factorial

One of the programming problems I have come across involves calculation of factorials of large numbers (of numbers up to 10^5). I have seen a simple Haskell code which goes like this factorial :: (Eq x, Num x) => x -> x factorial 0 = 1 factorial a = a * factorial (a - 1) which implicitly handles the huge numbers and somehow runs faster even without any caching that is involved in the code. When I tried to solve the problem using Java, I had to use BigInteger to hold the huge numbers and also use iterative version of factorial public static BigInteger factorialIterative(int n) { if(n == 0 || n

how can I combine these two functions in to one recursive function to have this result: factorial(6) 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 these are the codes def factorial( n ): if n <1: # base case return 1 else: return n * factorial( n - 1 ) # recursive call def fact(n): for i in range(1, n+1 ): print "%2d! = %d" % ( i, factorial( i ) ) fact(6) 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 as you see execution of these two gives a correct answer, I just want to make it to one recursive function. def factorial( n ): if n <1: # base case return 1 else: returnNumber = n * factorial(

Which function grows faster, exponential (like 2^n, n^n, e^n etc) or factorial (n!)? Ps: I just read somewhere, n! grows faster than 2^n. n! eventually grows faster than an exponential with a constant base (2^n and e^n), but n^n grows faster than n! since the base grows as n increases. n! = n * (n-1) * (n-2) * ... n^n = n * n * n * ... Every term after the first one in n^n is larger, so n^n will grow faster. n^n grows larger than n! -- for an excellent explanation, see the answer by @AlexQueue. For the other cases, read on: Factorial functions do asymptotically grow larger than exponential

I am trying to learn recursion and I am attempting to do factorial's via recursion instead of loops, but my program is causing "Process is terminated due to StackOverflowException class RecursionFactorial { public static int n; static void Main(string[] args) { string input; Console.WriteLine("Please enter a number to work out the factorial"); input = Console.ReadLine(); bool test = int.TryParse(input, out n); fact(n); } public static int fact(int y) { int count = n; if (y <= 1) { Console.WriteLine(y); } else { count = (count * y); Console.WriteLine(count); fact(y - 1); } } } fixed with this

Hi I need to print process of factorial calculation. E.g. If the user's input is 5, the system should print out "5 * 4 * 3 * 2 * 1 = 120" I have this code: public static void factorial() { Scanner sc = new Scanner(System.in); int factorial = 1; int count; System.out.println(me+", This is option 2: Calculate a factorial"); System.out.print("Enter your number: "); int number = sc.nextInt(); System.out.println(); if (number>0) { for (count=1; count<=number; count++) factorial = factorial*count; System.out.println(" = "+factorial); System.out.println(); } else { System.out.println("Enter a

I'm solving a factorial problem where the function takes a number and returns the factorial of that number. The problem I'm running into is that the code works but I don't know why. There are no loops to call it back after the code is executed and I'm not even sure where the current value is being stored.If I am correct the I assume the function is re-running every time it hit the return and it is running with a value of n-1 so one number less than the previous time it ran, however, I still do not see how the value is getting stored to multiple each number by the current value. Even if I log

long long x; for (int i = 1; i <= n; i++) { x = (x * i) % m; } cout << x; This is the trick to calculate (n!) mod m (assume m > n). However, I don't know why it's true. Can you explain the math mechanism behind this? The basic idea here is that you can take the modulus before, during, or after multiplication and get the same value after taking the modulus of the final result. As @Peter points out, (a * b) % m == ((a % m) * (b % m)) % m For the factorial, n! = 1 * 2 * 3 * ... * (n-1) * n so we have n! % m = (((((((1 * 2) % m) * 3) % m) * ... * n-1) % m) * n) % m taking the modulus after each

I am attempting to create a C code that finds the factorial of a integer so that I may convert my code to assembly language. My code seems to 'multiply' the second integer twice. i.e. 5*4*4*3... I cannot seem to find out why. Help please! #define N 5 int main() { int j = 0; int i = 0; int num1 = N; int num2 = N - 1; int sum = 0; while (num2 != 0) { while (j < num2) { sum += num1; j++; } j = 0; printf("%d\n", sum); printf("--------------\n"); --num2; num1 = sum; } printf("--->%d", sum); } Erroneous Output: 20 -------------- 80 -------------- 240 -------------- 480 -------------- 480 Here's the