10320 - Cow Trouble! Help Please!!

[Picard]

are you sure it's not just a compile error because of redefining M_PI (if you include math.h).
this is a good solution and it's very much a spoiler

[Joe Smith]

are you sure it's not just a compile error because of redefining M_PI (if you include math.h).
this is a good solution and it's very much a spoiler

This is strange -- I get AC now.

The only thing I remember changing from my most recent WA was I factored out a M_PI... but maybe that did it. Or maybe there was some bad data they changed. (I hate floating point.)

The reason I redefined M_PI (it's only a warning, not a compile error) is because I seem to remember that it wasn't defined in math.h on UVA. I may be wrong on that, but I know hypot isn't defined.

[Subeen]

my code gets WA. but i think the problem is simple and easy to solve. i find the area of 3/4 of the circle using the formula pi*R*R, and rest of the area using similar formula.
someone plz find what wrong with my code or the approach is wrong
[c]
#include <stdio.h>
#include <math.h>

#define pi 2*acos(0)

void main()
{
double l, w, R, area, area1, area2, area3;
while(3==scanf("%lf%lf%lf", &l, &w, &R))
{
area2 = area3 = 0;
area1 = 3 * pi * R * R / 4;
if(R >= l)
area2 = pi * (R - l) * (R - l) / 4;
if(R >= w)
area3 = pi * (R - w) * (R - w) / 4;
area = area1 + area2 + area3;
printf("%0.10lf\n", area);
}
}
[/c]
thanx...

[Dominik Michniewski]

Are yot sure that

Code: Select all

printf("%0.10lf" ....

is similar to

Code: Select all

printf("%.10lf" ....

I think (maybe wrong) that 0.10 prints in some systems something like 0001.00000000 - maybe it's your problem ?

I don't try to solve this question yet and I can't say any word about your algorithm

Best greetings

[Yarin]

Your solution only works when R<=l+w. This is the easy case, it gets a lot more complicated when R >l+w. Sadly the sample input doesn't have such an input case...

Try the input

7 3 9
9 5 2
18 2 7

The correct answer should be

222.2676802415
9.4247779608
135.0884841044

[Subeen]

Yarin, I don't understand what do u mean by

Your solution only works when R<=l+w. This is the easy case, it gets a lot more complicated when R >l+w.

after some modification ( area2 = area3 = 0) my program gives the same output as your's.

but still getting WA...

[Yarin]

I'm terribly sorry, I had the parameters the wrong way. The input should be

3 9 7
5 2 9
2 7 18

and the correct answer

128.0199006338
238.3325404965
992.2996351671

which your program does not output.

[choyon_buet]

Thanks to yarin cause i could realize my mistake with my calculation from the sample I/O given by u.And i got this problem accepted at last .

but i am so astonished to see that my output doesn't match with yours completely.it differs in the 9th and10th points after the decimal.


suppose for the
input : 5 2 9
my code gives the output: 238.3325404991

so here the last two digits are 91 when ur output is 65.

but both of us got accepted.
CAN U TELL ME HOW IS THAT POSSIBLE?

[saiqbal]

its possible bcoz its an special judge problem.
an special corrector program is written for this problem to ignore small precision error.

[hridoy]

Can anyone plz tell me How can I evaluate the formula when R>(l+w)??

[CMG]

im not exactly sure myself at the moment, but from observation I found a good approximation for when R>(l+w):

A = (pi*3*R^2)/4 + (pi*(R - l)^2)/4 + (pi*(R - w)^2)/4 - (pi*(R - l - w)^2)/4
= pi*(R^2 - l*w/2)

which gives 238.76104 for the 5 2 9 case above.

If I come up with anything better Ill post it.

[Sh0rty]

PLs can somebody help me?I have problem like others when R > W+L.I still don`t know resolution for this case.It`s very important for me - Thank you very much.

[DD]

The most tricky case happens when R > l + w. You need to solve some equations first to get the final answer.

PLs can somebody help me?I have problem like others when R > W+L.I still don`t know resolution for this case.It`s very important for me - Thank you very much.