如何找到阶乘？

Question

如何编写程序来查找任何自然数的阶乘？

Answer 1

这将适用于正整数的阶乘(尽管是一个非常小的子集):

unsigned long factorial(unsigned long f)
{
    if ( f == 0 ) 
        return 1;
    return(f * factorial(f - 1));
}

printf("%i", factorial(5));

由于您的问题的性质(以及您已承认的级别),此解决方案更多地基于解决此问题的概念,而不是将在下一个"置换引擎"中使用的函数.

Answer 2

这计算非负整数[*]的阶乘,直到ULONG_MAX,它将具有如此多的数字,即使它有时间计算它们,你的机器也不太可能存储更多.使用您需要链接的GNU多精度库.

#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <gmp.h>

void factorial(mpz_t result, unsigned long input) {
    mpz_set_ui(result, 1);
    while (input > 1) {
        mpz_mul_ui(result, result, input--);
    }
}

int main() {
    mpz_t fact;
    unsigned long input = 0;
    char *buf;

    mpz_init(fact);
    scanf("%lu", &input);
    factorial(fact, input);

    buf = malloc(mpz_sizeinbase(fact, 10) + 1);
    assert(buf);
    mpz_get_str(buf, 10, fact);
    printf("%s\n", buf);

    free(buf);
    mpz_clear(fact);
}

示例输出:

$ make factorial CFLAGS="-L/bin/ -lcyggmp-3 -pedantic" -B && ./factorial
cc -L/bin/ -lcyggmp-3 -pedantic    factorial.c   -o factorial
100
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

[*]如果你的意思是"数字",那么你必须更加具体.尽管Pascal通过使用Gamma函数扩展了域,但我还没有意识到其中任何其他数字被定义为阶乘.

Answer 3

为什么在Haskell中执行此操作时可以在C中执行此操作:

新生Haskell程序员

fac n = if n == 0 
           then 1
           else n * fac (n-1)

麻省理工学院的二年级学生Haskell程序员(作为新生学习计划)

fac = (\(n) ->
        (if ((==) n 0)
            then 1
            else ((*) n (fac ((-) n 1)))))

少年Haskell程序员(开始Peano球员)

fac  0    =  1
fac (n+1) = (n+1) * fac n

另一个初级Haskell程序员(读n + k模式是"Haskell的一个令人作呕的部分" 1 并加入了"Ban n + k模式" - 运动[2])

fac 0 = 1
fac n = n * fac (n-1)

高级Haskell程序员(投票给尼克松布坎南布什 - "向右倾斜")

fac n = foldr (*) 1 [1..n]

另一位高级Haskell程序员(投票给McGovern Biafra Nader - "向左倾斜")

fac n = foldl (*) 1 [1..n]

还有另一位高级Haskell程序员(到目前为止右倾他又回来了!)

-- using foldr to simulate foldl

fac n = foldr (\x g n -> g (x*n)) id [1..n] 1

记忆Haskell程序员(每天服用Ginkgo Biloba)

facs = scanl (*) 1 [1..]

fac n = facs !! n

无意义(咳咳)"无点"Haskell程序员(在牛津大学学习)

fac = foldr (*) 1 . enumFromTo 1

迭代Haskell程序员(前Pascal程序员)

fac n = result (for init next done)
        where init = (0,1)
              next   (i,m) = (i+1, m * (i+1))
              done   (i,_) = i==n
              result (_,m) = m

for i n d = until d n i

迭代单行Haskell程序员(前APL和C程序员)

fac n = snd (until ((>n) . fst) (\(i,m) -> (i+1, i*m)) (1,1))

累积Haskell程序员(快速达到高潮)

facAcc a 0 = a
facAcc a n = facAcc (n*a) (n-1)

fac = facAcc 1

继续传递Haskell程序员(早年养大RABBITS,然后搬到新泽西)

facCps k 0 = k 1
facCps k n = facCps (k . (n *)) (n-1)

fac = facCps id

童子军Haskell程序员(喜欢打结;总是"虔诚",他属于最低定点教会[8])

y f = f (y f)

fac = y (\f n -> if (n==0) then 1 else n * f (n-1))

组合Haskell程序员(避免变量,如果不是混淆;所有这些只是一个阶段,虽然它很少阻碍)

s f g x = f x (g x)

k x y   = x

b f g x = f (g x)

c f g x = f x g

y f     = f (y f)

cond p f g x = if p x then f x else g x

fac  = y (b (cond ((==) 0) (k 1)) (b (s (*)) (c b pred)))

