REALROOT  Real Roots
In this problem you are challenged to factor the real roots of polynomials.
You are rewarded points based on the number of testcases you solve.
Input
The first line contains the number of test cases, T.
The first line of each test case contains an integer N, the number of coeficients.
The second line contains the polynomial coeficients a_{N1}, ..., a_{1}, a_{0} such that (P(x) = a_{N1}x^{N1} + ... + a_{1}x^{1 }+ a_{0}x^{0}).
Output
The roots of the polynomial, in sorted order. That is all real numbers r_{0}, r_{1}, r_{2}, ... such that P(r) = 0.
Constrains
 T ≤ 20
 N ≤ 20
 All coeficients are in the inclusive range [10^{6},10^{6}]
 The output precision must be at least 2 decimal digits
 The roots must truly be real. No complex roots even if the imaginative part is very small.
Cases
 Same as the example
 ~3 random integer roots
 ~15 random integer roots
 ~10 Random interger coefficients
 ~10 Random floating point roots
 ~10 Random floating point coefficients
 Polynomials on the form p_{0}=x, p_{k+1}=(p_{k}a_{k})^2 e.g. ((x3)^21)^2
Example
Input: 4 3 1 0 2 x^{2}  2 3 1 0 1 x^{2} + 1 4 1 3 3 1 x^{3} + 3x^{2}  3x + 1 7 1 1 9 13 8 12 0 x^{6}  x^{5}  9x^{4} + 13x^{3} + 8x^{2}  12x
Output: 1.414214 1.4142135 sqrt(2), sqrt(2) No real roots 1 1 1 Triple root in x=1 3 1 0 1 2 2 Mixed integer roots
Notes
 All testcases are double checked using Mathematica with 30 digits of precision.
 Constrains are set such that most approaches should be fine using double working precision.
hide comments
Shashi Kant Prasad:
20140612 20:50:21
@Mitch Schwartz: Thanks for help. I also think, handling multiple roots is important here.


Mitch Schwartz:
20140612 20:50:21
@Shashi Kant Prasad: I think the 7th class is unclear, my assumption was that p_0=1 is a typo for p_0=x, and a recurrence is defined, e.g. if a=5 then p_1=(x5)^2, p_2=((x5)^25)^2, and any of these polynomials could be in the input if it meets the other constraints  also, maybe "a" is supposed to be a_k such that e.g. ((x5)^23)^2 is also possible (with a_0=5 and a_1=3). At any rate I think the main thing is to be able to handle multiple roots.


Shashi Kant Prasad:
20140612 20:50:21
In case 7: Polynomials on the form p0=1, pk+1=(pka)^2, what does pk mean?(kth derivative of p?) also Polynomials on the form p0=1, (pka)^2 is pk raised to power square(a).


Robert Gerbicz:
20140612 20:50:21
Thanks for the answers! Just observed that the points are multiple of 100/7, so we can get points after solving an input set, not for a single equation.


Thomas Dybdahl Ahle:
20140612 20:50:21
The judge is the standard SPOJ "Ignores FP rounding up to 10^2"


Mitch Schwartz:
20140612 20:50:21
It seems the custom judge suppresses runtime errors and also TLE according to infinite loop code I submitted just now (which got 0 score instead of TLE). I'm not sure if the author did this on purpose to make debugging harder, or just overlooked it.


Robert Gerbicz:
20140612 20:50:21
Is the input files totally deleted ?? A very simple


Thomas Dybdahl Ahle:
20140612 20:50:21
I reduced the output precision contrains, as some of the polynomials couldn't be evaluated to 6 digits using double precision. Last edit: 20120609 18:25:59 

Thomas Dybdahl Ahle:
20140612 20:50:21
Petar: Your solution misses certain duplicated roots. Last edit: 20120609 14:54:14 

Thomas Dybdahl Ahle:
20140612 20:50:21
Thank you, that was unclear. Hopefully it's better now. 
Added by:  Thomas Dybdahl Ahle 
Date:  20120606 
Time limit:  0.200s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 