大专宿舍:请教编程高手们，以下的程序如何运用？

以下是一段计算公式，请教编程高手们，该程序需放入什么环境中才能运行，即如何运用？

*para u_x1,u_y1,u_x2,u_y2
priv u_dxy,u_x,u_y,u_mx1,u_mx2,u_my1,u_my2,tmpany
if u_x1=0 .and. u_y1=0
u_xy=0
return u_xy
endif

u_dxy=3.141592654 && 圆周率
tmpany=str((u_x1-int(u_x1))*10000,4)
u_mx1=int(u_x1)+(val(left(tmpany,2))/60)+(val(right(tmpany,2))/3600)

tmpany=str((u_y1-int(u_y1))*10000,4)
u_my1=int(u_y1)+(val(left(tmpany,2))/60)+(val(right(tmpany,2))/3600)

tmpany=str((u_x2-int(u_x2))*1000000,6)
u_mx2=int(u_x2)+(val(left(tmpany,2))/60)+(val(substr(tmpany,3,2))/3600)+(val(right(tmpany,2))/360000)

tmpany=str((u_y2-int(u_y2))*1000000,6)
u_my2=int(u_y2)+(val(left(tmpany,2))/60)+(val(substr(tmpany,3,2))/3600)+(val(right(tmpany,2))/360000)

u_x=(round((111.3200144*cos(dtor(u_my1))+111.3200144*cos(dtor(u_my2)))*(u_mx1-u_mx2)/2,7))**2
u_y=(round(110.9481458*(u_my1-u_my2),7))**2
*u_x=((12756.334*u_dxy/360*cos(dtor(u_my1))+12756.334*u_dxy/360*cos(dtor(u_my2)))*(u_mx1-u_mx2)/2)**2
*u_y=(12713.721*u_dxy/360*(u_my1-u_my2))**2

tmpany=round(sqrt(u_x+u_y)*1000,0)
u_xy=iif(tmpany>=9999999,9999999,tmpany)
*u_xy=u_xy/1000
replace sv with u_xy
@ 13, 35 say u_xy
return

多谢！