列表编码Haskell程序员(更喜欢用一元计算)

arb = ()    -- "undefined" is also a good RHS, as is "arb" :)

listenc n = replicate n arb
listprj f = length . f . listenc

listprod xs ys = [ i (x,y) | x<-xs, y<-ys ]
                 where i _ = arb

facl []         = listenc  1
facl n@(_:pred) = listprod n (facl pred)

fac = listprj facl

解释性的Haskell程序员(从未"遇到过他不喜欢的语言")

-- a dynamically-typed term language

data Term = Occ Var
          | Use Prim
          | Lit Integer
          | App Term Term
          | Abs Var  Term
          | Rec Var  Term

type Var  = String
type Prim = String


-- a domain of values, including functions

data Value = Num  Integer
           | Bool Bool
           | Fun (Value -> Value)

instance Show Value where
  show (Num  n) = show n
  show (Bool b) = show b
  show (Fun  _) = ""

prjFun (Fun f) = f
prjFun  _      = error "bad function value"

prjNum (Num n) = n
prjNum  _      = error "bad numeric value"

prjBool (Bool b) = b
prjBool  _       = error "bad boolean value"

binOp inj f = Fun (\i -> (Fun (\j -> inj (f (prjNum i) (prjNum j)))))


-- environments mapping variables to values

type Env = [(Var, Value)]

getval x env =  case lookup x env of
                  Just v  -> v
                  Nothing -> error ("no value for " ++ x)


-- an environment-based evaluation function

eval env (Occ x) = getval x env
eval env (Use c) = getval c prims
eval env (Lit k) = Num k
eval env (App m n) = prjFun (eval env m) (eval env n)
eval env (Abs x m) = Fun  (\v -> eval ((x,v) : env) m)
eval env (Rec x m) = f where f = eval ((x,f) : env) m


-- a (fixed) "environment" of language primitives

times = binOp Num  (*)
minus = binOp Num  (-)
equal = binOp Bool (==)
cond  = Fun (\b -> Fun (\x -> Fun (\y -> if (prjBool b) then x else y)))

prims = [ ("*", times), ("-", minus), ("==", equal), ("if", cond) ]


-- a term representing factorial and a "wrapper" for evaluation

facTerm = Rec "f" (Abs "n" 
              (App (App (App (Use "if")
                   (App (App (Use "==") (Occ "n")) (Lit 0))) (Lit 1))
                   (App (App (Use "*")  (Occ "n"))
                        (App (Occ "f")  
                             (App (App (Use "-") (Occ "n")) (Lit 1))))))

fac n = prjNum (eval [] (App facTerm (Lit n)))

静态的Haskell程序员(他是用类做的,他有那个fundep Jones!在Thomas Hallgren的"功能依赖的乐趣"之后[7])

-- static Peano constructors and numerals

data Zero
data Succ n

type One   = Succ Zero
type Two   = Succ One
type Three = Succ Two
type Four  = Succ Three


-- dynamic representatives for static Peanos

zero  = undefined :: Zero
one   = undefined :: One
two   = undefined :: Two
three = undefined :: Three
four  = undefined :: Four


-- addition, a la Prolog

class Add a b c | a b -> c where
  add :: a -> b -> c

instance              Add  Zero    b  b
instance Add a b c => Add (Succ a) b (Succ c)


-- multiplication, a la Prolog

class Mul a b c | a b -> c where
  mul :: a -> b -> c

instance                           Mul  Zero    b Zero
instance (Mul a b c, Add b c d) => Mul (Succ a) b d


-- factorial, a la Prolog

class Fac a b | a -> b where
  fac :: a -> b

instance                                Fac  Zero    One
instance (Fac n k, Mul (Succ n) k m) => Fac (Succ n) m

-- try, for "instance" (sorry):
-- 
--     :t fac four

开始毕业的Haskell程序员(研究生教育倾向于从小问题中解放出来,例如,基于硬件的整数的效率)

-- the natural numbers, a la Peano

data Nat = Zero | Succ Nat


-- iteration and some applications

iter z s  Zero    = z
iter z s (Succ n) = s (iter z s n)

plus n = iter n     Succ
mult n = iter Zero (plus n)


-- primitive recursion

primrec z s  Zero    = z
primrec z s (Succ n) = s n (primrec z s n)


-- two versions of factorial

fac  = snd . iter (one, one) (\(a,b) -> (Succ a, mult a b))
fac' = primrec one (mult . Succ)


-- for convenience and testing (try e.g. "fac five")

int = iter 0 (1+)

instance Show Nat where
  show = show . int

(zero : one : two : three : four : five : _) = iterate Succ Zero


Origamist Haskell programmer
(always starts out with the “basic Bird fold”)

