Addition of Two polynomial using Structure

 
Logic: define a structure  "struct poly" for storing terms of polynomial i.e. two integers viz int coeff and int expo 

now define poly1 as a array of struct poly.."poly poly1[10]" which will store max 10 terms.
 same define poly2[10] and poly3[20]

write 3 seperate function for read poly ,display poly and addpoly
read poly- int readPoly(poly p[10])
write function  for scanning each term i.e. coeff and expo and save respective array of struct poly. (use for loop )
read function should return number of terms hence data type is int.

Code for scan poly:
printf("\n Enter the total number of terms in the polynomial:");
 scanf("%d",&t1);

 printf("\n Enter the COEFFICIENT and EXPONENT in DESCENDING ORDER\n");
 for(i=0;i<t1;i++)
 {
  printf("\n Enter the Coefficient and Exponent (%d): ",i+1);
  scanf("%d\t%d",&p[i].coeff,&p[i].expo);
   }

Display poly- void displayPoly( poly [],int terms);
Write Display function for display polynomial on outut window.
Run a for loop from 0 to terms-1 and display coefficient and  exponent of respective terms of polynomial.

Code for Display poly:
for(l=0;l<term-1;l++)
 printf("%dx^%d+",p[l].coeff,p[l].expo);

 printf("%dx^%d",p[l].coeff,p[l].expo);

Addition of two polynomial:
use three pointer 
i- for first poly
j-for second poly
k for third poly for storing result

use double for loop for i and j or use while loop to access terms of polynomial
compare exponent of poly1 and poly2...
1) if exponent of poly1 and poly2 are equal 
 if(poly1[i].expo==poly2[j].expo)

then copy exponent to poly3 also add coefficient of poly1 and poly 2 and assign it to coefficient of poly3.
in this case increment i,j,k pointer and move on to next terms of all poly.
poly3[k].coeff=poly1[i].coeff + poly2[j].coeff;
poly3[k].expo=poly1[i].expo;
i++;
j++;
k++;


2)if exponent of poly 1 is greater than poly2 
 if(poly1[i].expo>poly2[j].expo)

then assign poly3 term with poly 1 term and increment only i and k, as second term not copied to poly 3,dont increment poly 2 pointer.
poly3[k].coeff=poly1[i].coeff;
poly3[k].expo=poly1[i].expo;
i++;
k++;

3) last case will be exponent of poly2 is greater than poly1 
 if(poly1[i].expo>poly2[j].expo)

hence copy poly2 term into poly3 ,in this case increment j and k pointer .
poly3[k].coeff= poly2[j].coeff;
poly3[k].expo=poly2[j].expo;
j++;
k++;

Final Code for addition of polynomial using structure:

#include<stdio.h>

 /* declare structure for polynomial */
typedef struct poly
 {
 int coeff;
 int expo;
 }poly;
 // declare three arrays p1, p2, p3 of type structure poly.
 //each polynomial can have maximum of ten terms
 // addition result of p1 and p2 is stored in p3 

 poly poly1[10],poly2[10],poly3[20];
int m=0; //pointer to show output in each pass
 //function prototypes 
 int readPoly(poly []);
 int addPoly(poly [],poly [],int ,int ,poly []);
 void displayPoly( poly [],int terms);

 int main()
 {
 int t1,t2,t3;

 // read and display first polynomial 
 t1=readPoly(poly1);
 printf(" \n Entered First polynomial is : ");
 displayPoly(poly1,t1);
 //read and display second polynomial 
 t2=readPoly(poly2);
 printf(" \n Entered Second polynomial is : ");
 displayPoly(poly2,t2);

 //add two polynomials and display resultant polynomial 
 t3=addPoly(poly1,poly2,t1,t2,poly3);
 printf(" \n\n Resultant polynomial after addition : ");
 displayPoly(poly3,t3);
 printf("\n");

 return 0;
 }

 int readPoly(poly p[10])
 {
 int t1,i;

 printf("\n Enter the total number of terms in the polynomial:");
 scanf("%d",&t1);

 printf("\n Enter the COEFFICIENT and EXPONENT in DESCENDING ORDER\n");
 for(i=0;i<t1;i++)
 {
  printf("\n Enter the Coefficient and Exponent (%d): ",i+1);
  scanf("%d\t%d",&p[i].coeff,&p[i].expo);
   
 }
 return(t1);
 }

 int addPoly(poly poly1[10],poly poly2[10],int t1,int t2,poly poly3[10])
 {
 
 int i,j,k;
 i=0;
 j=0;
 k=0;

 while(i<t1 && j<t2)
 {m++;
  if(poly1[i].expo==poly2[j].expo)
  {
   poly3[k].coeff=poly1[i].coeff + poly2[j].coeff;
   poly3[k].expo=poly1[i].expo;
   printf("\n  addition of polynomial at %d pass:",m);
   displayPoly(poly3,m);

   i++;
   j++;
   k++;
  }
  else if(poly1[i].expo>poly2[j].expo)
  {
   poly3[k].coeff=poly1[i].coeff;
   poly3[k].expo=poly1[i].expo;
   printf("\n  addition of polynomial at %d pass:",m);
   displayPoly(poly3,m);

   i++;
   k++;
  }
  else
  {
   poly3[k].coeff=poly2[j].coeff;
   poly3[k].expo=poly2[j].expo;
   printf("\n  addition of polynomial at %d pass:",m);
   displayPoly(poly3,m);

   j++;
   k++;
  }
 }

 // for rest over terms of polynomial 1 
 while(i<t1)
 {
  m++;
  poly3[k].coeff=poly1[i].coeff;
  poly3[k].expo=poly1[i].expo;
  printf("\n  addition of polynomial at %d pass:",m);
  displayPoly(poly3,m);

  i++;
  k++;
 }
 // for rest over terms of polynomial 2 
 while(j<t2)
 {
  m++;
  poly3[k].coeff=poly2[j].coeff;
  poly3[k].expo=poly2[j].expo;
  printf("\n  addition of polynomial at %d pass:",m);
  displayPoly(poly3,m);

  j++;
  k++;
 }

 return(k); // k is number of terms in resultant polynomial
 }

 void displayPoly(poly p[10],int term)
 {
 int l;
 for(l=0;l<term-1;l++)
 printf("%dx^%d+",p[l].coeff,p[l].expo);

 printf("%dx^%d",p[l].coeff,p[l].expo);
}


/*
  Enter the total number of terms in the polynomial:3

 Enter the COEFFICIENT and EXPONENT in DESCENDING ORDER

 Enter the Coefficient and Exponent (1): 6
6

 Enter the Coefficient and Exponent (2): 4
4

 Enter the Coefficient and Exponent (3): 2
2

 Entered First polynomial is : 6x^6+4x^4+2x^2
 Enter the total number of terms in the polynomial:3

 Enter the COEFFICIENT and EXPONENT in DESCENDING ORDER

 Enter the Coefficient and Exponent (1): 6
6

 Enter the Coefficient and Exponent (2): 4
4

 Enter the Coefficient and Exponent (3): 1
1

 Entered Second polynomial is : 6x^6+4x^4+1x^1
  addition of polynomial at 1 pass:12x^6
  addition of polynomial at 2 pass:12x^6+8x^4
  addition of polynomial at 3 pass:12x^6+8x^4+2x^2
  addition of polynomial at 4 pass:12x^6+8x^4+2x^2+1x^1

 Resultant polynomial after addition : 12x^6+8x^4+2x^2+1x^1

--------------------------------
Process exited after 17.5 seconds with return value 0
Press any key to continue . . .

 */