `h':=0.25\n");
sum=sum+y[1]+y[n];
for(i=2;i<n;i++)
{
if(i%2==0)
sum=sum+y[i]*4;
else
sum=sum+y[i]*2;
}
res=sum*(0.25/3);
printf ("Integral from %3.2f to %3.2f is : = %f",x[1],x[n],res);
getch();
}
OUTPUT
Solution of 1/(1+x)dx using Simpson's 1/3 rule
Enter Number of data points : 5
Enter value of table value for x column
0 .25 .50 .75 1
X Y
0.000000 1.000000
0.250000 0.800000
0.500000 0.666667
0.750000 0.571429
1.000000 0.500000

Segment Interval `h':=0.25
Integral from 0.00 to 1.00 is : = 0.693254

C Program Of NA Using Simpson 3/8 Rule

2012-05-22T10:46:00.000+05:30

/*Simpson 3/8 rule */
# include <stdio.h>
# include <conio.h>
# define MAX 20
void main()
{
int j,n=0;
float x[MAX],y[MAX],sum=0,h,res,lb,ub,i;
clrscr();
printf("Solution of 1/(1+x)dx using Simpson's 3/8 rule \n");
printf ("Enter initial value : ");
scanf("%f",&lb);
printf("Enter final value : ");
scanf("%f",&ub);
printf("Enter no. of Subinterval : ");
scanf("%f",&h);
h=(ub-lb)/h;
for(i=lb;i<=ub;i=i+h)
{
x[n]=i;
y[n]=1/(1+x[n]);
n++;
}
printf("No.\t X\t\tY=1/(1+x)\n");
printf("---------------------------------------\n");
sum=y[0]+y[n-1];
for(j=0;j<n;j++)
printf("%d\t%f\t%f\n",j,x[j],y[j]);
printf("---------------------------------------\n");
for(j=1;j<n-1;j++)
{
if(j%3==0)
sum=sum+y[j]*2;
else
sum=sum+y[j]*3;
}
res=((3*h)/8)*sum;
printf("\nIntegral from %f to %f when h = %f is = %f",x[0],x[n-1],h,res);
getch();
}

OUTPUT

Solution of 1/(1+x)dx using Simpson's 3/8 rule
Enter initial value : 6
Enter final value : 15
Enter no. of Subinterval : 8
No. X Y=1/(1+x)
---------------------------------------
0 6.000000 0.142857
1 7.125000 0.123077
2 8.250000 0.108108
3 9.375000 0.096386
4 10.500000 0.086957
5 11.625000 0.079208
6 12.750000 0.072727
7 13.875000 0.067227
8 15.000000 0.062500
---------------------------------------

Integral from 6.000000 to 15.000000 when h = 1.125000 is = 0.817303

C Program Of NA Using Runga-Kutta Mthod Of 4th Order

2012-05-22T10:41:00.000+05:30

/* Runge-kutta method of 4th order to solve 10*dy/dx=(x*x)+(y*y)*/

# include <stdio.h>

# include <conio.h>

void main()

{

int c=0;

float x,y0,xp,h,i,y1,m1,m2,m3,m4;

float f(float,float);

clrscr();

printf("Solution by Runge kutta 2nd order Method\n");

printf("Enter initial Boundry condition x,y : ");

scanf("%f%f",&x,&y0);

printf("Enter Value of X at which Y is required : ");

scanf("%f",&xp);

printf("Enter Interval ,h : ");

scanf("%f",&h);

printf(" No. X\t m1 \t m2\t m3 \t m4 \t Y\n");

printf("-------------------------------------------------------------------\n");

for(i=x;i<=xp;i=i+h)

{

c++;

m1=h*f(i,y0);

m2=h*f(i+(h/2),y0+(m1/2));

m3=h*f(i+(h/2),y0+(m2/2));

m4=h*f(i+h,y0+m3);

y1=y0+((m1+2*m2+2*m3+m4)/6);

printf("%2d %5.6f %5.6f %5.6f %5.6f %5.6f %5.6f\n",c,i,m1,m2,m3,m4,y1);

y0=y1;

}

getch();

}

float f(float x,float y)

{

return ((x*x)+(y*y))/10;

}

OUTPUT

Solution by Runge kutta 2nd order Method

Enter initial Boundry condition x,y : 0 1

Enter Value of X at which Y is required : .09

Enter Interval ,h : .03

No. X m1 m2 m3 m4 Y

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

1 0.000000 0.003000 0.003010 0.003010 0.003021 1.003010

2 0.030000 0.003021 0.003033 0.003033 0.003047 1.006043

3 0.060000 0.003047 0.003062 0.003062 0.003079 1.009106

4 0.090000 0.003079 0.003097 0.003097 0.003117 1.012204

Program in C to insert a node at the last position in a given doubly linked list.

