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11042 - Complex, difficult and complicated
Posted: Sun Jun 04, 2006 6:40 pm
by temper_3243
hi,
Does anyone have testcases for the problem 11042. When i submit the code, the problem says received but nothing about status.
Can someone post good test cases.
Posted: Mon Jun 05, 2006 12:44 am
by temper_3243
HI,
I am getting WA. Can some one tell me where i am doiing things wrong .
Code: Select all
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
# define M_PI 3.14159265358979323846 /* pi */
double sqrt(double );
double atan(double );
double fabs(double );
int main()
{
int i,a,b,no,k;
double m,n,z;
scanf(" %d",&no);
while(no--)
{
scanf(" %d %d",&a,&b);
z=sqrt(a*a + b*b);
if(a==0)
m=M_PI/2;
else
m=atan((double)b/(double)a);
if(m==0)
{
printf("1\n");
continue;
}
k=fabs(M_PI/m);
if (fabs(fabs(k*m ) - M_PI) < 0.00009)
{
if(pow(z,k) <= pow(2,30))
printf("%d\n",k);
else
printf("TOO COMPLICATED\n");
}
else
printf("TOO COMPLICATED\n");
}
return 0;
}
Posted: Mon Jun 05, 2006 2:37 am
by mf
Here are some inputs for which your program gave wrong answer (at least, on my computer):
Code: Select all
5
0 0
-153 -265
-153 265
-97 -168
97 168
Correct answer for the first case is 1, for the rest -- "TOO COMPLICATED".
Posted: Mon Jun 05, 2006 3:45 am
by fernando
What's the approach or mathematical concept required to solve this problem?
Posted: Mon Jun 05, 2006 6:27 am
by Darko
This is the top Google link:
http://www.clarku.edu/~djoyce/complex/
The program above uses "polar coordinates", I solved it using cartesian coordinates ("complex plane"). (check the corresponding parts in that tutorial)
Posted: Mon Jun 05, 2006 4:51 pm
by Jan
I used brute force and got Accepted in 0.000 seconds.
Posted: Tue Jun 06, 2006 8:25 am
by shamim
Jan wrote:I used brute force and got Accepted in 0.000 seconds.
Is the solution simply to loop over N and stop when the number becomes real. If so, when should the looping stop if there is no solution.
Posted: Tue Jun 06, 2006 8:49 am
by arsalan_mousavian
shamim wrote:Jan wrote:I used brute force and got Accepted in 0.000 seconds.
Is the solution simply to loop over N and stop when the number becomes real. If so, when should the looping stop if there is no solution.
Not exactly , you nead a condition to stop the loop ( for the case that the result is TOO COMPLICATED ) but the rest is completely brute force
hope it helps
Posted: Tue Jun 06, 2006 8:49 am
by Darko
I picked n=1000 and it worked. One could probably come up with a case that breaks it. (hm, maybe not - a,b are integers and they'll end up being > 2^30 really fast... silly me.. why did I pick such a high number?)
Posted: Tue Jun 06, 2006 3:03 pm
by shamim
Darko wrote:I picked n=1000 and it worked. One could probably come up with a case that breaks it. (hm, maybe not - a,b are integers and they'll end up being > 2^30 really fast... silly me.. why did I pick such a high number?)
I choose 100 and got AC.
Is it always true, as n increases the abs(real) part will also increase. Since i*i is -1, shouldn't at times, the real part decrease or something.
Posted: Tue Jun 06, 2006 3:33 pm
by Jan
Spoiler
My method is -
Code: Select all
Suppose a and b given...
I considered x=1 and y=0
1. I have two complex numbers -
P = x + iy
Q = a + ib
2. Every time multiply P and Q and update x, y (x, y to srore the multiplication results. x contains real part and y contains complex part) .
3. If abs (x) > 2^30 then TOO COMPLICATED. End.
4. If y = 0, Got the Solution. End.
5. Goto step 1.
Posted: Tue Jun 06, 2006 5:40 pm
by Solaris
When you convert the number to polar form ... then
SPOILER
(a+ib)^n
= r^n (cos (n.t) + i sin (n.t) )
You can safely say that if r^n is greater than 2^30 then TOO COMPLICATED
Posted: Fri Jun 09, 2006 5:25 am
by mrahman
Hi guys,
Spoiler
I think there can not be any input which output is greater than 4.
There is only 4 Possible output for the problem
1. 1
2. 2
3. 4
4. TOO COMPLICATED
Try to categorize all the input into 4 partition and you can solve it using only few If-else.
No need to use any loop
No need to use any cos or sin function
Sorry for my poor english
Take care
Posted: Sun Jun 11, 2006 2:23 am
by Martin Macko
mrahman wrote:Try to categorize all the input into 4 partition and you can solve it using only few If-else.
No need to use any loop
No need to use any cos or sin function
You are absolutelly right... After experiencing a few numbers by hand one can easily find those categories.
Posted: Fri Sep 22, 2006 5:30 pm
by fh
I think there can not be any input which output is greater than 4.
There is only 4 Possible output for the problem
1. 1
2. 2
3. 4
4. TOO COMPLICATED
I don't think so... for the input:
1
96 32
the output is
12