Question

我需要一种方法来计算Python中长整数的第n个根。

我尝试了pow(m, 1.0/n)，但它不起作用：

OverflowError：long int太大而无法转换为float

有什么想法吗？

长整数我的意思是真正的长整数，如：

11968003966030964356885611480383408833172346450467339251 196093144141045683463085291115677488411620264826942334897996389 485046262847265769280883237649461122479734279424416861834396522 819159219215308460065265520143082728303864638821979329804885526 557893649662037092457130509980883789368448042961108430809620626 059287437887495827369474189818588006905358793385574832590121472 680866521970802708379837148646191567765584039175249171110593159 305029014037881475265618958103073425958633163441030267478942720 703134493880117805010891574606323700178176718412858948243785754 898788359757528163558061136758276299059029113119763557411729353 915848889261125855717014320045292143759177464380434854573300054 940683350937992500211758727939459249163046465047204851616590276 724564411037216844005877918224201569391107769029955591465502737 961776799311859881060956465198859727495735498887960494256488224 613682478900505821893815926193600121890632

Answer 1

如果这是一个非常大的数字。您可以使用二进制搜索。

def find_invpow(x,n):
    """Finds the integer component of the n'th root of x,
    an integer such that y ** n <= x < (y + 1) ** n.
    """
    high = 1
    while high ** n <= x:
        high *= 2
    low = high/2
    while low < high:
        mid = (low + high) // 2
        if low < mid and mid**n < x:
            low = mid
        elif high > mid and mid**n > x:
            high = mid
        else:
            return mid
    return mid + 1

例如：

>>> x = 237734537465873465
>>> n = 5
>>> y = find_invpow(x,n)
>>> y
2986
>>> y**n <= x <= (y+1)**n
True
>>>
>>> x = 119680039660309643568856114803834088331723464504673392511960931441>
>>> n = 45
>>> y = find_invpow(x,n)
>>> y
227661383982863143360L
>>> y**n <= x < (y+1)**n
True
>>> find_invpow(y**n,n) == y
True
>>>

Answer 2

Gmpy是一个C编码的Python扩展模块，它包装了GMP库，为Python代码提供了快速的多精度算法（整数，有理数和浮点数），随机数生成，高级数理论函数等等。

包含root功能：

x.root（n）：返回一个2元素的元组（y，m），这样y就是（可能被截断）x的第n个根; m，一个普通的Python int，如果根是精确的（x == y ** n）则为1，否则为0. n必须是普通的 Python int，＆gt; = 0。

例如，第20根：

>>> import gmpy
>>> i0=11968003966030964356885611480383408833172346450467339251 
>>> m0=gmpy.mpz(i0)
>>> m0
mpz(11968003966030964356885611480383408833172346450467339251L)
>>> m0.root(20)
(mpz(567), 0)

Answer 3

如果您正在寻找标准的东西，快速写入高精度。我会使用十进制并将精度（getcontext（）。prec）调整到至少x的长度。

代码（Python 3.0）

from decimal import *

x =   '11968003966030964356885611480383408833172346450467339251\
196093144141045683463085291115677488411620264826942334897996389\
485046262847265769280883237649461122479734279424416861834396522\
819159219215308460065265520143082728303864638821979329804885526\
557893649662037092457130509980883789368448042961108430809620626\
059287437887495827369474189818588006905358793385574832590121472\
680866521970802708379837148646191567765584039175249171110593159\
305029014037881475265618958103073425958633163441030267478942720\
703134493880117805010891574606323700178176718412858948243785754\
898788359757528163558061136758276299059029113119763557411729353\
915848889261125855717014320045292143759177464380434854573300054\
940683350937992500211758727939459249163046465047204851616590276\
724564411037216844005877918224201569391107769029955591465502737\
961776799311859881060956465198859727495735498887960494256488224\
613682478900505821893815926193600121890632'

minprec = 27
if len(x) > minprec: getcontext().prec = len(x)
else:                getcontext().prec = minprec

x = Decimal(x)
power = Decimal(1)/Decimal(3)

answer = x**power
ranswer = answer.quantize(Decimal('1.'), rounding=ROUND_UP)

diff = x - ranswer**Decimal(3)
if diff == Decimal(0):
    print("x is the cubic number of", ranswer)
else:
    print("x has a cubic root of ", answer)

答案

x是立方数22873918786185635329056863961725521583023133411 451452349318109627653540670761962215971994403670045614485973722724603798 107719978813658857014190047742680490088532895666963698551709978502745901 704433723567548799463129652706705873694274209728785041817619032774248488 2965377218610139128882473918261696612098418

Answer 4

你可以通过避免使用while循环来支持将低值设置为10 **（len（str（x））/ n）和从高到低* 10来使其运行得更快。可能更好的是替换len（ str（x））具有按位长度并使用位移。根据我的测试，我估计从第一次开始加速5％，从第二开始加速25％。如果整数足够大，这可能很重要（并且加速可能会有所不同）。如果不仔细测试，请不要相信我的代码。我做了一些基本测试，但可能错过了边缘情况。此外，这些加速比例随所选数量而变化。

