c programe of Newton forward interpolation formula

Implimentation of Newton’s Forward Interpolation Formula.

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<process.h>

#include<string.h>

void main()

{

  int n,x,p;

  int i,j;

  float ax[10];

  float ay[10];

  float y=0;

  float h;

  float d;

  float diff[20][20];

  float y1,y2,y3,y4;

  clrscr();

  printf("\n Enter the number of term -");

  scanf("%d",&n);

  printf("\n\n Enter the value in the form of x -");

  for(i=0;i<n;i++)

  {

     printf("\n\n Enter the value of x%d-",i+1);

     scanf("%f",&ax[i]);

   }

   printf("\n\n Enter the value in the form of y -");

    for(i=0;i<n;i++)

    {

      printf("\n\n Enter the value in the form y%d-",i+1);

     scanf("%f",&ay[i]);

    }

     printf("\n\n Enter the value of x for -");

     printf("\n which you want the value of y -");

     h=ax[1]-ax[0];

     for(i=0;i<n-1;i++)

     {

       diff[i][2]=ay[i+1]-ay[i];

     }

       for(j=2;j<=4;j++)

       {

             for(i=0;i<n-j;i++)

             {

               diff[i][j]=diff[i+1][j-1];

             }

             i=0;

             do

             {

               i++;

             }while(ax[i]<x);

             i--;

             p=(x-ax[i])/h;

             y1=p*diff[i-1][1];

             y2=p*(p+1)*diff[i-1][2]/2;

             y3=(p+1)*p*(p-1)*diff[i-2][3]/6;

             y4=(p+2)*(p+1)*p*(p-1)*diff[i-3][4]/24;

             y=ay[i]+y1+y2+y3+y4;

             printf("\n whenn=%6.4f,y=%6.8f",x,y);

             printf("\n\n\n Press Enter to Exit");

             }

             getch();

             }