Another update from ingydotnet^djgoku

Task/Hailstone-sequence/BASIC/hailstone-sequence-4.basic

@@ -1,37 +1,52 @@
-function Hailstone(sys *n)
-'=========================
-if n and 1
-  n=n*3+1
-else
-  n=n>>1
-end if
-end function
-
-function HailstoneSequence(sys n) as sys
-'=======================================
-count=1
-do
-  Hailstone n
-  Count++
-  if n=1 then exit do
-end do
-return count
-end function
-
-'MAIN
-'====
-
-maxc=0
-maxn=0
-e=100000
-for n=1 to e
- c=HailstoneSequence n
-  if c>maxc
-    maxc=c
-    maxn=n
-  end if
-next
-
-print e ", " maxn ", " maxc
-
-'result 100000, 77031, 351
+print "Part 1: Create a routine to generate the hailstone sequence for a number."
+print ""
+while hailstone < 1 or hailstone <> int(hailstone)
+    input "Please enter a positive integer: "; hailstone
+wend
+print ""
+print "The following is the 'Hailstone Sequence' for your number..."
+print ""
+print hailstone
+while hailstone <> 1
+    if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
+    print hailstone
+wend
+print ""
+input "Hit 'Enter' to continue to part 2...";dummy$
+cls
+print "Part 2: Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1."
+print ""
+print "No. in Seq.","Hailstone Sequence Number for 27"
+print ""
+c = 1: hailstone = 27
+print c, hailstone
+while hailstone <> 1
+    c = c + 1
+    if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
+    print c, hailstone
+wend
+print ""
+input "Hit 'Enter' to continue to part 3...";dummy$
+cls
+print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!"
+print ""
+print "Calculating result... Please wait... This could take a little while..."
+print ""
+print "Percent Done", "Start Number", "Seq. Length", "Maximum Sequence So Far"
+print ""
+for cc = 1 to 99999
+    hailstone = cc: c = 1
+    while hailstone <> 1
+        c = c + 1
+        if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
+    wend
+    if c > max then max = c: largesthailstone = cc
+    locate 1, 7
+    print "                                                                    "
+    locate 1, 7
+    print using("###.###", cc / 99999 * 100);"%", cc, c, max
+    scan
+next cc
+print ""
+print "The number less than 100,000 with the longest 'Hailstone Sequence' is "; largesthailstone;". It's sequence length is "; max;"."
+end