OJ Board

Do you consider a or b or c equal to 0 ?

Dominik

You mean one of them (a,b,c) or both ((a,b),(b,c),(a,c)) or the three of them are zeroes?
And what if these zeroes occur? should I ignore this input (since it can produce negative result in the sqrt which is undefined) ?

If a=0 OR b=0 OR c=0 (any of them is zero, all of them are zero, anything) the output would be zero.

Got AC now!!!!
But when either of the side is 0, would that be a triangle? It should be a line instead of a triangle!!
But THANK YOU VERY MUCH !!!!!
Have a nice life !!

Actually, the special case with only one of a, b and c equal to zero is only possible for the non-zero lengths being equal. (Because of the condition max{a,b,c}<=(a+b+c)/2). In this case the radius is zero by the used and already discussed Heron-formule.
Actually, the really bad case is the case a=b=0, c>0! Then sqrt(something<0) is not defined!!!

Am I right?

Stefan for the BiK Team

My code in C++ is:

#include <stdio.h>
#include <math.h>

int main()
{ double a,b,c;
double S;
double p;
while(scanf("%lf %lf %lf", &a, &b, &c)==3)
{
if(a!=0 && b!=0 && c!=0)
{
p = (a+b+c)/2;
S=sqrt((p-a)*(p-b)*(p-c)/p);
printf("The radius of the round table is: %.3lf\n",S);
}
else
printf("The radius of the round table is: 0\n");

}
return 0;
}

and I still get WA. Why ?

Thx for any response[cpp][/cpp]

My code in C++ is:

#include <stdio.h>
#include <math.h>

int main()
{ double a,b,c;
double S;
double p;
while(scanf("%lf %lf %lf", &a, &b, &c)==3)
{
if(a!=0 && b!=0 && c!=0)
{
p = (a+b+c)/2;
S=sqrt((p-a)*(p-b)*(p-c)/p);
printf("The radius of the round table is: %.3lf\n",S);
}
else
printf("The radius of the round table is: 0\n");

}
return 0;
}

and I still get WA. Why ?

Thx for any response[cpp][/cpp]

Hello, I 've tried to solve P10195 several times, but it gives WA, what is wrong with my code?

I have also checked the case that a == 0 or b == 0 or c == 0, here is my code:

[cpp]
#include <stdio.h>
#include <math.h>

int main()
{
double a, b, c, s;

while (scanf("%lf %lf %lf", &a, &b, &c) != EOF) {

if (a != 0 && b != 0 && c != 0) {

s = (a + b + c) / 2;
s = sqrt((s-a)*(s-b)*(s-c)/s);
printf("The radius of the round table is: %0.3lf\n", s);
}
else
printf("The radius of the round table is: 0\n");
}

return 0;
}
[/cpp]

I am frustrated

I just made some changes in my code, now got AC ...!!!
WA was given due to precission errors, I think...

However you may find this lines very useful:

while (scanf(

Hello, I 've tried to solve 10195 several times, but it gives wrong answer, what is wrong with my code?

I have also checked the case that a == 0 or b == 0 or c == 0, here is my code, and also check that the side are of made triangle or not i uses law of cos , i also check the limits of input

C++:
[/cpp]
#include<stdio.h>
#include<math.h>

double a,b,c,s,alpha,beta,gama;

void main(){

while(scanf("%lf %lf %lf",&a,&b,&c)!=EOF)
{
if(a>=1 && a<=1000000 && b >=1 && b<=1000000 && c>=1 &&c<=1000000)
{
beta = (double) acos(((c*c)+(a*a)-(b*b))/(2*a*c));
alpha = (double) acos(((b*b)+(c*c)-(a*a))/(2*b*c));
gama = (double) acos(((a*a)+(b*b)-(c*c))/(2*a*b));
if(beta+alpha+gama==180.00 || beta+alpha+gama >=179.60 ||beta+alpha+gama <=180.00 )
{
s=(a+b+c)/2;a=sqrt((s-a)*(s-b)*(s-c)/s);
printf("The radius of the round table is: %.3lf\n",a);
}
}
}
}

What am I missing here?:
[C++]

#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;

int main()
{
double a,b,c, radius, peri;
while(cin >> a >> b >> c)
{
if(a>0 && a<=1000000 && b >0 && b<=1000000 && c>0 &&c<=1000000)
{
peri = (a + b + c) / 2;
radius = sqrt((peri - a) * (peri - b) * (peri - c) / peri);
cout << setiosflags(ios::fixed) << setprecision(3) << "The radius of the round table is: " << radius << "\n";
}
else
exit(1);
}
return 0;
}

Thanks,
Matt

To mattapayne :

The values of a b or c can be zero...
and if either of them is zero then the answer is also zero...

.. but you seemed to return from the main function when any of the three
inputs is zero.

And this line is irrelevant..
[c]
if(a>0 && a<=1000000 && b >0 && b<=1000000 && c>0 &&c<=1000000)
[/c]

Since the problemset says No triangle size will be greater than 1000000 , you can safely assume they won't give you numbers more than this limit.

I have changed this two parts of your code and got AC.
Hope it helps.

You are right. Got AC. Thanks.

Matt

I think you were on the right track but made some unnecessary checking.....
It should be something like this
[/cpp]
//10195
// knights of the round table

#include <stdio.h>
#include <iostream.h>
#include <math.h>
void main()
{
double a,b,c,s,r;
while (cin>>a>>b>>c)
{

if(a==0.0||b==0.0||c==0.0)
printf("The radius of the round table is: 0.000\n");
else
{
s=(a+b+c)/2;
if(s<0)
s = s* -1;

r=(s-a)*(s-b)*(s-c)/s;
r=sqrt(r);
printf("The radius of the round table is: %.3lf\n", r);
}
}

}

Try this and see whether it works....... Take care and best of luck

is not the right process...?
perimeter, s=(a+b+c)/2;
the radius of the cicle, r=sqrt(s*(s-a)*(s-b)*(s-c))/s;

all are in double.