Add Basic solution to challenge 101

2 files changed, 163 insertions, 0 deletions

diff --git a/challenge-101/paulo-custodio/basic/ch-1.bas b/challenge-101/paulo-custodio/basic/ch-1.bas
new file mode 100644
index 0000000000..948f35eea7
--- /dev/null
+++ b/challenge-101/paulo-custodio/basic/ch-1.bas
@@ -0,0 +1,116 @@
+' You are given an array @A of items (integers say, but they can be anything).
+' 
+' Your task is to pack that array into an MxN matrix spirally counterclockwise, 
+' as tightly as possible.
+' 
+' ‘Tightly’ means the absolute value |M-N| of the difference has to be as small 
+' as possible.
+
+const max_numbers as integer = 100
+
+dim shared number_list(max_numbers) as integer
+dim shared number_rect(max_numbers,max_numbers) as integer
+dim shared num_width as integer, rect_width as integer, rect_height as integer
+dim shared num_numbers as integer
+
+' collect from command line
+sub collect_numbers() 
+    dim i as integer
+    i = 1
+    do while command(i)<>""
+        number_list(i) = val(command(i))
+        if len(command(i))>num_width then 
+            num_width = len(command(i))
+        end if
+        num_numbers = i
+        i = i + 1
+    loop
+end sub
+
+' get smallest rectangle
+sub smallest_rect(byref rect_width as integer, byref rect_height as integer)
+    dim i as integer
+    rect_width = 1
+    rect_height = num_numbers
+    for i=1 to num_numbers
+        if (num_numbers mod i) = 0 then
+            rect_width = i
+            rect_height = int(num_numbers / i)
+            if rect_width >= rect_height then exit for
+        end if
+    next
+end sub
+
+' pack the numbers in a spiral
+sub pack_numbers(rect_width as integer, rect_height as Integer)
+    dim row as integer, col as integer, idx as integer
+
+    ' fill matrix with -1 - to mark places that are free
+    for row=1 to rect_height
+        for col=1 to rect_width
+            number_rect(row, col) = -1
+        next
+    next
+    
+    idx = 1
+    row = rect_height
+    col = 1
+    do while idx <= num_numbers
+        ' go East
+        do while col <= rect_width 
+            if number_rect(row, col) >= 0 then exit do
+            number_rect(row, col) = number_list(idx)
+            idx = idx + 1
+            col = col + 1
+        loop
+        col = col - 1
+        row = row - 1
+        
+        ' go North
+        do while row >= 1
+            if number_rect(row, col) >= 0 then exit do
+            number_rect(row, col) = number_list(idx)
+            idx = idx + 1
+            row = row - 1
+        loop
+        row = row + 1
+        col = col - 1
+        
+        ' go West
+        do while col >= 1
+            if number_rect(row, col) >= 0 then exit do
+            number_rect(row, col) = number_list(idx)
+            idx = idx + 1
+            col = col - 1
+        loop
+        col = col + 1
+        row = row + 1
+        
+        ' go South
+        do while row <= rect_height 
+            if number_rect(row, col) >= 0 then exit do
+            number_rect(row, col) = number_list(idx)
+            idx = idx + 1
+            row = row + 1
+        loop
+        row = row - 1
+        col = col + 1
+    loop
+end sub
+
+' print the rectangle
+sub print_rect(rect_width as integer, rect_height as Integer)
+    dim row as integer, col as integer
+    for row=1 to rect_height
+        for col=1 to rect_width
+            print right(space(10) & str(number_rect(row, col)), num_width+1);
+        next
+        print
+    next
+end sub
+
+' main
+collect_numbers
+smallest_rect rect_width, rect_height
+pack_numbers rect_width, rect_height
+print_rect rect_width, rect_height
diff --git a/challenge-101/paulo-custodio/basic/ch-2.bas b/challenge-101/paulo-custodio/basic/ch-2.bas
new file mode 100644
index 0000000000..9a54875dc4
--- /dev/null
+++ b/challenge-101/paulo-custodio/basic/ch-2.bas
@@ -0,0 +1,47 @@
+' TASK #2 › Origin-containing Triangle
+' Submitted by: Stuart Little
+' You are given three points in the plane, as a list of six co-ordinates: 
+' A=(x1,y1), B=(x2,y2) and C=(x3,y3).
+' 
+' Write a script to find out if the triangle formed by the given three 
+' co-ordinates contain origin (0,0).
+' 
+' Print 1 if found otherwise 0.
+
+type Point
+    x as Double
+    y as Double
+end type
+
+function sign(P1 as Point, P2 as Point, P3 as Point) as Double
+sign = (P1.x - P3.x) * (P2.y - P3.y) _
+     - (P2.x - P3.x) * (P1.y - P3.y)
+end function
+
+function point_in_triangle(P0 as Point, P1 as Point, P2 as Point, P3 as Point) as Integer
+    dim d(3) as Double, has_neg as Boolean, has_pos as Boolean
+
+    d(1) = sign(P0, P1, P2)
+    d(2) = sign(P0, P2, P3)
+    d(3) = sign(P0, P3, P1)
+
+    has_neg = (d(1) < 0.0) or (d(2) < 0.0) or (d(3) < 0.0)
+    has_pos = (d(1) > 0.0) or (d(2) > 0.0) or (d(3) > 0.0)
+
+    if not (has_neg and has_pos) then
+        point_in_triangle = 1
+    else
+        point_in_triangle = 0
+    end if
+end function
+
+' main
+dim P(3) as Point, P0 as Point, i as Integer
+
+for i=1 to 3
+    P(i).x = val(Command(i*2-1))
+    P(i).y = val(Command(i*2))
+next
+P0.x = 0.0
+P0.y = 0.0
+print trim(str(point_in_triangle(P0, P(1), P(2), P(3))))