2012-04-09T21:43:00.002+05:30

#include<stdio.h>

#include<conio.h>

#include<alloc.h>

#define null 0

void main()

{

struct dnode

{

struct dnode *prev;

int data;

struct dnode *next;

} *temp,*start,*end,*n;

int i;

char c;

clrscr();

start=end=null;

for(i=0;i<=1;i++)

{

if(start==null)

{

printf("enter data for the node:");

scanf("%d",&temp->data);

temp->next=null;

temp->prev=null;

start=temp;

end=temp;

}

else

{

while(1)

{

printf("are you want to add another node(y/Y):");

fflush(stdin);

scanf("%c",&c);

if(c=='y'||c=='Y')

{

n=(struct dnode*)malloc(sizeof(struct dnode));

printf("enter data for the node:");

scanf("%d",&n->data);

n->next=null;

n->prev=temp;

temp->next=n;

temp=n;

end=n;

}

else

break;

}

}

}

printf("the list you created is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

printf("addition at last place\n");

{

n=(struct dnode *)malloc(sizeof(struct dnode));

printf("enter the data for the node:");

scanf("%d",&n->data);

n->prev=end;

n->next=null;

end->next=n;

end=n;

}

printf("after operation list is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

getch();

}

OUTPUT:

Enter data for the node:45

Are you want to create another node(y/Y):Y

Enter data for the node:65

Are you want to create another node(y/Y):n

The list you created

45 65

Addition at the first place

Enter data for the node:22

After operation list is: 45 65 22

Program in C to insert a node at first position in a given doubly linked list.

2012-04-09T21:33:00.000+05:30

#include<stdio.h>

#include<conio.h>

#include<alloc.h>

#define null 0

void main()

{

struct dnode

{

struct dnode *prev;

int data;

struct dnode *next;

} *temp,*start,*end,*n;

int i;

char c;

clrscr();

start=end=null;

for(i=0;i<=1;i++)

{

if(start==null)

{

printf("enter data for the node:");

scanf("%d",&temp->data);

temp->next=null;

temp->prev=null;

start=temp;

end=temp;

}

else

{

while(1)

{

printf("are you want to add another node(y/Y):");

fflush(stdin);

scanf("%c",&c);

if(c=='y'||c=='Y')

{

n=(struct dnode*)malloc(sizeof(struct dnode));

printf("enter data for the node:");

scanf("%d",&n->data);

n->next=null;

n->prev=temp;

temp->next=n;

temp=n;

end=n;

}

else

break;

}

}

}

printf("the list you created is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

printf("addition at first place\n");

{

n=(struct dnode *)malloc(sizeof(struct dnode));

printf("enter the data for the node:");

scanf("%d",&n->data);

n->prev=null;

n->next=start;

start->prev=n;

start=n;

}

printf("after operation list is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

getch();

}

OUTPUT:

Enter data for the node:45

Are you want to create another node(y/Y):Y

Enter data for the node:65

Are you want to create another node(y/Y):n

The list you created

45 65

Addition at the first place

Enter data for the node:22

After operation list is: 22 45 65

Program in C to subtract two polynomials.

2012-04-09T21:30:00.001+05:30

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#define NULL 0

void main()

{

struct poly

{

int coff;

int expo;

struct poly *link;

}*temp1,*start1,*temp2,*start2,*start3,*temp3;

int n,i,z=1,num;

char c;

clrscr();

start1=NULL;

printf("\tPOLYNOMIAL CREATION IN LINK LIST AND SUBTRACTION\n\n");

while(1)

{

if(start1==NULL)

{

temp1=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and Exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

start1=temp1;

temp1->link=NULL;

}

else

{

temp1->link=(struct poly*)malloc(sizeof(struct poly));

temp1=temp1->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp1->link=NULL;

break;

}

}

start2=NULL;

temp2=NULL;

z=1;

while(1)

{

if(start2==NULL)

{

temp2=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

start2=temp2;

temp2->link=NULL;

}

else

{

temp2->link=(struct poly*)malloc(sizeof(struct poly));

temp2=temp2->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp2->link=NULL;

break;

}

}

// TARVERSING

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

printf("Traversal of Polynomial Linked List 1\n");

while(temp1!=NULL)

