How to write a program in C programming language to multiply two polynomials?

Answer 1

#include <stdio.h>

#include <conio.h>

#define MAX 10

struct term

{

int coeff ;

int exp ;

} ;

struct poly

{

struct term t [10] ;

int noofterms ;

} ;

void initpoly ( struct poly *) ;

void polyappend ( struct poly *, int, int ) ;

struct poly polyadd ( struct poly, struct poly ) ;

void display ( struct poly ) ;

void main( )

{

struct poly p1, p2, p3 ;

clrscr( ) ;

initpoly ( &p1 ) ;

initpoly ( &p2 ) ;

initpoly ( &p3 ) ;

polyappend ( &p1, 1, 4 ) ;

polyappend ( &p1, 2, 3 ) ;

polyappend ( &p1, 2, 2 ) ;

polyappend ( &p1, 2, 1 ) ;

polyappend ( &p2, 2, 3 ) ;

polyappend ( &p2, 3, 2 ) ;

polyappend ( &p2, 4, 1 ) ;

p3 = polymul ( p1, p2 ) ;

printf ( "\nFirst polynomial:\n" ) ;

display ( p1 ) ;

printf ( "\n\nSecond polynomial:\n" ) ;

display ( p2 ) ;

printf ( "\n\nResultant polynomial:\n" ) ;

display ( p3 ) ;

getch( ) ;

}

/* initializes elements of struct poly */

void initpoly ( struct poly *p )

{

int i ;

p -> noofterms = 0 ;

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

{

p -> t[i].coeff = 0 ;

p -> t[i].exp = 0 ;

}

}

/* adds the term of polynomial to the array t */

void polyappend ( struct poly *p, int c, int e )

{

p -> t[p -> noofterms].coeff = c ;

p -> t[p -> noofterms].exp = e ;

( p -> noofterms ) ++ ;

}

/* displays the polynomial equation */

void display ( struct poly p )

{

int flag = 0, i ;

for ( i = 0 ; i < p.noofterms ; i++ )

{

if ( p.t[i].exp != 0 )

printf ( "%d x^%d + ", p.t[i].coeff, p.t[i].exp ) ;

else

{

printf ( "%d", p.t[i].coeff ) ;

flag = 1 ;

}

}

if ( !flag )

printf ( "\b\b " ) ;

}

/* adds two polynomials p1 and p2 */

struct poly polyadd ( struct poly p1, struct poly p2 )

{

int i, j, c ;

struct poly p3 ;

initpoly ( &p3 ) ;

if ( p1.noofterms > p2.noofterms )

c = p1.noofterms ;

else

c = p2.noofterms ;

for ( i = 0, j = 0 ; i <= c ; p3.noofterms++ )

{

if ( p1.t[i].coeff p2.t[j].exp )

{

p3.t[p3.noofterms].coeff = p1.t[i].coeff + p2.t[j].coeff ;

p3.t[p3.noofterms].exp = p1.t[i].exp ;

i++ ;

j++ ;

}

else

{

p3.t[p3.noofterms].coeff = p1.t[i].coeff ;

p3.t[p3.noofterms].exp = p1.t[i].exp ;

i++ ;

}

}

else

{

p3.t[p3.noofterms].coeff = p2.t[j].coeff ;

p3.t[p3.noofterms].exp = p2.t[j].exp ;

j++ ;

}

}

return p3 ;

}

Answer 2

The question is about writing a programme that will accept two polynomial equasions and add them. As an example:

3x^2 + 2x + 1

2x^2 - x +1

to yield:

3x^2 + x + 2

Further, the programme should represent the equasions as a linked list.

If you are still scratching your crust over this, then have a butcher's at the following code which reads two equasions from the command line and adds them together. A sample run is at the bottom. Best of luck.

#include <stdlib.h>

#include <unistd.h>

#include <string.h>

struct ele {

struct ele *next;

int coef;

int power;

};

/* read an equasion and build a linked list

equasion must be of the form nxp {+\-} nxp...

where n is the coefficent of x and p is the power (3x2 meaning

3-x-squared). The equasion must be oredered in decending power

and things like 0x3 can be omitted, and the last (constant)

does not need x0. 1x2 and x2 are the same.

*/

struct ele *build( char *buf )

{

struct ele *expr = NULL;

struct ele *tail = NULL;

struct ele *cep = NULL;

char *tok;

char *tp;

int sign = 1;

tok = strtok( buf, " \t" );

while( tok )

{

if( !isdigit( *tok ) && !*(tok+1) )

{

switch( *tok )

{

case '+': sign = 1; break;

case '-': sign = -1; break;

}

}

else

{

cep = (struct ele *) malloc( sizeof( struct ele ) );

memset( cep, 0, sizeof( *cep ) );

if( !tail )

expr = tail = cep;

else

{

tail->next = cep;

tail = cep;

}

if( *tok NULL e1->power > e2->power) )

{

cep->power = e1->power;

cep->coef = e1->coef;

e1 = e1->next;

}

else

{

cep->power = e2->power;

cep->coef = e2->coef;

e2 = e2->next;

}

}

}

return result;

}

void print( struct ele *ep )

{

while( ep )

{

if( ep->coef != 0 )

switch( ep->power )

{

default:

printf( "%+dx^%d ", ep->coef, ep->power );

break;

case 1:

printf( "%+dx ", ep->coef );

break;

case 0:

printf( "%+d", ep->coef );

break;

}

ep = ep->next;

}

printf( "\n" );

}

int main( int argc, char **argv )

{

struct ele *expr1 = NULL;

struct ele *expr2 = NULL;

struct ele *sum = NULL;

expr1 = build( argv[1] );

expr2 = build( argv[2] );

sum = add( expr1, expr2 );

print( expr1 );

print( expr2 );

print( sum );

}

Answer 3

Here's the program. Took me some time to figure it out, but I did manage to do it, and I am satisfied with my effort.

#include

main()

{

int i,j,k=0,m,l,n;

int a[80],b[80],c[80];

printf("Enter the degree of the first polynomial: ");

scanf("%d",&m);

printf("\nEnter its coefficients: ");

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

scanf("%d",&a[i]);

printf("\nEnter the degree of the second polynomial: ");

scanf("%d",&n);

printf("\nEnter its coefficients: ");

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

scanf("%d",&b[i]);

for (i=0;i<=(m+n);++i)

c[i]=0;

for (l=(m+n);l>=0;--l)

{

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

{

for (j=0;j<=n;++j)

if (((i+j)==l) && (i>=0) && (i<=m) && (j>=0) && (j<=n))

c[k]+= (a[m-i])*(b[n-j]);

}

++k;

}

printf("\nThe desired poynomial is: \n\n");

printf("%dx^%d",c[0],(m+n));

for (i=(m+n-1);i>=1;--i)

if (c[m+n-i]>=0)

printf("+%dx^%d",c[m+n-i],i);

else

printf("%dx^%d",c[m+n-i],i);

if (c[m+n]>=0)

printf("+%d",c[m+n]);

else

printf("%d",c[m+n]);

printf("\n\n");

}

Answer 4

Arrays may represent the coefficients, eg:

a(x) = 2+x+x3, b(x)=x+6x4, c(x)=a(x)+b(x)

a[0]= 2; a[1]=1; a[2]= 0; a[3]=1

b[0]= 1; b[1]=0; b[2]= 0; b[3]=0; b[4]=6 for (i=0; i<=4; ++i) c[i] = a[i]+b[i];

Answer 5

Write a algorithm to add two polynomials using aaray?