python 如何求N的阶乘

x = int(input())
a = 1
for i in range(1, x+1):
    a = a*i
print("product = %d" % float(a))

n = int(input())
ns = 1
for i in range(1,n+1):
    ns = ns*i
print(ns)

def factorial(n):
    if n==1:
        return n
    else:
        return n*factorial(n-1)
n = int(input())
res = factorial(n)
print(res)

n= int(input())
ns = [0 for i in range(10000) ]
length = 1
ns[0] = length = 1
if n>=2:
    for i in range(2,n+1):
        carry = 0
        for j in range(length):
            temp = ns[j] * i + carry
            carry = int(temp/10)
            ns[j] = temp % 10
        while carry>0:
            ns[length] += carry%10
            length+=1
            carry = int(carry/10)
while length>0:
    length -=1
    print(ns[length],end='')

    ns[0] = length =1 , carry = 0
    ∴j in range(1)

    temp = ns[j] * i + carry = ns[0] * i + carry =1*1+0=1  # temp为第j位数与i相乘并加上j-1位数与i相乘后进位的值的结果
    carry = int(temp/10) = 1/10 = 0      # carry=0所以不用进位
    ns[j] = temp % 10 即 ns[0] = 1 % 10 =1   #只取个位数值作为第j位的值

    ns[0] = 1, length =1 , carry = 0
    ∴j in range(1)

    temp = ns[j] * i + carry = ns[0] * i + carry =1*2+0=2  # temp为第j位数与i相乘并加上j-1位数与i相乘后进位的值的结果
    carry = int(temp/10) = 2 / 10 = 0      # carry=0所以不用进位
    ns[j] = temp % 10 即 ns[0] = 2 % 10 =2   #只取个位数值作为第j位的值
    #这样就已经的到2!的值了即2

    ns[0] = 2, length =1 , carry = 0
    ∴j in range(1)

    temp = ns[j] * i + carry = ns[0] * i + carry =2*3+0=6  # temp为第j位数与i相乘并加上j-1位数与i相乘后进位的值的结果
    carry = int(temp/10) = 6 / 10 = 0      # carry=0所以不用进位
    ns[j] = temp % 10 即 ns[0] = 6 % 10 =6   #只取个位数值作为第j位的值
    #这样就已经的到3!的值了即6

    ns[0] = 6, length =1 , carry = 0
    ∴j in range(1)

    temp = ns[j] * i + carry = ns[0] * i + carry =6*4+0=24  # temp为第j位数与i相乘并加上j-1位数与i相乘后进位的值的结果
    carry = int(temp/10) = 24 / 10 = 2      # carry=2>0所以需要向前进2
    ns[j] = temp % 10 即 ns[0] = 24 % 10 =4   #只取个位数值作为第j位的值

    while carry>0:        
            ns[length] += carry%10 即 ns[1] += 2 % 10 = 2  #carry = 2 所以向前进2
            length+=1 即 length =1+1=2 #位数加一
            carry = int(carry/10) = 2 / 10 = 0 # carry = 2<10所以不需要继续进位，while循环结束
            ∴length = 2 , ns[0] = 4 ,ns[1] = 2
    #这样就得到4!的值ns[1]*10+ns[0] 即 24,输出时可直接倒着打印然后end=''而不需要每位数乘10*n再相加

    ns[0] = 4, ns[1] = 2 length =2 , carry = 0
    ∴j in range(2)

    temp = ns[j] * i + carry = ns[0] * i + carry =4*5+0=20  # temp为第j位数与i相乘并加上j-1位数与i相乘后进位的值的结果
    carry = int(temp/10) = 20 / 10 = 2      # carry=2>0所以需要向前进2
    ns[j] = temp % 10 即 ns[0] = 20 % 10 =0   #只取个位数值作为第j位的值

    temp = ns[j] * i + carry = ns[1] * i + carry =2*5+2=12  # temp为第j位数与i相乘并加上j-1位数与i相乘后进位的值的结果
    carry = int(temp/10) = 12 / 10 = 1      # carry=1>0所以需要向前进1
    ns[j] = temp % 10 即 ns[1] = 12 % 10 =2   #只取个位数值作为第j位的值

    while carry>0:        
            ns[length] += carry%10 即 ns[2] += 1 % 10 = 1  #carry = 1 所以向前进2
            length+=1 即 length =2 +1 = 3 #位数加一
            carry = int(carry/10) = 1 / 10 = 0 # carry = 1<10所以不需要继续进位，while循环结束
            ∴length = 3 , ns[0] = 0 , ns[1] = 2 , ns[2] = 1
    # 这样就得到5!的值ns[2] ns[1] ns[0]即 120