{

printf("%2dx^%d-",temp1->coff,temp1->expo);

temp1=temp1->link;

}

printf("\b \b");

printf("\nTraversal of Polynomial Linked List 2\n");

while(temp2!=NULL)

{

printf("%2dx^%d -",temp2->coff,temp2->expo);

temp2=temp2->link;

}

printf("\b \b");

// subtraction

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

temp3=NULL;

start3=NULL;

while(1)

{

if((temp1!=NULL)||(temp2!=NULL))

{

if(start3==NULL)

{

if((temp1->expo)==(temp2->expo))

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff= (temp1->coff) - (temp2->coff);

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp2=temp2->link;

temp3->link=NULL;

}

else if(temp1->expo>temp2->expo)

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp3->link=NULL;

}

else

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

start3=temp3;

temp2=temp2->link;

temp3->link=NULL;

}

}

else

{

if(temp1->expo==temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff= (temp1->coff) - (temp2->coff);

temp3->expo=temp1->expo;

temp1=temp1->link;

temp2=temp2->link;

}

else if(temp1->expo>temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

temp1=temp1->link;

}

else

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

temp2=temp2->link;

}

}

}

else

break;

}

temp3->link=NULL;

//traversing temp3

temp3=NULL;

temp3=start3;

printf("\nTraversal of Polynomial Linked List after subtracting Temp1 & Temp2\n");

while(temp3!=NULL)

{

printf("%2dx^%d - ",temp3->coff,temp3->expo);

temp3=temp3->link;

}

printf("\b\b");

printf("\nTHE END\n");

getch();

}

OUTPUT:

POLYNOMIAL CREATION IN LINK LIST AND SUBTRACTION

Enter Coefficient and exponent for Node 1

6

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

8

2

Do u want to create another node

N

Enter Coefficient and exponent for Node 1

7

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

5

2

Do u want to create another node

N

Traversal of Polynomial Linked List 1

6x^3 + 8x^2

Traversal of Polynomial Linked List 2

7x^3 + 5x^2

Traversal of Polynomial Linked List after adding Temp1 & Temp2

-1X^3 + 3X^2 THE END

Program in C to add two polynomials.

2012-04-09T21:26:00.003+05:30

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#define NULL 0

void main()

{

struct poly

{

int coff;

int expo;

struct poly *link;

}*temp1,*start1,*temp2,*start2,*start3,*temp3;

int n,i,z=1,num;

char c;

clrscr();

start1=NULL;

printf("\tPOLYNOMIAL CREATION IN LINK LIST AND ADDITION\n\n");

while(1)

{

if(start1==NULL)

{

temp1=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and Exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

start1=temp1;

temp1->link=NULL;

}

else

{

temp1->link=(struct poly*)malloc(sizeof(struct poly));

temp1=temp1->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp1->link=NULL;

break;

}

}

start2=NULL;

temp2=NULL;

z=1;

while(1)

{

if(start2==NULL)

{

temp2=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

start2=temp2;

temp2->link=NULL;

}

else

{

temp2->link=(struct poly*)malloc(sizeof(struct poly));

temp2=temp2->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp2->link=NULL;

break;

}

}

// TARVERSING

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

printf("Traversal of Polynomial Linked List 1\n");

while(temp1!=NULL)

{

printf("%2dx^%d+",temp1->coff,temp1->expo);

temp1=temp1->link;

}

printf("\b\b");

printf("\nTraversal of Polynomial Linked List 2\n");

while(temp2!=NULL)

{

printf("%2dx^%d+",temp2->coff,temp2->expo);

temp2=temp2->link;

}

printf("\b\b");

// addition

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

temp3=NULL;

start3=NULL;

while(1)

{

if((temp1!=NULL)||(temp2!=NULL))

{

if(start3==NULL)

{

if((temp1->expo)==(temp2->expo))

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff= (temp1->coff) + (temp2->coff);

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp2=temp2->link;

temp3->link=NULL;

}

else if(temp1->expo>temp2->expo)

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp3->link=NULL;

}

else

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

start3=temp3;

temp2=temp2->link;

temp3->link=NULL;

}

}

else

{

if(temp1->expo==temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff= (temp1->coff) + (temp2->coff);

temp3->expo=temp1->expo;

temp1=temp1->link;

temp2=temp2->link;

}

else if(temp1->expo>temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

temp1=temp1->link;

}

else

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

temp2=temp2->link;

}

}

}

else

break;

}

temp3->link=NULL;