如果您使用的实际数据比您在此处发布的数据大得多，则此更改可能值得。

from timeit import Timer

def find_invpow(x,n):
    """Finds the integer component of the n'th root of x,
    an integer such that y ** n <= x < (y + 1) ** n.
    """
    high = 1
    while high ** n < x:
        high *= 2
    low = high/2
    while low < high:
        mid = (low + high) // 2
        if low < mid and mid**n < x:
            low = mid
        elif high > mid and mid**n > x:
            high = mid
        else:
            return mid
    return mid + 1

def find_invpowAlt(x,n):
    """Finds the integer component of the n'th root of x,
    an integer such that y ** n <= x < (y + 1) ** n.
    """
    low = 10 ** (len(str(x)) / n)
    high = low * 10

    while low < high:
        mid = (low + high) // 2
        if low < mid and mid**n < x:
            low = mid
        elif high > mid and mid**n > x:
            high = mid
        else:
            return mid
    return mid + 1

x = 237734537465873465
n = 5
tests = 10000

print "Norm", Timer('find_invpow(x,n)', 'from __main__ import find_invpow, x,n').timeit(number=tests)
print "Alt", Timer('find_invpowAlt(x,n)', 'from __main__ import find_invpowAlt, x,n').timeit(number=tests)

Norm 0.626754999161

Alt 0.566340923309

Answer 5

哦，对于数字大，你会使用十进制模块。

ns：您的数字为字符串

ns = "11968003966030964356885611480383408833172346450467339251196093144141045683463085291115677488411620264826942334897996389485046262847265769280883237649461122479734279424416861834396522819159219215308460065265520143082728303864638821979329804885526557893649662037092457130509980883789368448042961108430809620626059287437887495827369474189818588006905358793385574832590121472680866521970802708379837148646191567765584039175249171110593159305029014037881475265618958103073425958633163441030267478942720703134493880117805010891574606323700178176718412858948243785754898788359757528163558061136758276299059029113119763557411729353915848889261125855717014320045292143759177464380434854573300054940683350937992500211758727939459249163046465047204851616590276724564411037216844005877918224201569391107769029955591465502737961776799311859881060956465198859727495735498887960494256488224613682478900505821893815926193600121890632"
from decimal import Decimal
d = Decimal(ns)
one_third = Decimal("0.3333333333333333")
print d ** one_third

，答案是：2.287391878618402702753613056E + 305

TZ指出这不准确......他是对的。这是我的考试。

from decimal import Decimal

def nth_root(num_decimal, n_integer):
    exponent = Decimal("1.0") / Decimal(n_integer)
    return num_decimal ** exponent

def test():
    ns = "11968003966030964356885611480383408833172346450467339251196093144141045683463085291115677488411620264826942334897996389485046262847265769280883237649461122479734279424416861834396522819159219215308460065265520143082728303864638821979329804885526557893649662037092457130509980883789368448042961108430809620626059287437887495827369474189818588006905358793385574832590121472680866521970802708379837148646191567765584039175249171110593159305029014037881475265618958103073425958633163441030267478942720703134493880117805010891574606323700178176718412858948243785754898788359757528163558061136758276299059029113119763557411729353915848889261125855717014320045292143759177464380434854573300054940683350937992500211758727939459249163046465047204851616590276724564411037216844005877918224201569391107769029955591465502737961776799311859881060956465198859727495735498887960494256488224613682478900505821893815926193600121890632"
    nd = Decimal(ns)
    cube_root = nth_root(nd, 3)
    print (cube_root ** Decimal("3.0")) - nd

if __name__ == "__main__":
    test()

关闭大约10 ** 891

Answer 6

可能是为了你的好奇心：

http://en.wikipedia.org/wiki/Hensel_Lifting

这可能是Maple用来实际找到大数字的第n个根的技术。

提出x^n - 11968003.... = 0 mod p的事实，并从那里开始......

Answer 7

在旧版本的Python中，1/3等于0.在Python 3.0中，1/3等于0.33333333333（并且1//3等于0）。

因此，要么将代码更改为使用1/3.0，要么切换到Python 3.0。

Answer 8

我提出了自己的答案，这需要@Mahmoud Kassem的想法，简化代码，并使其更具可重用性：

def cube_root(x):
    return decimal.Decimal(x) ** (decimal.Decimal(1) / decimal.Decimal(3))

我在Python 3.5.1和Python 2.7.8中进行了测试，它似乎运行良好。

结果将包含运行函数的十进制上下文指定的位数，默认情况下为28位小数。根据{{1}}模块中power功能的文档，＆＃34; The result is well-defined but only “almost always correctly-rounded”.＆＃34;。如果您需要更准确的结果，可以按如下方式完成：

decimal

Answer 9

尝试将指数转换为浮点数，因为Python中/的默认行为是整数除法

<磷>氮**（1 /浮子（3））

Answer 10

好吧，如果你不是特别担心精确度，你可以把它转换为刺痛，砍掉一些数字，使用指数函数，然后将结果乘以你切断多少的根。

E.g。 32123大约等于32 * 1000，立方根大约等于立方根32 *立方根1000.后者可以通过将0的数除以3来计算。

这样就无需使用扩展模块。

如何计算一个非常大的整数的第n个根

10 个答案:

代码（Python 3.0）

答案