views:

170

answers:

3

Trying out a problem of finding the first k digits of a num^num I wrote the same program in C++ and Python

C++

long double intpart,num,f_digit,k;
cin>>num>>k;
f_digit= pow(10.0,modf(num*log10(num),&intpart)+k-1);
cout<<f_digit;

Python

(a,b) = modf(num*log10(num))
f_digits = pow(10,b+k-1)
print f_digits

Input

19423474 9

Output

C++    > 163074912
Python > 163074908

I checked the results the C++ solution is the accurate one. Checked it at http://www.wolframalpha.com/input/?i=19423474^19423474

Any idea how can I get the same precision in Python ???

EDIT : I know about the external library packages to obtain this precision but any NATIVE solution ???

+2  A: 

Python floats are doubles under the hood, as you discovered. You will have to resort to C code, or an external library, to get better floating-point precision.

The GMP library is a good one, and it has a python wrapper called 'GMPY', available on PyPI

Bill Gribble
Any of the coding competition don't allow any external API's and i prefer coding in Python for obvious reasons. So i wanted a possible native solution !!!
jknair
+7  A: 

Decimal is a built in python class that handles floating points correctly (as base 10, not as IEEE 7somethingsomething standard). I don't know if it supports logarithms and all that though.

Edit: It does indeed support logarithms "and all that".

You can set the precision of it as well. Default is 28 places, but it can be as large as you want. Think of it as a BigInt for decimals.

Tor Valamo
Decimal('163074912.1616735983662131415') !!! :) thanks
jknair
A: 

In general, I would do it this way. However, it doesn't seem to perform anywhere near fast enough for your example numbers.

num = 453
k = 9
result = num ** num

print str(result)[:k]
# Prints: '163111849'
sharth