您需要将多少字节存储2^57,885,161 - 1为整数？
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Assuming we're doing two's complement and that 8 bits is equal to one byte; we'd need at least (57,885,161+7)/8 bytes.
If you needed a simple way to possibly explain it is by using mathematical induction that says 2^32 - 1 is the maximum number that a 32-bit integer would represent, and 32 is a base of 2 that is divisible by 8, our assumed number of bits per byte. 2^32 - 1 would be 4 bytes.
Extending this definition of assumptions you have the number 2^57885161 which isn't divisible by 8, but adding 7 to it is. So you're left with 2^57885168, and when you divide it by 8 you get the resultant 7235646 bytes.
This is just an explanation of GregS's comment.
于 2013-02-10T00:26:31.303 回答
3
答案很大程度上取决于您将如何处理它。如果您正在编写一个专门使用梅森素数的程序，您可能只需要四个字节来存储它，并理解它代表梅森素数指数。
如果要将其存储为典型的未压缩“大整数”，则大约需要 7235646 个字节（ceildiv(57885161, 8)）。有些格式比其他格式更有效。例如，由于格式开销（因为 Python 使用 30 位数字） ，Pythonlong格式使用 7718048 字节将此数字存储在我的机器 ( ) 上。(2**57885161 - 1).__sizeof__()
于 2013-02-10T00:18:03.710 回答