Newton fwd interpolation+ output

#include<stdio.h>

float fact(int x)
{
    if(x==1)
        return 1;
    else
        return x*fact(x-1);

}


int main()
{
    float y[10][10],x[10],u,t,a,ans,l,xp,sum;
    int n,i,j;

    printf("enter n ");
    scanf("%d",&n);
    printf("\n enter the x and y values");
    for(i=0;i<n;i++)
    {
        scanf("%f %f",&x[i],&y[0][i]);
    }

    for(i=0;i<n;i++)
        printf("\nx[%d]= %f \t y[%d]=%f",(i+1),x[i],(i+1),y[0][i]);

    i=0;
    printf("\n");
    printf("\n");

    do
    {
        i++;

        for(j=0;j<n-i;j++)
        {
            y[i][j]=y[i-1][j+1]-y[i-1][j];
            printf("y=%f\t",y[i][j]);
        }
        printf("\n");

    }while(i<n);

    printf("Enter the value to be calculated");
    scanf("%f",&xp);

    u=(xp-x[0])/(x[1]-x[0]);
    l=u;
    sum=0;
    a=0;

    for(i=1;i<n-1;i++)
    {
        u=u*(l-i);
        t=fact(i+1);
        a=(u*y[i+1][0])/t;
        sum+=a;
    }

    ans=y[0][0]+l*y[1][0]+sum;
    printf("\n\nthe ans is %f",ans);
    return 0;
}



OUTPUT:
enter n 4

 enter the x and y values20
.3420
23 .3907
26 .4384
29 .4848

x[1]= 20.000000          y[1]=0.342000
x[2]= 23.000000          y[2]=0.390700
x[3]= 26.000000          y[3]=0.438400
x[4]= 29.000000          y[4]=0.484800

y=0.048700      y=0.047700      y=0.046400
y=-0.001000     y=-0.001300
y=-0.000300

Enter the value to be calculated21


the ans is 0.358326Press any key to continue