factorial

问题 I need to find the factorial in java without using loop or recursion ? So if there is any way then please help . Thanks 回答1: Slightly impractical but no explicit loop anywhere. import javax.swing.Timer; import java.awt.event.*; import java.util.concurrent.ArrayBlockingQueue; public class Fac { public static int fac(final int _n) { final ArrayBlockingQueue<Integer> queue = new ArrayBlockingQueue<Integer>(1); final Timer timer = new Timer(0, null); timer.addActionListener(new ActionListener() {

问题 I have a TABLE "elements" with one COLUMN "number", type SMALLINT that contains numbers 1 thru 56. How can I generate unique sets of 5 numbers of every possible combination from 1 to 56, using an SQL statement? In APL (programming language) a simple dyadic function 5!56 does the trick! EDIT: In good ole MS-DOS QBASIC, I accomplished it like this: 10 OPEN "C:\5NUMBERS.OUT" FOR OUTPUT ACCESS READ WRITE AS #1 12 LET SER = 0 15 LET E = 56 30 FOR B5 = 5 TO E 40 FOR B4 = 4 TO E 50 FOR B3 = 3 TO E

问题 This is a rather theoretical question, so while the language is specifically Java, any general solution will suffice. Suppose I wanted to write a trivial factorial function: long factorial(int n) { //handle special cases like negatives, etc. long p = 1; for(int i = 1; i <= n; i++) { p = p * n; } return p; } But now, I also want to check if the factorial overflows (without simply hard coding a MAX_FACTORIAL_PARAMETER or something of the like). In general checking for overflow during

问题 I'm trying to improve the running time of the factorial calculation of the large number. The first Code which simply loop over and multiplies. def calculate_factorial_multi(number): ''' This function takes one agruments and returns the factorials of that number This function uses the approach successive multiplication like 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 ''' ''' If 0 or 1 retrun immediately ''' if number == 1 or number == 0: return 1 result = 1 # variable to hold the result for x in xrange

问题 So consider the following program-segment! I've tried to use the basic recursion function to determine the factorial of a number, but now using the BigInteger class. public static BigInteger fact(int a) { BigInteger factorial = BigInteger.ONE; BigInteger factz = BigInteger.ONE; if(a == 1) { return factorial; } else { return factz.multiply(fact(a-1)); } } So when I try implementing this in a program, it returns the output as 1. Is it because BigInteger objects are immutable? Or am I missing

问题 var lookup = {}; function memoized(n) { if(n <= 1) { return 1; } if(lookup[n]) { return lookup[n]; } lookup[n] = n * memoized(n - 1); return lookup[n]; } vs. function fact(n) { if(n <= 1) { return 1; } return n * fact(n-1); } If we call fact(3) With the second method we get --> 3 * (2 * (1)) What is the efficiency gain of storing the result in a hash. Is it only for subsequent calls to the same function? I can't see how you would gain anything if you are only calling the function once. With

问题 I have a problem, then given some input number n, we have to check whether the no is factorial of some other no or not. INPUT 24, OUTPUT true INPUT 25, OUTPUT false I have written the following program for it:- int factorial(int num1) { if(num1 > 1) { return num1* factorial(num1-1) ; } else { return 1 ; } } int is_factorial(int num2) { int fact = 0 ; int i = 0 ; while(fact < num2) { fact = factorial(i) ; i++ ; } if(fact == num2) { return 0 ; } else { return -1; } } Both these functions, seem

问题 I tried a recursive factorial algorithm in Rust. I use this version of the compiler: rustc 1.12.0 (3191fbae9 2016-09-23) cargo 0.13.0-nightly (109cb7c 2016-08-19) Code: extern crate num_bigint; extern crate num_traits; use num_bigint::{BigUint, ToBigUint}; use num_traits::One; fn factorial(num: u64) -> BigUint { let current: BigUint = num.to_biguint().unwrap(); if num <= 1 { return One::one(); } return current * factorial(num - 1); } fn main() { let num: u64 = 100000; println!("Factorial {}!

问题 I want to calculate the multinomial coefficient: where it is satisifed n=n0+n1+n2 The Matlab implementation of this operator can be easily done in the function: function N = nchooseks(k1,k2,k3) N = factorial(k1+k2+k3)/(factorial(k1)*factorial(k2)*factorial(k3)); end However, when the index is larger than 170, the factorial would be infinite which would generate NaN in some cases, e.g. 180!/(175! 3! 2!) -> Inf/Inf-> NaN . In other posts, they have solved this overflow issue for C and Python.

问题 I've compaired the scala version (BigInt(1) to BigInt(50000)).reduce(_ * _) to the python version reduce(lambda x,y: x*y, range(1,50000)) and it turns out, that the scala version took about 10 times longer than the python version. I'm guessing, a big difference is that python can use its native long type instead of creating new BigInt-objects for each number. But is there a workaround in scala? 回答1: The fact that your Scala code creates 50,000 BigInt objects is unlikely to be making much of a