What's test 8 case?

 0 -18 0 81
 (x-3)(x-3)(x+3)(x+3)
 5 10 15 21
 (x2+3)(x2+5x+7)
 0 6 0 9
 (x2+3)(x2+3)
 0 -6 0 9
 (x2-3)(x2-3)
 0 10 0 9
 (x2+1)(x2+9)
 1 3 2 2
 (x2+x+1)(x2+2)
 8 22 46 55
 (x+5)(x3+3x2+7x+11)
 -2 -8 -24 -55
 (x-5)(x3+3x2+7x+11)
 2 -13 -14 24
 (x-1)(x+2)(x-3)(x+4)
 1 0 1 0 // helps with WA 6
 (x)(x3+x2+1)
 1 0 -1 0
 (x)(x3+x2-1)
 1 0 -2 0
 (x)(x-1)(x2+2x+2)

Edited by author 05.05.2012 02:10

Maybe this one will help you:

1 1 1 0

(x)(x+1)(x2+1)

hi i dont know why i got WA for 5 test case ... can anyone say about the test case ...

Edited by author 30.12.2008 21:50

then check this case:

0 0 0 -4

(in fact any polynomial which can be only factored into two quadratic polynomials and has negative free term will work)

Edited by author 10.07.2010 00:58

If you solve this problem with a help of finding roots of the initial polynom, use long long or __int64 type in this function! This fact costed me about 5 submissions...
This is the part of my code:

long long PolynomValue (vector<int>PolynomCoeffs, int x0)
{
long long value;
...
return value;
}

int FindRoot (vector<int> PolynomCoeffs)
{
...
if (!PolynomValue(PolynomCoeffs,a)) return a;
...
}
If I changed long long to int I would not find a root in several cases.

To understand me more clearly try these tests:

-20000 19998 20000 -19999
(x-1)(x-1)(x+1)(x-19999)

-19998 -20000 20000 -19999
(x-19999)(x3+x2-x+1)

-19999 -19999 -19999 -20000
(x+1)(x-20000)(x2+1)

I guess 13th and 14th tests are something like these ones.
Good luck!

Edited by author 22.01.2010 21:07

Could somebody give me a hand?

I got too many wa on it..

Please write what method or algorithm you used. Thanks.

Can we output polynoms in any order?