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SQRT2 - Digits of SQRT(2) |
In this task you are to find as many digits of the square root of 2 as possible. You have to make it within the limit of time and source code length.
Input
There is no input for this problem
Output
The output must contain as many digits of the square root of 2 as possible (max = 2000000)
Score
The score awarded to your program will be the first position of the digit where the first difference occurs.
Example
Output:1.41421356237309504
will be awarded with 19 points.
| Added by: | Roman Sol |
| Date: | 2007-01-07 |
| Time limit: | 2.75s |
| Source limit: | 4096B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: CLPS LISP sbcl LISP clisp ERL JS-RHINO NODEJS PERL6 VB.NET |
| Resource: | ZCon 2007 |
hide comments
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2024-05-20 11:08:28
haha, using the Newton iteration to calculate 1/sqrt(2), with the function f(x)=1/x^2-2 |
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2020-06-22 06:53:45
lesser the score means more the points?? please help =(Francky)=> No. You need to output as many digits as you can. Better the score means more the points. Those points are not given to total ranking, but rather others computed according to your rank here. Last edit: 2020-06-22 08:12:57 |
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2018-02-01 14:30:33
600003 using python 3. Just 6 lines of code :) Last edit: 2018-02-02 21:26:26 |
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2016-06-16 00:14:10
43002 using python!! |
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2016-04-15 21:41:13 thierry
hum ... Time limit is said to be 2.75 but i can't seem to go past 2.50 w/o getting TLE ... weird |
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2016-01-22 10:11:36
got 50000 ! |
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2015-10-27 17:24:56 rahul_verma
got 101 point |
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2015-08-26 05:56:17 Siddharth Singh
i got 3712 points but they arent credited in my profile till now :( |
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2015-08-15 10:28:30
Got 50000 using Python |
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2015-08-01 07:24:17
got 4000 points :D |
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