Problem Statement 
 A polygon is a closed plane figure with n>2 sides. In this problem, a polygon will be given as int[]s xs and ys. The kth point of the polygon will have, as x and y coordinates, the kth elements of xs and ys respectively. There will be an edge between every two consecutive points given. In addition, there will be an edge between the first and last points given.
A dilation is a transformation in plane geometry that takes every point of a figure, and multiplies its x and y coordinates by a given factor. A dilation by a factor greater than 1 acts to "blowup" a figure, increasing its size proportionally.
An orthogonal shift is a transformation applied to a plane figure that moves it along a line that is perpendicular to its face. Consider a figure drawn in the xy plane. An orthogonal shift would change this figure's zcoordinate by the given shift amount.
In this problem we are going to form a 3dimensional solid object using a polygon, a dilation, and an orthogonal shift. The base of the solid will be the polygon given by xs and ys. The top face will be a dilated copy of the base that has been shifted in a direction orthogonal to the plane of the base. In other words, the bottom and top of the solid will be parallel polygons separated by an integral amount. The top face will be a proportionate copy of the bottom face. To complete the solid, edges will be drawn from every vertex in the base to its corresponding vertex in the dilated top face. For example:xs = {0,1,1,0}
ys = {0,0,1,1}
factor = 3
shift = 5
The base of the figure corresponding to the input shown above will be a square including the points (0,0),(1,0),(1,1),(0,1). Dilating this square by a factor of three will produce the polygon with points (0,0),(3,0),(3,3),(0,3). Note, to produce this dilation we multiplied every coordinate by the given factor. This larger square will be shifted 5 units in the zdirection (orthogonal to the base). The resultant figure is finished by drawing edges between each point and its corresponding point in the shifted and dilated copy.
If thought about in 3dimensions, the base would have points each having a zcoordinate of 0, whereas the top would have points each having a zcoordinate of 5. The edges would connect (0,0,0) to (0,0,5), (1,0,0) to (3,0,5), (1,1,0) to (3,3,5), and (0,1,0) to (0,3,5). Had factor been 1, the resulting object would instead have been a rectangular solid.
Your method will return the surface area of the resulting solid rounded down. The surface area is calculated by adding together the areas of the base, the top, and the faces generated by the edges connecting the base and top. 

Definition 
 Class:  SolidArea  Method:  totalArea  Parameters:  int[], int[], int, int  Returns:  int  Method signature:  int totalArea(int[] xs, int[] ys, int factor, int shift)  (be sure your method is public) 




Constraints 
  xs will contain between 3 and 50 elements inclusive 
  xs and ys will contain the same number of elements 
  factor will be between 1 and 50 inclusive 
  shift will be between 1 and 50 inclusive 
  Each element of xs will be between 100 and 100 inclusive 
  Each element of ys will be between 100 and 100 inclusive 
  There will be no repeated points 
  No two nonconsecutive edges in the resulting polygon may intersect with each other 
  No two consecutive edges in the resulting polygon may have the same slope 
  The resulting surface area prior to rounding will not be within .001 of an integer 

Examples 
0)  
  Returns: 51  The example from above. Given areas are approximates rounded for brevity:
Base area = 1.0
Top area = 9.0
Face areas = 10.0, 10.7703, 10.7703, 10.0



1)  
  Returns: 36  The base is a square centered at the origin. The top has been dilated by a factor of 2 and shifted 1 unit. Given areas are approximates rounded for brevity:
Base area = 4.0
Top area = 16.0
Face areas = 4.243, 4.243, 4.243, 4.243



2)  
 {0,2,6,6,8,8,6,2,2,0}  {10,10,6,10,10,0,0,4,0,0}  9  9 
 Returns: 14993  This base is a polygon resembling the letter N. The top has been dilated by a factor of 9 and shifted 9 units. Given areas are approximates rounded for brevity:
Base area = 64.0
Top area = 5184.0
Face areas =
805.047, 1936.801, 976.729, 805.047, 3231.486,
90.0, 993.177, 367.151, 90.0, 450.0



3)  
 {100,100,100,100}  {100,100,100,100}  50  50 
 Returns: 200005203  

4)  
 {0,1,3,1,3,0,3,1,3,1}  {3,1,1,1,3,2,3,1,1,1}  10  9 
 Returns: 3601  
