factorial

问题 A factorial of a natural number (any number greater or equal than 0 ) is that number multiplied by the factorial of itself minus one, where the factorial of 0 is defined as 1 . For example: 0! = 1 1! = 1 * 0! 2! = 2 * 1! 3! = 3 * 2! 4! = 4 * 3! 5! = 5 * 4! Another way of writing this is to multiply all natural numbers between 1 and n for n! : 5! = 1 * 2 * 3 * 4 * 5 How can I express this with a recursive function in F#? And should I do it with a recursive function? //Factorials! let factorial

问题 I coded 3 factorial algorithms: First, I expect to fail by Stack Overflow. No problem. Second, I try tail recusive call , convert previous algorithm from recursive to iterative. It doesn't work but I don't understand why . Third, I use trampoline() method and works fine as I expect. def factorial factorial = { BigInteger n -> if (n == 1) return 1 n * factorial(n - 1) } factorial(1000) // Stack Overflow factorial = { Integer n, BigInteger acc = 1 -> if (n == 1) return acc factorial(n - 1, n *

问题 How can I make a function that calculates the factorial (or the gamma function) of decimal numbers in JavaScript? For example, how could I calculate 2.33! ? 回答1: I might have found an existing solution... It's an implementation of Lanczos method, I found it at the swedish wikipedia (http://sv.wikipedia.org/wiki/Gammafunktionen). It was written in python and says to be correct up to 15 decimals. I ported it to js, cross checked some random values against (http://www.efunda.com/math/gamma

问题 If I calculate factorial(100) then I get an answer of [1] 9.332622e+157 but when I try to calculate a larger factorial, say factorial(1000) I get an answer of [1] Inf Is there a way to use arbitrary precision when calculating factorials such that I can calculate say factorial(1000000) ? 回答1: For arbitrary precision you can use either gmp or Rmpfr . For specifically factorial gmp offers factorialZ and Rmpfr has factorialMpfr . So you can run something like below > Rmpfr::factorialMpfr(200) 1

问题 How would you write a non-recursive algorithm to compute n! ? 回答1: Since an Int32 is going to overflow on anything bigger than 12! anyway, just do: public int factorial(int n) { int[] fact = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600}; return fact[n]; } 回答2: in pseudocode ans = 1 for i = n down to 2 ans = ans * i next 回答3: public double factorial(int n) { double result = 1; for(double i = 2; i<=n; ++i) { result *= i; } return result; } 回答4: In the interests

问题 I've been using this factorial program for Java: public static long factorial(int a) { if(a<1) { return 1; } long result=1; long x=a; while(x>1) { result*=x; x--; } return result; } However, it seems to "break" and return a negative number after the factorial of 25. It returns a negative number for a while then just returns "0." Am I doing anything wrong that is causing this? 回答1: You've overflowed long . Use BigInteger instead. 回答2: 25! = 15511210043330985984000000 The maximum value of a

问题 I am new to Python, I am trying to write a Python program to calculate and print factorials up to n! as shown using recursion . However, I only get the following result: Here is my code right now: def factorial( n ): print("n n!") if n <=1: print(str(n) + ' ' + str(n)) return 1 else: r = n * factorial( n - 1 ) print(str(n) + ' ' + str(r)) return r What is wrong in my code? 回答1: With recursion: factorial = lambda n: n * factorial(n - 1) if (n > 1) else 1 This is similar to your function,

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I am solving a problem on calculation of factorial and the challenge is as follows! You are asked to calculate factorials of some small positive integers. Input An integer t, 1<=t<=100, denoting the number of testcases, followed by t lines, each containing a single integer n, 1<=n<=100. Output For each integer n

问题 public class Car { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); System.out.println(n+"!="+factorial(n)); } public static int factorial(int num) { return (num == 0) ? 1 : num * factorial (num - 1); } } how make this code to text in console 3! = 1*2*3 = 6 ? 回答1: Don't use recursion for this. Besides, it isn't really efficient or necessary. Scanner in = new Scanner(System.in); int n = in.nextInt(); int fact = 1; String s = n + "! = 1"; for

问题 I'm working on a scorecard at work which has columns representing 4 possible outcomes, for example: Successful, unsuccessful, Exceptional, Other Each staff member is assessed 5 times in the month against those ratings. So 1 person might have 3 successful , 2 exceptional , 0 unsuccessful , 0 other . So the max instances of each outcome is 5 but the total sum of instances can't be more than 5. I could try to type out all the combinations (and get them wrong) - is there any function/formula/VBA