exponentiation

问题 what is the fastest method to calculate this, i saw some people using matrices and when i searched on the internet, they talked about eigen values and eigen vectors (no idea about this stuff)...there was a question which reduced to a recursive equation f(n) = (2*f(n-1)) + 2 , and f(1) = 1, n could be upto 10^9.... i already tried using DP, storing upto 1000000 values and using the common fast exponentiation method, it all timed out im generally weak in these modulo questions, which require

问题 Is there a function for integer exponentiation in OCaml? ** is only for floats. Although it seems to be mostly accurate, isn't there a possibility of precision errors, something like 2. ** 3. = 8. returning false sometimes? Is there a library function for integer exponentiation? I could write my own, but efficiency concerns come into that, and also I'd be surprised if there isn't such a function already. 回答1: Regarding the floating-point part of your question: OCaml calls the underlying

问题 This question already has answers here : Strange behaviour of the pow function (5 answers) Closed 4 years ago . I was simply writing a program to calculate the power of an integer. But the output was not as expected. It worked for all the integer numbers except for the power of 5. My code is: #include <stdio.h> #include <math.h> int main(void) { int a,b; printf("Enter the number."); scanf("\n%d",&a); b=pow(a,2); printf("\n%d",b); } The output is something like this: "Enter the number. 2 4

问题 I have a need to create an integer value to a specific power (that's not the correct term, but basically I need to create 10, 100, 1000, etc.) The "power" will be specified as a function parameter. I came up with a solution but MAN does it feel hacky and wrong. I'd like to learn a better way if there is one, maybe one that isn't string based? Also, eval() is not an option. Here is what I have at this time: function makeMultiplierBase(precision) { var numToParse = '1'; for(var i = 0; i <

问题 Like we use fermat's little theorem for modular exponentiation , I was just wondering that is there any such method for fast matrix exponentiation ? Can we use fermat's theorem for matrix exponentiation ? If no then is there any faster method than divide and conquer method of exponentiation ? 回答1: I would look for Octave's implementation of that. There it gives you some references. Here some links: http://epubs.siam.org/doi/abs/10.1137/1020098 http://www.cs.cornell.edu/cv/researchpdf/19ways+

问题 What is the best (most efficient) algorithm for finding all integer power roots of a number? That is, given a number n , I want to find b (base) and e (exponent) such that n = b e I want to obtain all the possible value pairs of b and e Ps: n b and e are to be positive integers . 回答1: I think brute force approach should work: try all e s from 2 (1 is a trivial solution) and up, taking r = n ^ 1/e , a double . If r is less than 2, stop. Otherwise, compute ceil(r)^e and floor(r)^e , and compare

问题 I don't see an exponentiation operator defined in the base arithmetic operators in the Swift language reference. Is there really no predefined integer or float exponentiation operator in the language? 回答1: There isn't an operator but you can use the pow function like this: return pow(num, power) If you want to, you could also make an operator call the pow function like this: infix operator ** { associativity left precedence 170 } func ** (num: Double, power: Double) -> Double{ return pow(num,

问题 Refer to this, the exponentiation operator returns the result of raising first operand to the power second operand, like the exponentiation operator in Python, which is part of the ECMAScript 2016 (ES7) proposal. We know the result of Boolean with exponentiation operator in Python as following: >>> False ** False == True True >>> False ** True == False True >>> True ** False == True True >>> True ** True == True True I want to know whether the Boolean could be used in the exponentiation

问题 So I've been doing some work recently with the modpow function. One of the forms I needed was Modular Exponentiation when the Modulus is a Power of 2. So I got the code up and running. Great, no problems. Then I read that one trick you can make to get it faster is, instead of using the regular exponent, takes it's modulus over the totient of the modulus. Now when the modulus is a power of two, the answer is simply the power of 2 less than the current one. Well, that's simple enough. So I

问题 I recently came across a C code (working by the way) where I found freq_xtal = ((622.08E6 * vcxo_reg_val->hiv * vcxo_reg_val->n1)/(temp_rfreq)); From my intuition it seems that 622.08E6 should mean 622.08 x 10^6 . Is this assumption correct? I tried googling for a similar example where E notation is used in a C program. Surprisingly couldn't find any result 回答1: Yes, that is correct. It works just like an old-school scientific calculator. In this case, it looks like you're dealing with a 622