See “Hacker’s Delight”, multiword division (pages 140-145). The basic concept (going back to Knuth) is to think of your problem in base-65536 terms. Then you have a 4 digit by 2 digit division problem, with 2/1 digit division as a primitive. The C code is here: https://github.com/hcs0/Hackers-Delight/blob/master/divmnu.c.txt
The current answers all focus on decimal digits, when applying the “add all digits and see if that divides by 3”. That trick actually works in hex as well; e.g. 0x12 can be divided by 3 because 0x1 + 0x2 = 0x3. And “converting” to hex is a lot easier than converting to decimal. Pseudo-code: … Read more
On x86 the remainder is a by-product of the division itself so any half-decent compiler should be able to just use it (and not perform a div again). This is probably done on other architectures too. Instruction: DIV src Note: Unsigned division. Divides accumulator (AX) by “src”. If divisor is a byte value, result is … Read more
I’ve been tinkering with an approach that (1) multiplies a and b with the school algorithm on 21-bit limbs (2) proceeds to do long division by c, with an unusual representation of the residual a*b – c*q that uses a double to store the high-order bits and a long to store the low-order bits. I … Read more
They are being divided in integer arithmetics. So dividing integer a by integer b you get how many times b fits into a. Also a % b will give you a remainder of a division. So (a / b ) * b + a % b = a
Just multiply one of the numbers by 1.0: SELECT something*1.0/total FROM somewhere That will give you floating point division instead of integer division.
If you are storing the value (128-bits) using the largest possible native representation your architecture can handle (64-bits) you will have problems handling the intermediate results of the division (as you already found 🙂 ). But you always can use a SMALLER representation. What about FOUR numbers of 32-bits? This way you could use the … Read more
Short Answer: NO Long Answer: NO. Explanation: The compiler is already optimizing statements like this for you. If there is a technique for implementing this quicker than an integer division then the compiler already knows about it and will apply it (assuming you turn on optimizations). If you provide the appropriate architecture flags as well … Read more
Do floating point division then convert to an int. No extra modules needed. Python 3: >>> int(-1 / 2) 0 >>> int(-3 / 2) -1 >>> int(1 / 2) 0 >>> int(3 / 2) 1 Python 2: >>> int(float(-1) / 2) 0 >>> int(float(-3) / 2) -1 >>> int(float(1) / 2) 0 >>> int(float(3) / … Read more
When you are using a binary operator, both arguments should be of a same type and the result will be in their type too. When you want to divide (int)/(long) it turns into (long)/(long) and the result is (long). you shouldmake it (double)/(long) or (int)/(double) to get a double result. Since double is greater that … Read more