A double 不能代表这样的数字，它太大了.范围大约在-10 ³⁰⁸ 和10 ³⁰⁸ 之间.

A double as defined by IEEE-754 can't represent such a number, it's too large. The bounds are approximately between -10³⁰⁸ and 10³⁰⁸.

您需要使用 BigDecimal 来表示它:一个具有任意字节数的数字来表示数字.

You need to use a BigDecimal to represent it: a number with an arbitrary number of bytes to represent numbers.

实现此目标的更好方式:

Better way to implement this:

double c = 0.0005d;//initialize c
double r = 0.01d;  //initialize r
double a = 0.0006d;//initialize a
BigDecimal abd = new BigDecimal(a); //BigDecimal for a
BigDecimal cbd = new BigDecimal(c); //BigDecimal for c
BigDecimal rbd = new BigDecimal(r); //BigDecimal for r
for (int counter = 1; counter <= t; counter++) {//perhaps other offset for counter?
   abd = abd.multiply(new BigDecimal(counter));
   cbd = cbd.multiply(rbd).add(abd);
}

这种方法的潜在问题是精度太高:Java将精确地计算所有操作，从而得出具有数千位数字的数字.由于每个运算都会炸掉位数，因此在几次迭代中，简单的加法和乘法运算变得不可行.

A potential problem with this approach is that the precision is too high: Java will calculate all operations exactly resulting in numbers that have thousands of digits. Since every operation blows up the number of digits, within a few iterations, simple addition and multiplication operations become unfeasible.

您可以通过使用可选的 MathContext 参数定义精度来解决此问题:它确定结果的精度.例如，您可以使用 MathContext.DECIMAL128 :

You can solve this by defining a precision using the optional MathContext parameter: it determines on how precise the result should be. You can for instance use MathContext.DECIMAL128:

int t = 101835;
double c = 0.0005d;//initialize c
double r = 0.01d;  //initialize r
double a = 0.0006d;//initialize a
BigDecimal abd = new BigDecimal(a); //BigDecimal for a
BigDecimal cbd = new BigDecimal(c); //BigDecimal for c
BigDecimal rbd = new BigDecimal(r); //BigDecimal for r
for (int counter = 1; counter <= t; counter++) {//perhaps other offset for counter?
    abd = abd.multiply(new BigDecimal(counter),MathContext.DECIMAL128);
    cbd = cbd.multiply(rbd,MathContext.DECIMAL128).add(abd,MathContext.DECIMAL128);
}
System.out.println(abd);
System.out.println(cbd);

这给出了:

abd = 3.166049846031012773846494375835059E+465752
cbd = 3.166050156931013454758413539958330E+465752

这大概是正确的，毕竟 a 的结果应该是:

This is approximately correct, after all the result of a should be:

根据 Wolfram Alpha ，这大约是正确的.

Which is approximately correct according to Wolfram Alpha.

此外，如果是 for 循环，我建议使用 for ，而不要使用 while .由于 while 倾向于创建另一种类型的无穷大:无限循环;).

Furthermore I would advice to use a for and not a while if it is a for loop. Since while tends to create another type of infinity: an infinite loop ;).