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| Show all threads Hide all threads Show all messages Hide all messages | | TEST 13 | Mortus | 1666. Vasya Ferrari | 28 Jul 2024 14:25 | 1 | -280 19326 6488 -15295
(x2-156x+115)(x2-124x-133) | | What's test 8 case? | OPyshkin | 1666. Vasya Ferrari | 2 Apr 2018 01:41 | 1 | | | some tests | Levenets (ONPU) | 1666. Vasya Ferrari | 5 May 2012 02:11 | 5 | 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) Re: some tests Maxim Dvoynishnikov (Dnipropetrovsk SU) 2 Jan 2009 16:33 May be there are good tests but they didn't help me with WA13. For 0 -18 0 81 (first test) (x-3)(x+3)(x-3)(x+3) the correct answer?
Edited by author 22.01.2010 21:03 Test:
0 0 0 4
answer:
(x2+2x+2)(x2-2x+2) | | No subject | IgorKoval(from Pskov) | 1666. Vasya Ferrari | 5 May 2012 02:09 | 1 |
Edited by author 05.05.2012 02:10 | | If you WA at #20 | SerailHydra | 1666. Vasya Ferrari | 28 Apr 2012 14:12 | 2 | Maybe this one will help you:
1 1 1 0
(x)(x+1)(x2+1) It didn't help me. Can anyone say about other test case? | | why WA on test case 5 | shankar | 1666. Vasya Ferrari | 24 Apr 2012 13:39 | 2 | 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 i've got WA5 too, but all tests in this forum are good
PS. I find my mistake:
5 12 15 9 -> (x2+2x+3)(x2+3x+3)
But now I have WA19 ((
Edited by author 25.04.2012 22:08 | | If you have WA #10... | 2rf [Perm School #9] | 1666. Vasya Ferrari | 10 Jul 2010 00:55 | 1 | 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 | | Hint: if you have WA#13 or WA#14 | bsu.mmf.team | 1666. Vasya Ferrari | 22 Jan 2010 20:55 | 1 | 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 | | why WA on test 14 | Churchill | 1666. Vasya Ferrari | 27 Dec 2008 17:11 | 2 | Could somebody give me a hand? I had WA #14 when forgot to check the case when
Polynomial = (Polynomial of 1st degree)*(Polynomial of 3rd degree) | | what is test case 6? | demo | 1666. Vasya Ferrari | 23 Dec 2008 13:52 | 5 | I got too many wa on it.. I got WA, too.
Can you tell me this case ? Got it.
Here is the case that helped me:
9998 -20000 0 0
Result:
(x)(x)(x-2)(x+10000)
Edited by author 23.12.2008 13:51 #include<iostream.h>
int max1(int *x,int *y,int i,int j)
{
int k,m1=0,r=-1;bool p=true;
for(k=i+1;k<=j&&p&&r==-1;k++)
if(x[k]>=x[i])
{
m1=y[k];
r=k;
p=false;
}
if(r!=-1){
for(k=r;k<=j;k++)
if(x[k]>=x[i] && y[k]>=m1)
m1=y[k];
return m1+1;
}
return -1;
}
int main()
{
int n,a[1005],i,l[1005],max;
cin>>n;
for(i=1;i<=n;i++)
cin>>a[i];
l[n]=1;
if(a[n-1]>a[n])
l[n-1]=1;
else
l[n-1]=2;
for(i=n-2;i>=1;i--)
l[i]=max1(a,l,i,n);
max=l[1];
for(i=2;i<=n;i++)
if(l[i]>max)
max=l[i];
cout<<max;
return 0;
} | | What metod who use? | Nurbek_[KTMU MANAS] | 1666. Vasya Ferrari | 21 Dec 2008 09:28 | 2 | Please write what method or algorithm you used. Thanks. if the polynomial can be factorized to the form (x2+mx+n)*(x2+px+q),by expanding it,we can get four equation of coefficients,one of them is nq=d,notice |d|<=20000,we can enumerate its factor and check whether all the equation are meeted.the same to other forms | | Multiple orders | partisan | 1666. Vasya Ferrari | 20 Dec 2008 14:08 | 2 | Can we output polynoms in any order? Yes (-) Sandro (USU) 20 Dec 2008 14:08 |
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