在Python中使用math模块的sqrt函数来处理长数字[重复]

Question

我正在Python中处理200位数字。当使用 math.sqrt(n) 求数字的平方根时，我得到了错误的答案。

In[1]: n=9999999999999999999999999999999999999999999999999999999999999999999999
       999999999999999999999999998292000000000000000000000000000000000000000000
       0000000000000000000000000000000000000000000000000000726067

In[2]: x=int(math.sqrt(n))

In[3]: x
Out[1]: 10000000000000000159028911097599180468360808563945281389781327
        557747838772170381060813469985856815104L

In[4]: x*x
Out[2]: 1000000000000000031805782219519836346574107361670094060730052612580
        0264077231077619856175974095677538298443892851483731336069235827852
        3336313169161345893842466001164011496325176947445331439002442530816L

In[5]: math.sqrt(n)
Out[3]: 1e+100

x 的值比预期的要大，因为 x*x（201 位数字）大于 n（200 位数字）。这里发生了什么？我在这里弄错了一些概念吗？我还能怎样找到非常大的数的根？

Answer 1

使用小数模块：

import decimal
D = decimal.Decimal
n = D(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)

with decimal.localcontext() as ctx:
    ctx.prec = 300
    x = n.sqrt()
    print(x)
    print(x*x)
    print(n-x*x)

产量

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999145.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999983754999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998612677

99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

0E-100

Answer 2

math.sqrt

返回 IEEE-754 64 位结果，大约为 17 位数字。还有其他库可以处理高精度值。除了上面提到的

decimal

和

mpmath

库之外，我还维护

gmpy2

库 (https://code.google.com/p/gmpy/)。

>>> import gmpy2
>>> n=gmpy2.mpz(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)
>>> gmpy2.get_context().precision=2048
>>> x=gmpy2.sqrt(n)
>>> x*x
mpfr('99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0',2048)
>>>

gmpy2

库还可以返回整数平方根（

isqrt

）或快速检查整数是否是精确平方（

is_square

）。

Answer 3

这是我不久前写的一个使用英雄方法的整数平方根程序。对于初始近似，它使用输入值位长度一半的数字，因此它很快开始收敛。然而，我还没有对其进行计时，看看它在 Python 中是否比仅使用更简单的初始近似更快。 :)

#! /usr/bin/env python

''' Long integer square roots. Newton's method.

    Written by PM 2Ring. Adapted from C to Python 2008.10.19
'''

import sys

def root(m):
    # Get initial approximation
    n, a, k = m, 1, 0
    while n > a:
        n >>= 1
        a <<= 1
        k += 1
        #print k, ':', n, a

    # Go back one step & average
    a = n + (a>>2)
    #print a

    # Apply Newton's method
    while k:
        a = (a + m // a) >> 1
        k >>= 1
        #print k, ':', a
    return a

def main():
    m = len(sys.argv) > 1 and int(sys.argv[1]) or 2*10L**100
    print "The Square Root of", m
    print root(m)

if __name__ == '__main__':
    main()