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
Python uses the classic Algorithm D from Knuth’s ‘The Art of Computer Programming’. The running time is (generally) proportional to the product of lengths of the two numbers. Space is proportional to the sum of the lengths of the two numbers. The actual division occurs in Objects/longobject.c, see x_divrem(). For background on the internal representation … Read more
In addition to Luchian‘s answer, this is the corresponding part from the C++11 standard: The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined. For integral … 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
Use modulus: // Will evaluate to true if the variable is divisible by 2 variable % 2 === 0
The segfault you are seeing is coming from next checking the range for you is an assertion in your Debug implementation to check against undefined behavior. The behavior of iterators and pointers is not defined beyond the their allocated range, and the “one past-the-end” element: Are iterators past the “one past-the-end” iterator undefined behavior? This … Read more
Wiki says: Given two positive numbers, a (the dividend) and n (the divisor), a modulo n (abbreviated as a mod n) is the remainder of the Euclidean division of a by n. …. When either a or n is negative, the naive definition breaks down and programming languages differ in how these values are defined. … Read more
Since .Net 4.0+ implements BigInteger with a special modular arithmetics function ModPow (which produces “X power Y modulo Z”), you don’t need a third-party library to emulate ModInverse. If n is a prime, all you need to do is to compute: a_inverse = BigInteger.ModPow(a, n – 2, n) For more details, look in Wikipedia: Modular … Read more
Here’s an answer from the MSDN documentation. When you divide two integers, the result is always an integer. For example, the result of 7 / 3 is 2. To determine the remainder of 7 / 3, use the remainder operator (%). int a = 5; int b = 3; int div = a / b; … Read more
Edit Apr 19, 2013: Modulo function updated to handle boundary cases as noted by aka.nice and arr_sea: static const double _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348; static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696; // Floating-point modulo // The result (the remainder) has same sign as the divisor. // Similar to matlab’s mod(); Not similar to fmod() – Mod(-3,4)= 1 fmod(-3,4)= -3 … Read more