Search Results

Project Euler #2: Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... … 000; public static void main(String[] args) { long time = System.nanoTime(); double max = MAX_NUM * SQRT_5; long sum = 0; for (int i = 3, result = (int) Math.round(2 …

Here is my Python code for Project Euler #2: Each new term in the Fibonacci sequence is generated by adding the previous two terms. … By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... …

I just learned Go (yesterday in fact) and today I made this solution to Project Euler #2. The problem is to sum the even Fibonacci numbers. … matrix calculation is raising a matrix to the power of 4, which I did using this Go matrix library which I believe uses BLAS under the hood. /* p2.go * by hacatu * * This program solves Project Euler …

This is my implementation of Project Euler #2 in C#. It takes 0 milliseconds to run, and I believe it is correct - I have seen nothing to indicate otherwise! … static void GetFibos(ref List<int> nums, int maxVal) { nums.Add(1); int lastNum = 2; while (lastNum <= maxVal) { nums.Add(lastNum); lastNum = nums[nums.Count - 2] + nums …

Project Euler Problem 2 asks for the sum of all even Fibonacci numbers below 4 million. My first attempt at this problem using functional programming with F#. … open System let rec fib n = match n with | 0 | 1 | 2 -> n | _ -> fib (n - 1) + fib (n - 2) let rec nOfFibTermsUpUntil x y = if fib (y + 1) < x then nOfFibTermsUpUntil …

I lately spend a lot of time at Project Euler programming challenges. … By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... …

I'm trying to learn javascript using project euler and I decided to try to force myself to learn about generators in javascript using problem number 2 which asks us to: Find the sum of the even-valued … terms in the fibonacci (f(n)) sequence such that f(n) < 4M , beginning with n=1 is 1, n=2 is 2, n=3 is 3 and so on My code below presently works but I'm hoping to get some input as I began learning javascript …

This is a question about Project Euler number 2, a relatively easy one. The question is: Each new term in the Fibonacci sequence is generated by adding the previous two terms. … By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... …

I figured working through Project Euler problems in Swift would be a good way to learn any tips or tricks. For example, tuples are something I'm not used to using, but they proved useful here. … By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... …

I was hoping you could give me a feedback on the solution I have come up with for the second challenge of Project Euler, and maybe explain me why many of the other solutions I have come across were wrapped … By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... …

By starting with 1 and 2, the first 10 terms will be: 1,2,3,5,8,13,21,34,55,89,... … fib2 = new BigInteger("1"); BigInteger sum = new BigInteger("0"); BigInteger fib = new BigInteger("0"); while (input.compareTo(fib)>0) { if (fib.doubleValue() % 2 …

This is my solution to Project Euler problem 2: By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms. … (self,limit_value): sequence = 2 sum = 0 while True: result = self.F(sequence) if result > limit_value …

Just solved Euler Project Problem 2 using JavaScript and I would like to know how I could improve my code. I do think it looks good, but you never know. … By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... …

By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. …

I am currently working on Project Euler problem #2 which requires you to print the sum of all even fibonacci numbers below 4 million. … second_last_n = last_n last_n = current_n yield current_n def filter_even(nums): # yields all numbers in nums (a generator) which are even for n in nums: if (n % 2 …