//traversing temp3

temp3=NULL;

temp3=start3;

printf("\nTraversal of Polynomial Linked List after adding Temp1 & Temp2\n");

while(temp3!=NULL)

{

printf("%2dx^%d + ",temp3->coff,temp3->expo);

temp3=temp3->link;

} printf("\b \b");

getch();

}

OUTPUT:

POLYNOMIAL CREATION IN LINK LIST AND ADDITION

Enter Coefficient and exponent for Node 1

6

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

8

2

Do u want to create another node

N

Enter Coefficient and exponent for Node 1

7

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

5

2

Do u want to create another node

N

Traversal of Polynomial Linked List 1

6x^3 + 8x^2

Traversal of Polynomial Linked List 2

7x^3 + 5x^2

Traversal of Polynomial Linked List after adding Temp1 & Temp2

13X^3 +13X^2

program in C to store any polynomial using linked list.

2012-04-09T21:11:00.000+05:30

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#define NULL 0

void main()

{

struct poly

{

int coff;

int expo;

struct poly *link;

}*temp,*start,*n1,*prev;

int n,i,z=1,num;

char c;

clrscr();

start=NULL;

n1=NULL;

printf("\tPOLYNOMIAL CREATION IN LINK LIST AND TRAVERSAL\n\n");

while(1)

{

if(start==NULL)

{

temp=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp->coff,&temp->expo);

start=temp;

temp->link=NULL;

}

else

{

temp->link=(struct poly*)malloc(sizeof(struct poly));

temp=temp->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp->coff,&temp->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp->link=NULL;

break;

}

// TARVERSING

n1=start;

printf("Traversal\n");

while(n1!=NULL)

{

printf(" %2dx^%d + ",n1->coff ,n1->expo);

n1=n1->link;

}

printf("\b\b\nTHE END\n");

getch();

}

Output:

POLYNOMIAL CREATION IN LINK LIST AND TRAVERSAL

Enter Coefficient and exponent for Node 1

Do you want to create another node

Enter Coefficient and exponent for Node 2

Do u want to create another node

Traversal

6x^3 + 8x^2

THE END

idea free caller tune for one month..

2012-04-05T11:58:00.002+05:30

dial 54444 from ur idea mobile to get 1 month free caller tune................

njoy..............................

Get all latest updates from google for free

2012-04-02T23:27:00.003+05:30

Now u can get all the information's
to get just dial 180041999999(toll free) and say your needs then you will receive an sms relevant to your needs.

Eg: if you want meaning for a particular word just call and say the word the relevant meaning will be sent through sms

how to get puk number of airtel

2012-04-02T23:24:00.002+05:30

Sim locked?
need to unlock it?
just type your mobile number and sent to 785
immediately your will receive your PUK code and PIN code directly into your inbox.,

Enjoy.,!!

AIRTEL FREE MISS CALL ALERTS

2012-04-02T23:13:00.002+05:30

Friends here is the new trick for airtel miss call alert...

Just Type OFFER and send it to 54321

You will receive a message from Airtel

Thank you for subscribing to MCA service. Your activation request is under process You will receive a confirmation SMS shortly. After some time you will get a activation message .

change ur windows XP password using command

2012-04-02T22:58:00.000+05:30

Below are the steps how to change the XP password using command

-Start-> Run
-Open Cmd
-Type "net user "
-ENTER
-Ask for New Password
-Give Password Which you Want -Twice ENTER

Thats it
Your password gets changed
Try and post your comments here........................

C Program for Circular Queue implementation through Array

2012-04-02T22:42:00.000+05:30

//Program for Circular Queue implementation through Array
#include <stdio.h>
#include<conio.h>
#define MAXSIZE 5
int cq[MAXSIZE]={0};
int front,rear;
void main()
{
void add(int);
void del();
void display();
int ch,i,num;
front = -1;
rear = -1;
clrscr();
printf("\nProgram for Circular Queue demonstration through array");
while(1)
{
printf("\n\nMAIN MENU\n1.INSERTION\n2.DELETION\n3.DISPLAY\n4.EXIT");
printf("\n\nENTER YOUR CHOICE : ");
scanf("%d",&ch);
switch(ch)
{
case 1:
printf("\n\nENTER THE QUEUE ELEMENT : ");
scanf("%d",&num);
add(num);
break;
case 2:
del();
break;
case 3:
display();
break;
case 4:
exit(0);
default: printf("\n\nInvalid Choice . ");
}

