计算机如何评估大量数字? [英] How do computers evaluate huge numbers?

问题描述

如果我在Wolfram Alpha中输入一个值，例如1234567 ^ 98787878，它可以为我提供许多详细信息.这包括十进制近似值，总长度，最后一位数字等.您如何评估这么大的数字?据我了解，一种编程语言必须具有特殊的数据类型才能存储数字，更不用说将其添加到其他东西了.虽然我可以看到如何处理两个非常大的数字的加法运算，但我看不到如何计算出巨大的数字.

If I enter a value, for example 1234567^98787878 into Wolfram Alpha it can provide me with a number of details. This includes decimal approximation, total length, last digits etc. How do you evaluate such large numbers? As I understand it a programming language would have to have a special data type in order to store the number, let alone add it to something else. While I can see how one might approach the addition of two very large numbers, I can't see how huge numbers are evaluated.

10 ^ 2可以通过重复加法来计算.但是，如上面的示例这样的数字将需要一个巨大的循环.有人可以解释如何评估这么大的数字吗?另外，例如，有人如何创建自定义的大型数据类型来支持C#中的大量数字?

10^2 could be calculated through repeated addition. However a number such as the example above would require a gigantic loop. Could someone explain how such large numbers are evaluated? Also, how could someone create a custom large datatype to support large numbers in C# for example?

推荐答案

很简单，您可以自己完成

Well it's quite easy and you can have done it yourself

1. 位数可以通过对数获得:

自A^B = 10 ^ (B * log(A, 10))

在这种情况下，我们可以计算(A = 1234567; B = 98787878)

we can compute (A = 1234567; B = 98787878) in our case that

B * log(A, 10) = 98787878 * log(1234567, 10) = 601767807.4709646...

integer part + 1 (601767807 +1 = 601767808 )是位数

integer part + 1 (601767807 + 1 = 601767808) is the number of digits

第一个，例如，五个，个数字也可以通过对数获得； 现在我们应该分析

First, say, five, digits can be gotten via logarithm as well; now we should analyze fractional part of the

B * log(A, 10) = 98787878 * log(1234567, 10) = 601767807.4709646...

f = 0.4709646...

第一个数字为10^f(删除了小数点)= 29577 ...

first digits are 10^f (decimal point removed) = 29577...

最后一个，例如，五个，个数字可以作为对应的剩余:

Last, say, five, digits can be obtained as a corresponding remainder:

最后五位数字= A^B rem 10^5

A rem 10^5 = 1234567 rem 10^5 = "34567

A rem 10^5 = 1234567 rem 10^5 = `34567

A ^ B rem 10 ^ 5 **=**(((A rem 10 ^ 5)^ B)rem 10 ^ 5 **=** (34567 ^ 98787878)rem 10 ^ 5 = 45009`

A^B rem 10^5**=**((A rem 10^5)^B) rem 10^5**=** (34567^98787878) rem 10^5=45009`

最后五位数字为 45009

您可能会在这里发现BigInteger.ModPow(C#)非常有用

You may find BigInteger.ModPow (C#) very useful here

最后

1234567 ^ 98787878 = 29577 ... 45009(601767808位)

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