in digital 1 is +5 or gnd?!!!?
156153
This depends on the logic family employed. For "standard CMOS" which is common today, +5V is 1 and 0V or ground is 0. However, other systems use other voltages. Lower voltages are common today for a few reasons. 3.3V logic, for example, is widely used where 3.3V is a 1 (and 0V is 0). However, it is possible to have negative logic. In addition, some high speed systems like ECL use voltages like -1.75V for a 0 and -0.9V for a 1 (this allows transistors to stay out of saturation which makes for fast switching but poor noise immunity and difficult design).
314000
since they are all stored as binary all you need to do is display the results as either octal or decimal.\n//**************************************\n//INCLUDE files for :Convert Hex,dec,oct\n//**************************************\n#include <iostream.h>\n#include <conio.h> //for clrscr() & getche()func\n#include <stdlib.h> // for exit() function...\n//**************************************\n// Name: Convert Hex,dec,oct\n// Description:Converts an inputed number to Hex, dec, or oct.\nThe code was mainly written for education purposes.\n// By: Vagabond\n//\n//\n// Inputs:Inputs are from a main menu screen\n//\n// Returns:None\n//\n//Assumes:None\n//\n//Side Effects:None\n//This code is copyrighted and has limited warranties.\n//Please see http://www.Planet-Source-Code.com/xq/ASP/txtCodeId.10360/lngWId.3/qx/vb/scripts/ShowCode.htm\n//for details.\n//**************************************\n\n//****************************************************************************//\n// //\n// Program Name : ConvDec.cpp //\n// //\n// Programmer : Vagabond//\n//: Novice C++ //\n// //\n// Description : The user is asked to input a number in either Hex, Octal or //\n//Decimal, which is then converted to Hex, Octal & Decimal. //\n//The user can exit the program from the MAIN MENU or by//\n//the Escape key after each convertion.//\n// //\n//Compiler : Borland Turbo C++ v4.5.//\n// Date : 16/03/2006 //\n// //\n//****************************************************************************//\n#include <iostream.h>\n#include <conio.h>// for clrscr() & getche() functions...\n#include <stdlib.h> // for exit() function...\nconst int ESC = 27; // Used for escape sequince\nint main()\n\n\n    {\n    int key = 0;\n    int choise = 0;\n    long number;\n    while(key != ESC)\n\n\n        {\n        cout << "Choose Your Number Base&#92;n";\n        cout << "&#92;n&#92;t1. Hexadecimal.";\n        cout << "&#92;n&#92;t2. Octal.";\n        cout << "&#92;n&#92;t3. Decimal.";\n        cout << "&#92;n&#92;t4. Exit.&#92;n&#92;t";\n        cin >> choise;\n        switch(choise)\n\n\n            {\n            case 1: clrscr();\n             cout << "Enter a Hex number: ";\n             cin >> hex >> number;\n             cout << "&#92;n value in octal = "\n             << oct << number << endl;\n             cout << " value in decimal = "\n             << dec << number << endl;\n             break;\n            case 2: clrscr();\n             cout << "Enter a Octal number: ";\n             cin >> oct >> number;\n             cout << "&#92;nvalue in hex = "\n             << hex << number << endl;\n             cout << " value in decimal = "\n             << dec << number << endl;\n             break;\n            case 3: clrscr();\n             cout << "Enter a dec number: ";\n             cin >> dec >> number;\n             cout << "&#92;n value in octal = "\n             << oct << number << endl;\n             cout << "value in hex = "\n             << hex << number << endl;\n             break;\n            case 4: clrscr();\n             cout << "Program terminated by user...";\n             exit(0);\n             break; //Unreachable code.... \n            default : clrscr();\n             cout << "ERROR ~ Invalid selection&#92;n&#92;n";\n             break;\n            }\n            cout << "Press any key to continue or 'Esc' to exit";\n            key = getche();\n            clrscr();\n            }\n            cout << "Program terminated by user...";\n            return 0;\n
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
479514
It printed this for me 0♦8\n\nHope this Helps.\nGood Luck!!!
164812
If it does not contain a decimal place I guess:\n5?
139403
in decimal its 2\nin binary its  0\nin other like octal hexa etc its 2\nin logical or its 1\nnow wat do u want????
187527
+= is a C standard, not specific to C#. Such as...\n\nb=9;\na=1;\na+=b; // a=10
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.
908586
GB >= 0.6*(CS + GB)\n\n"cushion" variable U introduced : \n\nU >= 0 \nGB + U = 0.6 * (CS + GB)
320902
You can go to the copyright page.  There are sometimes more then one way to determine though.  The string of numbers that has 1-10 is usually a good indicator but not always a difinitive one.  Where it looks like:\n\n1  2  3  4  5  6  7  8  9  0\n\nIf it looks like \n\n2  3  4  5  6  7  8  9  0 \n\n\nThe lack of the number one in the second string may indicate FIRST EDITION : SECOND PRINTING\nit is not a first edition.  It will also sometimes denote on the CP (copyright page) that it is the first edition first printing.
421674
y = (x mod 5) + 1.\n\nThough that does seem a bit like cheating :)
199659
D has to be 1 because when you do your addition by hand:\n\n CROSS\n+ROADS\n------\nDANGER\n\nYou see that the D in DANGER must have been carried over from the addition of C and R in CROSS and ROADS, respectively. Thus, we have:\n\n CROSS\n+ROA1S\n------\n1ANGER\n\nThis, by some fairly weird algebra and modular arithmetic, leads to the following relationships:\n\nR = 2S mod 10\nE = (3S + 1 - (2S mod 10)) mod 10\nA = (C + 2R - (R + O mod 10)) mod 10\nA = G - O mod 10\nN = (R + 2O - (O + A mod 10)) mod 10\nC = A - R mod 10\nO = N - R mod 10\nO = G - A mod 10\nG = (O + A + S + 1 - (S + 1 mod 10)) mod 10\n\nObviously, this is of no help whatsoever to anyone, so we do trial and error.\n\nSo now we try things.\n\n 12344\n+25614\n------\n3NO! D has to be 1\n\n 92344\n+25614\n------\n11NO! A cannot be 1 because D is 1\n\n 96233\n+62513\n------\n158746 Yes! it works.
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
892832
These are the coordinates in the form (x,y) where the first number represent x value and the second number represent y value\nDomain means the spread of x values. In your case the x values start at 1 and continue on till infinity, so your domain will be Natural numbers or N.\nJim Burnell is partially correct because he ignored ... which means the coordinates continue beyond listed and the domain is not just 1,2,3,4 but it is 1,2,3,4,5,6, and so on
121013
if it is in binary, 1 + 1 = 10
278871
To the computer hardware, 1's complement is a one-step action.  Adding each bit would be many steps.\n\n1's complement works just as well and is much faster.
1016721
actually, I use BCD for decimals\n\n0.264\n0010 0110 0100 (BCD)