}
}

void add(int item)
{
if(front ==(rear+1)%MAXSIZE)
{
printf("\n\nCIRCULAR QUEUE IS OVERFLOW");
}
else
{
if(front==-1)
front=rear=0;
else
rear=(rear+1)%MAXSIZE;
cq[rear]=item;
printf("\n\nRear = %d Front = %d ",rear,front);
}
}

void del()
{
int a;
if(front == -1)
{
printf("\n\nCIRCULAR QUEUE IS UNDERFLOW");
}
else
{
a=cq[front];
cq[front]=0;
if(front==rear)
front=rear=-1;
else
front = (front+1)%MAXSIZE;
printf("\n\nDELETED ELEMENT FROM QUEUE IS : %d ",a);
printf("\n\nRear = %d Front = %d ",rear,front);

}
}

void display()
{
int i;
for(i=0;i<MAXSIZE;i++)
printf("%d\t",cq[i]);
}

C Program to implement NEWTON'S BACKWARD METHOD OF INTEROLATION

2012-04-01T01:11:00.004+05:30

/*
Program to implement NEWTON'S BACKWARD METHOD OF INTEROLATION.
----------------------------------------

*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n; // no. of terms.
int i,j,k; // Loop variables
float mx[10]; // 'X' array limit 9
float my[10]; // 'Y' array limit 9
float x; // User Query for what value of X
float x0=0; //
float y0; // Value coressponding to x0.
float sum=0; // Calculated value for coressponding X.
float h; // Calc. Section
float fun=0; //
float p; // Calc. Section
float diff[20][20]; // to store Y
float y1,y2,y3,y4; // Formulae variables.

clrscr();
// Input section.

printf("\t\t !! NEWTON'S GREGORY BACKWARD INTERPOLATION FORMULA !! ");
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
{
printf("\n Enter the value of x%d -> ",i+1);
scanf("%f",&mx[i]);
}

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
{
printf("\n Enter the value of y%d -> ",i+1);
scanf("%f",&my[i]);
}

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=mx[1]-mx[0];
for(i=0;i<n-1;i++)
diff[i][1]=my[i+1]-my[i];

for(j=2;j<=4;j++)
for(i=0;i<n-j;i++)
diff[i][j]=diff[i+1][j-1]-diff[i][j-1];
i=0;

while(!mx[i]>x)
i++;

x0=mx[i];
sum=0;

y0=my[i];
fun=1;

p=(x-x0)/h;
sum=y0;

for(k=1;k<=4;k++)
{
fun=(fun*(p-(k-1)))/k;
sum=sum+fun*diff[i][k];
}

// Outut Section
printf("\n When x = %6.4f , y = %6.8f",x,sum);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");
getch();
}

C Program to implement NEWTON'S FORWARD METHOD OF INTEROLATION

2012-04-01T01:10:00.001+05:30

/*
Program to implement NEWTON'S FORWARD METHOD OF INTEROLATION.
--------------------------------------
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n; // no. of terms.
int i,j; // Loop variables
float ax[10]; // 'X' array limit 9
float ay[10]; // 'Y' array limit 9
float x; // User Query for what value of X
float y=0; // Calculated value for coressponding X.
float h; // Calc. Section
float p; // Calc. Section
float diff[20][20]; // to store Y
float y1,y2,y3,y4; // Formulae variables.

clrscr();
printf("\t\t !! NEWTON'S GREGORY FORWARD INTERPOLATION FORMULA !! ");
// Input section.
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
{
printf("\n Enter the value of x%d -> ",i+1);
scanf("%f",&ax[i]);
}

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
{
printf("\n Enter the value of y%d -> ",i+1);
scanf("%f",&ay[i]);
}

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=ax[1]-ax[0];

// for 1.
for(i=0;i<n-1;i++)
diff[i][1]=ay[i+1]-ay[i];
// for 2, 3, 4
for(j=2;j<=4;j++)
for(i=0;i<n-j;i++)
diff[i][j]=diff[i+1][j-1]-diff[i][j-1];

do
{
i++;
}
while(ax[i]<x);
i--;

p=(x-ax[i])/h;
y1=p*diff[i-1][1]; // 1
y2=p*(p+1)*diff[i-1][2]/2; // 2
y3=p*(p+1)*(p-1)*diff[i-2][3]/6; // 3
y4=(p+2)*(p+1)*p*(p-1)*diff[i-3][4]/24;

// Taking sum
y=ay[i]+y1+y2+y3+y4;

// Outut Section
printf("\n When x = %6.4f , y = %6.8f",x,y);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");

getch();
}