-- (curried, list) fold and an application

fold c n []     = n
fold c n (x:xs) = c x (fold c n xs)

prod = fold (*) 1


-- (curried, boolean-based, list) unfold and an application

unfold p f g x = 
  if p x 
     then [] 
     else f x : unfold p f g (g x)

downfrom = unfold (==0) id pred


-- hylomorphisms, as-is or "unfolded" (ouch! sorry ...)

refold  c n p f g   = fold c n . unfold p f g

refold' c n p f g x = 
  if p x 
     then n 
     else c (f x) (refold' c n p f g (g x))


-- several versions of factorial, all (extensionally) equivalent

fac   = prod . downfrom
fac'  = refold  (*) 1 (==0) id pred
fac'' = refold' (*) 1 (==0) id pred

笛卡尔倾向于Haskell程序员(喜欢希腊食物,避免辛辣印度的东西;灵感来自Lex Augusteijn的"Sorting态射"[3])

-- (product-based, list) catamorphisms and an application

cata (n,c) []     = n
cata (n,c) (x:xs) = c (x, cata (n,c) xs)

mult = uncurry (*)
prod = cata (1, mult)


-- (co-product-based, list) anamorphisms and an application

ana f = either (const []) (cons . pair (id, ana f)) . f

cons = uncurry (:)

downfrom = ana uncount

uncount 0 = Left  ()
uncount n = Right (n, n-1)


-- two variations on list hylomorphisms

hylo  f  g    = cata g . ana f

hylo' f (n,c) = either (const n) (c . pair (id, hylo' f (c,n))) . f

pair (f,g) (x,y) = (f x, g y)


-- several versions of factorial, all (extensionally) equivalent

fac   = prod . downfrom
fac'  = hylo  uncount (1, mult)
fac'' = hylo' uncount (1, mult)

博士 Haskell程序员(吃了很多香蕉,他的眼睛出了问题,现在他需要新的镜片!)

-- explicit type recursion based on functors

newtype Mu f = Mu (f (Mu f))  deriving Show

in      x  = Mu x
out (Mu x) = x


-- cata- and ana-morphisms, now for *arbitrary* (regular) base functors

cata phi = phi . fmap (cata phi) . out
ana  psi = in  . fmap (ana  psi) . psi


-- base functor and data type for natural numbers,
-- using a curried elimination operator

data N b = Zero | Succ b  deriving Show

instance Functor N where
  fmap f = nelim Zero (Succ . f)

nelim z s  Zero    = z
nelim z s (Succ n) = s n

type Nat = Mu N


-- conversion to internal numbers, conveniences and applications

int = cata (nelim 0 (1+))

instance Show Nat where
  show = show . int

zero = in   Zero
suck = in . Succ       -- pardon my "French" (Prelude conflict)

plus n = cata (nelim n     suck   )
mult n = cata (nelim zero (plus n))


-- base functor and data type for lists

data L a b = Nil | Cons a b  deriving Show

instance Functor (L a) where
  fmap f = lelim Nil (\a b -> Cons a (f b))

lelim n c  Nil       = n
lelim n c (Cons a b) = c a b

type List a = Mu (L a)


-- conversion to internal lists, conveniences and applications

list = cata (lelim [] (:))

instance Show a => Show (List a) where
  show = show . list

prod = cata (lelim (suck zero) mult)

upto = ana (nelim Nil (diag (Cons . suck)) . out)

diag f x = f x x

fac = prod . upto


Post-doc Haskell programmer
(from Uustalu, Vene and Pardo’s “Recursion Schemes from Comonads” [4])

-- explicit type recursion with functors and catamorphisms

newtype Mu f = In (f (Mu f))

unIn (In x) = x

cata phi = phi . fmap (cata phi) . unIn


-- base functor and data type for natural numbers,
-- using locally-defined "eliminators"

data N c = Z | S c

instance Functor N where
  fmap g  Z    = Z
  fmap g (S x) = S (g x)

type Nat = Mu N

zero   = In  Z
suck n = In (S n)

add m = cata phi where
  phi  Z    = m
  phi (S f) = suck f

mult m = cata phi where
  phi  Z    = zero
  phi (S f) = add m f


-- explicit products and their functorial action

data Prod e c = Pair c e

outl (Pair x y) = x
outr (Pair x y) = y

fork f g x = Pair (f x) (g x)

instance Functor (Prod e) where
  fmap g = fork (g . outl) outr


-- comonads, the categorical "opposite" of monads

class Functor n => Comonad n where
  extr :: n a -> a
  dupl :: n a -> n (n a)

instance Comonad (Prod e) where
  extr = outl
  dupl = fork id outr


-- generalized catamorphisms, zygomorphisms and paramorphisms

gcata :: (Functor f, Comonad n) =>
           (forall a. f (n a) -> n (f a))
             -> (f (n c) -> c) -> Mu f -> c

gcata dist phi = extr . cata (fmap phi . dist . fmap dupl)

zygo chi = gcata (fork (fmap outl) (chi . fmap outr))

para :: Functor f => (f (Prod (Mu f) c) -> c) -> Mu f -> c
para = zygo In


-- factorial, the *hard* way!

