我正在努力提高大数阶乘计算的运行时间。
第一个简单循环和相乘的代码。
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(1, number + 1, 1):
result *= x
return result
此函数的分析结果:
For n = 1000 -- Total time: 0.001115 s
for n = 10000 -- Total time: 0.035327 s
for n = 100000 -- Total time: 3.77454 s.
从 n = 100000 的行分析器我可以看到大部分时间都花在乘法步骤上,即“98.8”
31 100000 3728380 37.3 98.8 result *= x
所以试图减少阶乘的乘法 减半,对于偶数,因此进行强度降低。
后半部分乘法代码:
def calculate_factorial_multi_half(number):
if number == 1 or number == 0:
return 1
handle_odd = False
upto_number = number
if number & 1 == 1:
upto_number -= 1
print upto_number
handle_odd = True
next_sum = upto_number
next_multi = upto_number
factorial = 1
while next_sum >= 2:
factorial *= next_multi
next_sum -= 2
next_multi += next_sum
if handle_odd:
factorial *= number
return factorial
此函数的分析结果:
For n = 1000 -- Total time: 0.00115 s
for n = 10000 -- Total time: 0.023636 s
for n = 100000 -- Total time: 3.65019 s
这在中等范围内显示出一些改进,但在缩放方面没有太大改进。
在这个函数中,大部分时间都花在了乘法上。
61 50000 3571928 71.4 97.9 factorial *= next_multi.
所以我厌倦了删除尾随零然后相乘。
没有尾随零代码。
def calculate_factorial_multi_half_trailO(number):
'''
Removes the trailling zeros
'''
if number == 1 or number == 0:
return 1
handle_odd = False
upto_number = number
if number & 1 == 1:
upto_number -= 1
handle_odd = True
next_sum = upto_number
next_multi = upto_number
factorial = 1
total_shift = 0
while next_sum >= 2:
factorial *= next_multi
shift = len(str(factorial)) - len(str(factorial).rstrip('0'))
total_shift += shift
factorial >>= shift
next_sum -= 2
next_multi += next_sum
if handle_odd:
factorial *= number
factorial <<= total_shift
return factorial
此函数的分析结果:
For n = 1000 -- Total time: 0.061524 s
for n = 10000 -- Total time: 113.824 s
因此,由于字符串转换,“96.2%”的时间花在了字符串转换上,而不是减少时间,而是增加了时间
22 500 59173 118.3 96.2 shift = len(str(factorial)) - len(str(factorial).rstrip('0')).
所以我的问题是如何在不影响时间的情况下获取尾随零并有效地使用 shift。
所有分析都已完成。 基本操作系统(Linux):64 位,内存:6GB
最佳答案
没有尾随零似乎效率不高。
首先,我建议使用 prime decomposition减少乘法总数,因为小于 x
的素数约为 x/lnx
。
def calculate_factorial_multi(number):
prime = [True]*(number + 1)
result = 1
for i in xrange (2, number+1):
if prime[i]:
#update prime table
j = i+i
while j <= number:
prime[j] = False
j += i
#calculate number of i in n!
sum = 0
t = i
while t <= number:
sum += number/t
t *= i
result *= i**sum
return result
for n = 10000, total time : 0.017s
for n = 100000, total time : 2.047s
for n = 500000, total time : 65.324s
(PS。在你的第一个程序中,n = 100000,在我的机器上总时间是 3.454s。)
现在让我们测试它是否有效而不尾随零。尾随零的数量等于 n!
中质因数 5
的数量。
程序是这样的
def calculate_factorial_multi2(number):
prime = [True]*(number + 1)
result = 1
factor2 = 0
factor5 = 0
for i in xrange (2, number+1):
if prime[i]:
#update prime table
j = i+i
while j <= number:
prime[j] = False
j += i
#calculate the number of i in factors of n!
sum = 0
t = i
while t <= number:
sum += number/t
t *= i
if i == 2:
factor2 = sum
elif i == 5:
factor5 = sum
else:
result *= i**sum
return (result << (factor2 - factor5))*(10**factor5)
for n = 10000, total time : 0.015s
for n = 100000, total time : 1.896s
for n = 500000, total time : 57.101s
只是比以前快了一点。所以没有尾随零似乎不是很有用
关于python - 有效地计算没有尾随零的阶乘?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33340158/