如何计算一个很大整数的n次根

Question

我需要一种在Python中计算长整数第n个根的方法。

我试过pow(m, 1.0/n)，但没用：

OverflowError: long int too large to convert to float

有什么想法吗？

我指的是很长的整数，比如：

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

Answer 1

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

包括一个root函数：

x.root(n): returns a 2-element tuple (y,m), such that y is the (possibly truncated) n-th root of x; m, an ordinary Python int, is 1 if the root is exact (x==y**n), else 0. n must be an ordinary Python int, >=0.

例如，20根：

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

Answer 2

通过避免while循环，将low设置为10**（len（str（x））/n）和high设置为low*10，可以使它运行得稍微快一点。最好是用按位长度替换len（str（x））并使用位移位。根据我的测试，我估计第一次加速5%，第二次加速25%。如果整数足够大，这可能很重要（加速可能会有所不同）。如果不仔细测试代码，就不要相信我的代码。我做了一些基本测试，但可能漏掉了一个边缘案例。此外，这些加速值随所选的数字而变化。

如果您使用的实际数据比您在这里发布的数据大得多，那么这个更改可能是值得的。

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)

标准值0.626754999161

海拔0.566340923309

Answer 3

如果真的是个大数字。你可以用二进制搜索。

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
>>>