How to print descending numbers in php using FOR?  (e.g.   10 9 8 7...0)?
348563
for ($i=10; $i >= 0; $i--) {\n  print "$i ";\n}
101541
This is kinda wierd without drawing it out, but I'll try to explain and you can come up with the picture.\n\n1)  Start Program\n2)  Accept Input\n3)  Add 1 to counter\n4)  If counter is not equal to 3; go to #2\n5)  If N2 > N3, do #6-8; if not, go to #9\n6)  Move N3 to a hold field\n7)  Move N2 to N3\n8)  Move the hold field to N2\n9)  if N1 > N2, do #6-8 with the N1 & N2 replacing each other\n10) if N2 > N3, do #6-8\n11) print them
361979
if the last # is in the tenths place put it over 10
604143
here is a program for 10 numbers in ascending order\n\n#include <iostream.h>\n#include <conio.h>\n\nvoid main()\n{\nclrscr();\nint array[10],t;\nfor(int x=0;x<10;x++)\n{\ncout << "Enter Integer No. " << x+1 << " : " << endl;\ncin>>array[x];\n}\nfor (x=0;x<10;x++)\n{\nfor(int y=0;y<9;y++)\n{\nif(array[y]>array[y+1])\n{\nt=array[y];\narray[y]=array[y+1];\narray[y+1]=t;\n}\n}\n}\ncout << "Array in ascending order is : ";\nfor (x=0;x<10;x++)\ncout << endl << array[x];\ngetch();\n}\n\nThis program takes in the ten integers array[x] to be stored in the single-dimensional array as a screen input from the user.\nIt then sorts out these ten integers into ascending order and prints them out using the 'cout' command.
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
228224
have a loop going thru numbers form 65 to to 90, then stick that number into chr() function, which gives you a character that you can print on the screen.
248038
I will give your in binary :\n0000\n0001\n0010\n0011\n0100\n0101\n0110\n0111\n1000\n1001\n1010\n1011\n1100\n1101\n1110\n1111
962762
Here is how to reverse a number - it doesn't matter how many digits.\nYou'll have to figure out the rest of your program yourself - there isn't much left anyway.\n\nint reverse (int n)\n{\n   int r = 0;\n   for (;n>0;n/=10) r=r*10+n%10;\n   return r;\n}
392646
You usually do this by using a programming language and comparing the input to a regular expression.\n\nLook up "regular expression tutorial" in Yahoo search and you'll find more information that you probably want  :)
241489
Try any of this code\nDim i as integer=Convert.ToInt32("your number as string")\nor\nDim i as integer=Val("your number as string")\nor\nDim i as integer=CInt("your number as string")
520311
http://www.tenthumbstypingtutor.com/try.php
946903
Numbers 1 - 9 are spelled out, once you get to10 and above, you can just use the numerals. Here's a web site to prove my point...\n\nhttp://www.grammarbook.com/numbers/numbers.asp
808564
printf("$%i.%i",dollars,cents);\n\nor if you have a floating point value you can do\n\nprintf("%5.2f",money);
457195
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
926416
your getting a little anxious in your coding:\ntry changing the section below:\n\ncout << "Enter a number Negative, Positve, or Zero.";\ncout << '&#92;n';\ncin >> num;\n\nif (num > 0)\n{\n    pos(num);\n}\nelse if (num < 0)\n{\n  neg(num);\n}\nelse\n{\n  zero(num);\n}
461539
printf("%-15u&#92;n",40000);\n\nprovided I understood it correctly.
874354
$number = 0.98; \n$formatted_number = number_format ($number, 3);\nif (abs($number) < 1) {\n  $formatted_number = str_replace('0.', '.', $formatted_number);\n}\n\nThis should set $formatted_number to ".980"...