Typically, we do hashing by calculating the integer or string according to a ru开发者_JS百科le, then return hash(int-or-str) % m as the index in the hash table, but how do we choose the modulo m? Is t
Recently I\'ve been confused about t开发者_运维百科he modulo operator, %. It\'s known that a % b == a-a/b*b when we have integers a and b where a > b, and we can do this calculation by hand if a a
For unsigned int x, is it possible to calculate x % 255 (or 2^n - 1 in general) using only the following开发者_JAVA百科 operators (plus no loop, branch or function call)?
I know this will seem like a really stupid question, but I just don\'t get why this isn\'t working. This:
I am trying to understand how to repeat loops using the mod operator. If you have two strings, \"abc\" and \"defgh\", how can % be开发者_如何学Python used to loop through abc, repeating it until the
Is the开发者_运维问答re a built in function that would allow me to calculate the modular inverse of a(mod n)?
Why is X % 0开发者_JAVA技巧 an invalid expression? I always thought X % 0 should equal X. Since you can\'t divide by zero, shouldn\'t the answer naturally be the remainder, X (everything left over)?T
I want to overload the % operator in c++, in order to avoid editing a huge block of code by hand. I tried this:
(51^43)Mod77 in scientific calculator gives 2 as the output however, (int)(Ma开发者_StackOverflow中文版th.pow(51,43)%(double)77) gives 12 which should be 2 instead.
When I put in Python interpreter a ** b % c with large a (20 figures) b (4 figures) c (20 figures) I saw that Python calculates it pretty fast, almost like pow (a,b,c). I expect another behavior that
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