fac = para phi where
  phi  Z             = suck zero
  phi (S (Pair f n)) = mult f (suck n)


-- for convenience and testing

int = cata phi where
  phi  Z    = 0
  phi (S f) = 1 + f

instance Show (Mu N) where
  show = show . int

终身教授(向新生传授Haskell)

fac n = product [1..n]

Answer 4

感谢Christoph,一个适用于不少"数字"的C99解决方案:

#include <math.h>
#include <stdio.h>

double fact(double x)
{
  return tgamma(x+1.);
}

int main()
{
  printf("%f %f\n", fact(3.0), fact(5.0));
  return 0;
}

产生6.000000 120.000000

Answer 5

对于大n你可能会遇到一些问题,你可能想要使用斯特林的近似值:

这是:

Answer 6

如果您的主要目标是一个有趣的功能:

int facorial(int a) {
   int b = 1, c, d, e;
   a--;
   for (c = a; c > 0; c--)
   for (d = b; d > 0; d--)
   for (e = c; e > 0; e--)
   b++;
   return b;
}

(不推荐作为实际使用的算法.)

Answer 7

尾递归版本:

long factorial(long n)
{
    return tr_fact(n, 1);
}
static long tr_fact(long n, long result)
{
    if(n==1)
        return result;
    else
        return tr_fact(n-1, n*result);
}

Answer 8

在C99(或Java)中,我会像这样迭代地编写阶乘函数:

int factorial(int n)
{
    int result = 1;
    for (int i = 2; i <= n; i++)
    {
        result *= i;
    }
    return result;
}

• C不是函数式语言,您不能依赖尾调优化.所以除非你需要,否则不要在C(或Java)中使用递归.

• 仅仅因为阶乘通常被用作递归的第一个例子,它并不意味着你需要递归来计算它.

• 如果n太大,这将无声地溢出,就像C(和Java)中的自定义一样.

• 如果int表示的数字对于您想要计算的因子而言太小,则选择另一个数字类型.如果你需要更大一点,浮动或加倍,如果n不是太大,你不介意一些不精确,或者如果你想要真正大的因子的确切值,那么大整数.

Answer 9

这是一个使用OPENSSL的BIGNUM实现的C程序,因此对学生不是特别有用.(当然接受BIGNUM作为输入参数是疯狂的,但有助于演示BIGNUM之间的交互).

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <openssl/crypto.h>
#include <openssl/bn.h>

BIGNUM *factorial(const BIGNUM *num)
{
    BIGNUM *count = BN_new();
    BIGNUM *fact = NULL;
    BN_CTX *ctx = NULL;

    BN_one(count);
    if( BN_cmp(num, BN_value_one()) <= 0 )
    {
        return count;
    }

    ctx = BN_CTX_new();
    fact = BN_dup(num);
    BN_sub(count, fact, BN_value_one());
    while( BN_cmp(count, BN_value_one()) > 0 )
    {
        BN_mul(fact, count, fact, ctx);
        BN_sub(count, count, BN_value_one());
    }

    BN_CTX_free(ctx);
    BN_free(count);

    return fact;
}

此测试程序显示如何为输入创建数字以及如何处理返回值:

int main(int argc, char *argv[])
{
    const char *test_cases[] =
    {
        "0", "1",
        "1", "1",
        "4", "24",
        "15", "1307674368000",
        "30", "265252859812191058636308480000000",
        "56", "710998587804863451854045647463724949736497978881168458687447040000000000000",
        NULL, NULL
    };
    int index = 0;
    BIGNUM *bn = NULL;
    BIGNUM *fact = NULL;
    char *result_str = NULL;

    for( index = 0; test_cases[index] != NULL; index += 2 )
    {
        BN_dec2bn(&bn, test_cases[index]);

        fact = factorial(bn);

        result_str = BN_bn2dec(fact);
        printf("%3s: %s\n", test_cases[index], result_str);
        assert(strcmp(result_str, test_cases[index + 1]) == 0);

        OPENSSL_free(result_str);
        BN_free(fact);
        BN_free(bn);
        bn = NULL;
    }

    return 0;
}

用gcc编译:

gcc factorial.c -o factorial -g -lcrypto

Answer 10

int factorial(int n){
    return n <= 1 ? 1 : n * factorial(n-1);
}