calculation 2^17000 and also adittion of each terms

Question

1.00/5 (4 votes)

See more:

Hello
Can someone gimme the code to calculate 2^17000(2 to the power 17000),also simultaneously add all the terms.By terms I mean 2^17000 terms.The program should execute within 1 sec.my problem is i am not understanding how to store such a large value.It has to be done only in c++ or java(not in .net or any other platforms) keeping in mind the execution time.

Posted 16-Aug-13 19:39pm

aghori

Updated 19-Aug-13 18:57pm

v2

Add a Solution

Comments

CPallini 19-Aug-13 3:56am

What 'terms'?

[no name] 20-Aug-13 1:26am

You say "2^17000 terms". That does not make sense. Perhaps you need to give an example of "term".

pasztorpisti 20-Aug-13 8:19am

Exactly, I was also confused reading that part about the sum. I silently assumed that he has to add exponents of 2.

3 solutions

Solution 1

.Net includes a BigInteger structure, which includes a Pow method:
http://msdn.microsoft.com/en-us/library/system.numerics.biginteger.pow.aspx[^]

Posted 16-Aug-13 19:47pm

OriginalGriff

Comments

Sergey Alexandrovich Kryukov 17-Aug-13 1:53am

OP asks about C++ implementation, not C++/CLI. I answered...
—SA

Solution 3

= 2^17001 - 1
a 2125 byte file with FF
so long for your homework.

Posted 18-Aug-13 21:13pm

mbue

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Sergey Alexandrovich Kryukov · Accepted Answer · 2013-08-16T19:52:00

Solution 2

First of all, what's so big in storage of such big integers. This is a little more than 2K you need for storage, plus some overhear for implementation. Not too big, really. But of course, standard CPU numeric types won't help here.

What you need is called "big integer class". Please see, for example, this open-source implementation: http://sourceforge.net/projects/cpp-bigint/[^].

You can find some more: http://bit.ly/14mQCGc[^].

—SA

Posted 16-Aug-13 19:52pm

Sergey Alexandrovich Kryukov

Comments

pasztorpisti 17-Aug-13 8:36am

5ed, but I would note that calculating a base 2 exponent with computers is also useful, a base 2 exponent is a big byte array with only a single bit set. :-) If OP has to add only 2 exponents that have this bits set on different non-overlapping positions then instead of adding its enough to use binary OR between the arrays...

Sergey Alexandrovich Kryukov 17-Aug-13 11:39am

Thank you.
—SA