Question

我有两个大号a和b，长度大约为10000，这样a <= b。现在，我必须找到c = a / b，最多10位小数，如何在不失去精度的情况下完成？

Answer 1

decimal模块应该有效。如TigerhawkT3的链接所示，您可以选择商数应该是小数位数。

from decimal import *
getcontext().prec = 6
a = float(raw_input('The first number:'))       #Can be int() if needed
b = float(raw_input('The second number:'))
c = Decimal(a) / Decimal(b)
print float(c)

Answer 2

您可以使用decimal模块：

from decimal import localcontext, Decimal

def foo(a, b):
    with localcontext() as ctx:
        ctx.prec = 10   # Sets precision to 10 places temporarily
        c = Decimal(a) / Decimal(b) # Not sure if this is precise if a and b are floats, 
                                    # str(a) and str(b) instead'd ensure precision i think.
    return float(c)

Answer 3

您可以使用此函数计算任何类型的长度浮点数

def longdiv(divisor,divident):
    quotient,remainder=divmod(divisor,divident)
    return (str(quotient)+str(remainder*1.0/divident)[1:])

在python中浮动大数字的划分

3 个答案: