如何计算一个非常大的整数的第n个根

OverflowError：long int太大而无法转换为float

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

10 个答案:

``````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
``````

``````>>> 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
>>>
``````

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。

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。

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

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)

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

``````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

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()
``````

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

``decimal``

<磷>氮**（1 /浮子（3））

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。