如何在Java中对BigDecimal进行分数处理？ [英] How to do a fractional power on BigDecimal in Java?

问题描述

在我的小项目中，我需要做一些像Math.pow（7777.66,5555.44）这样的东西只有非常大的数字。我遇到了一些解决方案：

In my little project I need to do something like Math.pow(7777.66, 5555.44) only with VERY big numbers. I came across a few solutions:

• 使用双倍 - 但数字太大


• 使用BigDecimal.pow但不支持分数


• 使用X ^（A + B）= X ^ A * X ^ B公式（B是第二个数字的余数），但是不支持大X或大A，因为我仍然转换为双


• 使用某种泰勒系列算法或类似的东西 - 我不是很擅长数学所以这一个是我的最后一个选项，如果我找不到任何解决方案（某些库或（A + B）^（C + D）的公式）。

• Use double - but the numbers are too big
• Use BigDecimal.pow but no support for fractional
• Use the X^(A+B)=X^A*X^B formula (B is the remainder of the second num), but again no support for big X or big A because I still convert to double
• Use some kind of Taylor series algorithm or something like that - I'm not very good at math so this one is my last option if I don't find any solutions (some libraries or a formula for (A+B)^(C+D)).

任何人都知道图书馆或简单的解决方案？我想很多人都处理同样的问题...

Anyone knows of a library or an easy solution? I figured that many people deal with the same problem...

p.s。
我发现了一些名为ApFloat的图书馆声称大约是这样做的，但我得到的结果非常接近甚至8 ^ 2给了我60 ...

p.s. I found some library called ApFloat that claims to do it approximately, but the results I got were so approximate that even 8^2 gave me 60...

推荐答案

1.7976931348623157E308（Double.MAX_VALUE）下的参数解决方案，但支持数字为MILLIONS的结果：

The solution for arguments under 1.7976931348623157E308 (Double.MAX_VALUE) but supporting results with MILLIONS of digits:

由于double支持数字高达MAX_VALUE（例如，100！in double看起来像这样：9.332621544394415E157），使用BigDecimal.doubleValue（）没有问题。但你不应该只做Math.pow（double，double），因为如果结果大于MAX_VALUE，你将获得无穷大。 SO：使用公式X ^（A + B）= X ^ A * X ^ B将计算分离为两个幂，使用BigDecimal.pow，使用Math将小（第二个参数的其余部分）分开。战争，然后繁殖。 X将被复制到DOUBLE - 确保它不大于MAX_VALUE，A将是INT（最大值2147483647但BigDecimal.pow不支持超过10亿的整数），B将是double，总是小于1。这样你可以执行以下操作（忽略我的私有常量等）：

Since double supports numbers up to MAX_VALUE (for example, 100! in double looks like this: 9.332621544394415E157), there is no problem to use BigDecimal.doubleValue(). But you shouldn't just do Math.pow(double, double) because if the result is bigger than MAX_VALUE you will just get infinity. SO: use the formula X^(A+B)=X^A*X^B to separate the calculation to TWO powers, the big, using BigDecimal.pow, and the small (remainder of the 2nd argument), using Math.pow, then multiply. X will be copied to DOUBLE - make sure it's not bigger than MAX_VALUE, A will be INT (maximum 2147483647 but the BigDecimal.pow doesn't support integers more than a billion anyway), and B will be double, always less than 1. This way you can do the following (ignore my private constants etc):

    int signOf2 = n2.signum();
    try {
        // Perform X^(A+B)=X^A*X^B (B = remainder)
        double dn1 = n1.doubleValue();
        // Compare the same row of digits according to context
        if (!CalculatorUtils.isEqual(n1, dn1))
            throw new Exception(); // Cannot convert n1 to double
        n2 = n2.multiply(new BigDecimal(signOf2)); // n2 is now positive
        BigDecimal remainderOf2 = n2.remainder(BigDecimal.ONE);
        BigDecimal n2IntPart = n2.subtract(remainderOf2);
        // Calculate big part of the power using context -
        // bigger range and performance but lower accuracy
        BigDecimal intPow = n1.pow(n2IntPart.intValueExact(),
                CalculatorConstants.DEFAULT_CONTEXT);
        BigDecimal doublePow =
            new BigDecimal(Math.pow(dn1, remainderOf2.doubleValue()));
        result = intPow.multiply(doublePow);
    } catch (Exception e) {
        if (e instanceof CalculatorException)
            throw (CalculatorException) e;
        throw new CalculatorException(
            CalculatorConstants.Errors.UNSUPPORTED_NUMBER_ +
                "power!");
    }
    // Fix negative power
    if (signOf2 == -1)
        result = BigDecimal.ONE.divide(result, CalculatorConstants.BIG_SCALE,
                RoundingMode.HALF_UP);

结果示例：

50!^10! = 12.50911317862076252364259*10^233996181

50!^0.06 = 7395.788659356498101260513

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