Question

I m试图缩小一组数字,以输入DP子集-总算法。 (如果数字太大,就会打击。) 具体来说,我需要找到10人的最大力量,能够在不失去精确性的情况下将人数分成几个部分。我的工作是例行的,但由于工作往往在 lo中进行,我希望工作比我所介绍的简陋武力方法更快。我的数字是 Dec。

from decimal import Decimal
import math

def largest_common_power_of_10(numbers: list[Decimal]) -> int:
    """
    Determine the largest power of 10 in list of numbers that will divide into all numbers
    without losing a significant digit left of the decimal point
    """
    min_exponent = float( inf )
    for num in numbers:
        if num != 0:
            # Count the number of trailing zeros in the number
            exponent = 0
            while num % 10 == 0:
                num //= 10
                exponent += 1
            min_exponent = min(min_exponent, exponent)

    # The largest power of 10 is 10 raised to the min_exponent
    return int(min_exponent)


decimal_numbers = [Decimal("1234"), Decimal("5000"), Decimal("200")]
result = largest_common_power_of_10(decimal_numbers)
assert(result == 0)
decimal_numbers = [Decimal(470_363_000.0000), Decimal(143_539_000.0000), Decimal(1_200_000.0000)]
result = largest_common_power_of_10(decimal_numbers)
assert(result == 3)
divisor = 10**result
# Later processing can use scaled_list
scaled_list = [x/divisor for x in decimal_numbers]
assert(scaled_list == [Decimal( 470363 ), Decimal( 143539 ), Decimal( 1200 )])
reconstituted_list = [x * divisor for x in scaled_list]
assert(reconstituted_list == decimal_numbers)

Answer 1

如果您的名单上所有的人都感到愤怒,就可以立即与数学图书馆一起这样做。

import math

def largest_common_power_of_10(numbers):
    
    # get the greatest power of 10 that divides all numbers in list
    gcd_nums = math.gcd(*numbers)
    gcd = [x for x in range(1, len(str(gcd_nums))) if gcd_nums % math.pow(10, x) == 0]

    if len(gcd) == 0:
        return 0
    else:
        gcp = max(gcd)
        return gcp

Examples:

numbers = [11230000,125,44500000]
largest_common_power_of_10(numbers)
# 0

numbers = [11230000,1540,44500000]
largest_common_power_of_10(numbers)
# 1

numbers = [11230000,1540000,44500000]
largest_common_power_of_10(numbers)
# 4

注:

这只可能与3.9和更大范围的工作,因为在释放时,数学.gcd()明显接受名单。

Answer 2

您可使用https://docs.python.org/3/library/math.html#math.gcd” rel=“nofollow noreferer”math.gcd获取特定数字的最普通分母,然后使用

import math
from decimal import Decimal

def largest_common_power_of_10(numbers):
    return Decimal(math.gcd(*numbers)).normalize().as_tuple().exponent

因此:

print(largest_common_power_of_10([1234, 5000, 200]))
print(largest_common_power_of_10([470_363_000, 143_539_000, 1_200_000]))
print(largest_common_power_of_10([11_230_000, 1_540_000, 44_500_000]))

产出:

0
3
4

Demo:

马达加斯加 (从@KellyBundy借用的最新代码):

  0.11 ± 0.01 ms  blhsing
  0.15 ± 0.00 ms  JesseSealand
 85.91 ± 0.50 ms  MikeP

友情链接