I'm using the GMP library to make a Pi program, that will calculate about 7 trillion digits of Pi. Problem is, I can't figure out how many bits are needed to hold that many decimal places.
7 trillion digits can represent any of 10^(7 trillion) distinct numbers.
x bits can represent 2^x distinct numbers.
So you want to solve:
2^x = 10^7000000000000
Take the log-base-2 of both sides:
x = log2(10^7000000000000)
Recall that log(a^b) = b * log(a):
x = 7000000000000 * log2(10)
I get 23253496664212 bits. I would add one or two more just to be safe. Good luck finding the petabytes to hold them, though.
I suspect you are going to need a more interesting algorithm.
I wanna just correct one thing about what was written in the response answer:
Recall that log(a^b) = a * log(b)
well it is the opposite :
log(a^b) = b * log(a)
2^10 = 1024, so ten bits will represent slightly more than three digits. Since you're talking about 7 trillion digits, that would be something like 23 trillion bits, or about 3 terabytes, which is more than I could get on one drive from Costco last I visited.
You may be getting overambitious. I'd wonder about the I/O time to read and write entire disks for each operation.
(The mathematical way to solve it is to use logarithms, since a number that takes 7 trillion digits to represent has a log base 10 of about 7 trillion. Find the log of the number in the existing base, convert the base, and you've got your answer. For shorthand between base 2 and base 10, use ten bits==three digits, because that's not very far wrong. It says that the log base 10 of 2 is .3, when it's actually more like .301.)
来源：https://stackoverflow.com/questions/6414714/arbitrary-precision-arithmetic-with-gmp