NEWTON DIVIDED DIFFERENCE INTERPOLATION PROGRAM IN C

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

void main()
{
  float x[10],y[10],d[10][10];
  float a,u,sum=0,prod;
  int i,j,n,k;

  clrscr();
  printf("n enter the no of points");
  scanf("%d",&n);
  printf("nenter the value for interpolation");
  scanf("%f",&a);
  printf("n enter the value for x");
  for(i=1;i<=n;i++)
  {
    scanf("%f",&x[i]);
  }
  printf("n enter the value for y");
  for(i=1;i<=n;i++)
  {
    scanf("%f",&y[i]);
  }
  /*for(i=1;i<=n;i++)
  {
    printf("n x=%f",x[i]);
    printf("n y=%f",y[i]);
  } */
  if((a<x[1])||(a>x[n]))
  {
    printf("n values lies outside the range");
  }
  //i=1;
  /*while(a>x[i])
  //{
    i++;
  }
  k=i-1;
  printf("n k=%d",k);*/
  for(j=1;j<=n-1;j++)
  {
    for(i=1;i<=(n-j);i++)
    {
      if(j==1)
      {
        d[i][j]=((y[i+1]-y[i])/(x[i+1]-x[i]));
        printf("n 1d[i][j]=%f",d[i][j]);
      }
      else
        d[i][j]=((d[i+1][j-1]-d[i][j-1])/(x[i+j]-x[i]));
      printf("n d[i][j]=%f",d[i][j]);
    }

  }
  sum=y[i];
  printf("n sum=%f",sum);
  for(i=1;i<=n;i++)
  {
    prod=1.0;
    for(j=0;j<=i-1;j++)
    // prod=prod*(u-j);
    prod=prod*(a-x[i]);
    sum=sum+d[i][j]*prod;
  }
  printf("n interpolated value=%f",sum);
  getch();
}