pseudo-code for the hanoi Tower when the number of discs is ZERO?
88527
If you have no disks, you have nothing to do so there can't be any pseudocode.
794630
http://www.7is7.com/otto/countdclock.html
315839
keep a table of binary 4 bit equivalents 0 1 2 ... F of hexadecimal degits.\nconvert each 4 bits of your binary no. to equivalent hex. no looking the table.\n\nyou got the answer
212031
Well, the question doesn't have all the required details, but why not:\n- make sure that you ar eusing the correct type of vaiable (string/number/etc)\n- make a simple if statement that says, IF the number of tries is less than 10 then continue (where number of tries is lets say "x")\nOnce an answer is chosen, if the answer is corret, then thats it, ELSE, if x < 10 then x = x + 1 \n(this makes sure that x is less than 10...and adds 1)\nIF x = 10 THEN MsgBox "HARD LUCK" (like that, if the previous number was 9, by adding 1 it became 10...this line finishes the game up)\n\nAt the end, make sur eto reset the value of "x"\nx = 0
130921
Ah, love these kinds of problems.\n\nOK, there is a very efficient solution if you restrict yourself to integer types. Try this code:\n\nint Answer = 0;\nint i;\nfor (i=0; i < 2*n + 1; i++)\n{\n   Answer ^= Array[i];\n}\nprintf ("Single entry is %d&#92;n", Answer);\n\nThe clincher here is that every integer when exclusive ORed with itself yields zero, so all of the duplicates simply disappear - you will only be left with the unpaired entry.
900433
Discounting rotations and mirror images, there is only one way to make a magic square using the digits 1 through 9.  \n\nFirst, we can prove the 5 has to go in the middle:\n\nLet the numbers in the top row from left to right be a, b, and c.  Let the numbers in the middle row from left to right be d, e, and f.  (Note e is the middle box.)  Let the numbers in the bottom row from left to right be g, h, and i.     \n\nWe know a + b + c = 15 and g + h + i = 15\n\nWe also know a + e + i = 15, b + e + h = 15, and c + e + g =15.  Therefore, (a + e + i) + (b + e + h) + (c + e + g) = 15 * 3 = 45.  We can rearrange this to (a + b + c) + (g + h + i) + 3e = 45.  Since a + b + c = 15 and g + h + i = 15, that reduces to 15 + 15 + 3e = 45, or 3e = 15.  Therefore e (the middle box) must equal 5.   \n\nKnowing the middle box is 5, try 9 in a corner and you'll notice a problem - the 6, 7 and 8 can't all be placed in the square without the row or column with the 9 exceeding 15.  So, the 9 has to occupy a side box.  Since we are discounting rotations and mirror images, we can arbitrarily select box b for the 9.  This forces the placement of the 1 in box h.   \n\nWith the 9 and 1 now placed, the 8 and 6 have to go in the row with the 1.  So, we'll put the 8 in box g and the 6 in box i.  (We still haven't done anything to create a second square if rotations and mirror images are discounted.)  That forces the placement of the final four numbers.
1110789
This is the old school Konami Contra code, dunno if it'll help\n\nup, up, down, down, left, right, left, right, B, A, START
698509
Sounds impossible.  Could you post the code you already have?
139522
I've included a link below to a short C program (developed by Jack Klein) which generates and checks (12,8) hamming codes.
106927
Pseudocode:\n\n// in a single statement\n(home_number AND (NOT office_number)) OR ((NOT home_number) AND office_number)\n\nNote: the above statement is the equivalent of an XOR logic statement, using AND and NOT
1171538
go to towercodes.com and you can get some codes. also, from there you can get to other websites like that!
238200
set the number of decimal places to 0
22692
Scott's code won't work if the numbers can be negative.\n\nYou can find the upper and lower bound as you read the file into the array.\n\nConst MAXFILESIZE = 1000\nDim lowBound as double, hiBound as double\nDim val as double\nDim ln as String, i as Integer\nDim dataTable as String(MAXFILESIZE)\n\ni = 0\n' Get initial values from first line of file\nln = Console.Readline()\n' Should check ln != null\ndataTable(i) = ln\nval = CDbl(ln)\nlowBound = hiBound = val\nwhile (ln != null)\n   i = i + 1\n   ' Read a line\n   ln = Console.Readline()\n   ' Put data into the table\n   dataTable(i) = ln\n   ' Adjust bounds if necessary\n   val = Cdbl(ln)\n   if val > hiBound then\n      hiBound = val        ' remember i?\n   elseif val < lowBound then\n      lowBound = val       ' remember i?\n      end if\n   ln = Console.Readline()\n   end while\n\n(Sorry about the formatting -- that's Yahoo's doing.)
957617
2d point of no return
578454
I can do all that, but except for small numbers I wold never bother.\n\nThe "easy" way, such as it is, is to always go through binary on your way from decimal to hex. \nFor the others, list the bits and add/partition.
940580
check koders.com\n\nor google.com&#92;codesearch
431775
Using the odd/even integer syntax\n\ncreate a grid, numbering each of the blocks, and for all Odd number blocks, insert the command to add a star.\n\nAre you learning C++ in school?