exponentiation

问题 I bombed an interview (phone screen with collabedit) recently. Here is the question: Write an interative O(lg n) algorithm for finding the power of x^y (x is a double, y>0 is an int). I first did the recursive divide and conquer one and tried to convert it to iterative... and I couldn't :S Is there a method to convert recursion to iterative (it is easy for tail recursion, but how about recursive functions with two possible recursive calls which depend on conditions to decide which call will

问题 I bombed an interview (phone screen with collabedit) recently. Here is the question: Write an interative O(lg n) algorithm for finding the power of x^y (x is a double, y>0 is an int). I first did the recursive divide and conquer one and tried to convert it to iterative... and I couldn't :S Is there a method to convert recursion to iterative (it is easy for tail recursion, but how about recursive functions with two possible recursive calls which depend on conditions to decide which call will

问题 This question already has answers here : What do these operators mean (** , ^ , %, //)? [closed] (3 answers) Closed 2 years ago . Why is ^ not squaring in Python? I know exponentiation is ** instead, but what exactly is ^ and why wasn't that operator used instead? For example 2^2=0 , 3^2=1 . 回答1: The ^ operator was already used for bitwise xor. >>> x = 42; format(x, '08b') '00101010' >>> y = 137; format(y, '08b') '10001001' >>> z = x ^ y; format(z, '08b') '10100011' That leaves the old

问题 This question already has answers here : What do these operators mean (** , ^ , %, //)? [closed] (3 answers) Closed 2 years ago . Why is ^ not squaring in Python? I know exponentiation is ** instead, but what exactly is ^ and why wasn't that operator used instead? For example 2^2=0 , 3^2=1 . 回答1: The ^ operator was already used for bitwise xor. >>> x = 42; format(x, '08b') '00101010' >>> y = 137; format(y, '08b') '10001001' >>> z = x ^ y; format(z, '08b') '10100011' That leaves the old

问题 How does gfortran handle exponentiation with a integer vs a real? I always assumed it was the same, but consider the example: program main implicit none integer, parameter :: dp = selected_real_kind(33,4931) real(kind=dp) :: x = 82.4754500815524510_dp print *, x print *, x**4 print *, x**4.0_dp end program main Compiling with gfortran gives 82.4754500815524510000000000000000003 46269923.0191143410452125643548442147 46269923.0191143410452125643548442211 Now clearly these numbers almost agree -

问题 What is the benefit to using the ECMAScript 2016 exponentiation operator over the current Math.pow() ? In other words, besides reducing key strokes, what is the difference between Math.pow(2, 2) => 4 and 2 ** 2 => 4 回答1: None. As you can read in the ES7 spec, both Math.pow and the ** exponentation operator cast their arguments/operands to numbers and use the very same algorithm to determine the result. Addendum : this changed with the introduction of the BigInt type in ES2020, whose values

问题 What is the benefit to using the ECMAScript 2016 exponentiation operator over the current Math.pow() ? In other words, besides reducing key strokes, what is the difference between Math.pow(2, 2) => 4 and 2 ** 2 => 4 回答1: None. As you can read in the ES7 spec, both Math.pow and the ** exponentation operator cast their arguments/operands to numbers and use the very same algorithm to determine the result. Addendum : this changed with the introduction of the BigInt type in ES2020, whose values

来源： https://stackoverflow.com/questions/49960020/with-which-algorithm-exponential-functions-are-computed-in-the-gnu-c-standard

来源： https://stackoverflow.com/questions/49960020/with-which-algorithm-exponential-functions-are-computed-in-the-gnu-c-standard

问题 I need to remove logarithms of my data and thus am taking e to the power of the values which are logarithmed. My issue is that when I have e to the power of more than 709 R returns the value of infinity. How can I surpass this? e^710 [1] Inf Thanks :) 回答1: If you really want to work with numbers that big you can use a Rmpfr package. library('Rmpfr') x <- mpfr(710, precBits = 106) exp(x) 1 'mpfr' number of precision 106 bits [1] 2.233994766161711031253644458116e308 来源： https://stackoverflow