modulo

问题 This question already has answers here : Is & faster than % when checking for odd numbers? (8 answers) Closed 2 years ago . which is faster to find the even number if(n%2==0) or if(n&1==0) ? def isEven(n): if n%2 == 0: return True return False OR def isEven(n): if n&1 == 1: return False return True 回答1: This looks like Python, so I tried it in Python(3.6): from timeit import timeit a = timeit(stmt='for i in range(100): i%2') b = timeit(stmt='for i in range(100): i&1') print(a, b) Times vary

问题 This question already has answers here : Is & faster than % when checking for odd numbers? (8 answers) Closed 2 years ago . which is faster to find the even number if(n%2==0) or if(n&1==0) ? def isEven(n): if n%2 == 0: return True return False OR def isEven(n): if n&1 == 1: return False return True 回答1: This looks like Python, so I tried it in Python(3.6): from timeit import timeit a = timeit(stmt='for i in range(100): i%2') b = timeit(stmt='for i in range(100): i&1') print(a, b) Times vary

问题 This question already has answers here : Is & faster than % when checking for odd numbers? (8 answers) Closed 2 years ago . which is faster to find the even number if(n%2==0) or if(n&1==0) ? def isEven(n): if n%2 == 0: return True return False OR def isEven(n): if n&1 == 1: return False return True 回答1: This looks like Python, so I tried it in Python(3.6): from timeit import timeit a = timeit(stmt='for i in range(100): i%2') b = timeit(stmt='for i in range(100): i&1') print(a, b) Times vary

问题 I am attempting to write a regular expression to accept any binary strings with the only criteria being that the number of 0s is not a factor of 3 ([number of 0s] % 3 != 0). How can this be achieved? 回答1: You can use .match() to achieve this. .match() returns an array of all occurrences matching the regex. Using modulo on the returned array's .length will tell you if the number of 0s is divisible by 3. var someString = '012345167891abcd1efghi1jklmn'; var numOfOnes = someString.match(/(1)/g) /

问题 I am attempting to write a regular expression to accept any binary strings with the only criteria being that the number of 0s is not a factor of 3 ([number of 0s] % 3 != 0). How can this be achieved? 回答1: You can use .match() to achieve this. .match() returns an array of all occurrences matching the regex. Using modulo on the returned array's .length will tell you if the number of 0s is divisible by 3. var someString = '012345167891abcd1efghi1jklmn'; var numOfOnes = someString.match(/(1)/g) /

问题 Consider the following function that computes the integral or floating-point modulo depending on the argument type, at compile-time: template<typename T> constexpr T modulo(const T x, const T y) { return (std::is_floating_point<T>::value) ? (x < T() ? T(-1) : T(1))*((x < T() ? -x : x)-static_cast<long long int>((x/y < T() ? -x/y : x/y))*(y < T() ? -y : y)) : (static_cast<typename std::conditional<std::is_floating_point<T>::value, int, T>::type>(x) %static_cast<typename std::conditional<std:

问题 I'm stuck on this cryptography problem using multiplication of a whole number and a fraction mod 10. Here is the equation: 7 * (4/11) mod 10 =? I know I am supposed to convert this to an integer since the mod operator does not work with fractions, but I cannot figure this one out. Obviously, 7 * (4/11) = 28/11, but I cannot get the mod 10 of a fraction. The instructor wants the exact answer, not a decimal. Any help would be greatly appreciated! 回答1: 8 8 is the correct answer indeed. 7*4/11

问题 i am basically coming from java background and struggling to understand the modulo operation in Ruby. (5 % 3) (-5 % 3) (5 % -3) (-5 % -3) The above operation in Java yields, 2 -2 2 -2 But in Ruby, the same expression yields 2 1 -1 -2 . How logically ruby is good at this? How the module operation is implemented in Ruby ? If the same operation is defined as a web services, how both services can match the logic. 回答1: In Java, the result of the modulo operation has the same sign as the dividend.

来源： https://stackoverflow.com/questions/29168283/how-do-i-find-factorials-of-large-numbers-modulo-1000000007-in-c

来源： https://stackoverflow.com/questions/43302709/how-to-make-sense-of-modulo-in-c