factorial

问题 This question already has answers here : How would you program Pascal's triangle in R? (2 answers) How to work with large numbers in R? (1 answer) Closed 3 years ago . For a class assignment, I need to create a function that calculates n Choose k. I did just that, and it works fine with small numbers (e.g. 6 choose 2), but I'm supposed to get it work with 200 choose 50, where it naturally doesn't. The answer is too large and R outputs NaN or Inf, saying: > q5(200, 50) [1] "NaN" Warning

问题 This question already has answers here : How would you program Pascal's triangle in R? (2 answers) How to work with large numbers in R? (1 answer) Closed 3 years ago . For a class assignment, I need to create a function that calculates n Choose k. I did just that, and it works fine with small numbers (e.g. 6 choose 2), but I'm supposed to get it work with 200 choose 50, where it naturally doesn't. The answer is too large and R outputs NaN or Inf, saying: > q5(200, 50) [1] "NaN" Warning

问题 0/10 test cases are passing. Here is the challenge description: (to keep formatting nice - i put the description in a paste bin) Link to challenge description: https://pastebin.com/UQM4Hip9 Here is my trial code - PYTHON - (0/10 test cases passed) from math import factorial from collections import Counter from fractions import gcd def cycle_count(c, n): cc=factorial(n) for a, b in Counter(c).items(): cc//=(a**b)*factorial(b) return cc def cycle_partitions(n, i=1): yield [n] for i in range(i,

问题 I want to prove that two factorial functions are equivalent in Coq using induction. The base case n = 0 is easy, however, the induction case is more complicated. I see, that if I could rewrite (visit_fac_v2 n' (n * a)) to n * (visit_fac_v2 n' a) , I would be done. However, translating this idea into Coq causes me troubles. How would one go about proving this in Coq? Fixpoint fac_v1 (n : nat) : nat := match n with | 0 => 1 | S n' => n * (fac_v1 n') end. Fixpoint visit_fac_v2 (n a : nat) : nat

问题 I want to prove that two factorial functions are equivalent in Coq using induction. The base case n = 0 is easy, however, the induction case is more complicated. I see, that if I could rewrite (visit_fac_v2 n' (n * a)) to n * (visit_fac_v2 n' a) , I would be done. However, translating this idea into Coq causes me troubles. How would one go about proving this in Coq? Fixpoint fac_v1 (n : nat) : nat := match n with | 0 => 1 | S n' => n * (fac_v1 n') end. Fixpoint visit_fac_v2 (n a : nat) : nat

问题 I want to prove that two factorial functions are equivalent in Coq using induction. The base case n = 0 is easy, however, the induction case is more complicated. I see, that if I could rewrite (visit_fac_v2 n' (n * a)) to n * (visit_fac_v2 n' a) , I would be done. However, translating this idea into Coq causes me troubles. How would one go about proving this in Coq? Fixpoint fac_v1 (n : nat) : nat := match n with | 0 => 1 | S n' => n * (fac_v1 n') end. Fixpoint visit_fac_v2 (n a : nat) : nat

问题 This question already has answers here : Understanding recursion in Python (4 answers) Closed 5 months ago . Found this solution to make a factorial() function in python, but I am having trouble with understanding ' why ' it works. The function is : def factorial(x): if x <= 1: return 1 else: return x * factorial(x-1) I'm having trouble understanding where the actual multiplication happens? It would seem to me, that the function would keep calling itself until it gets to 1, and returns 1. Can

问题 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

问题 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

问题 I'm trying to write an if statement like this if(denominator([(i-1)! + 1] / i)-1,print(hi),print(ho)) i can be any integer for example 10, when I set i to 10 it gives this error. ? [(x-1)! + 1] / x *** this should be an integer: [(x-1)!+1]/x ^----------- I really only need to check if [(x-1)! + 1] / x is an integer or not the denominator thing is what I came up with, I also tried Mod but couldn't get that working either. 回答1: It seems that you are confused with the names x and i . Please, see