10573 - Geometry Paradox

[shahriar_manzoor]

Sorry! Although the problem statement says floating point, all the judge inputs are integers. I think in my defense I can say "All integers are floating point numbers" like "All alkalies are base" in chemistry days.

[hito]

Hi,I want to say something first...my English is pool...
But I got A.C. and also think the problem is beautiful and clear enough...

if r1=r2,
then r1+r2=R=t/2
so R^2=t^2/4
the gray area=(t^2/4)*pi-2*(t^2/16)*pi=(t^2/8)*pi

else r1!=r2
then (r1+r2)^2=(r1-r2)^2+(t/2)^2
we can get 4*r1*r2=(t/2)^2
by the problem,t value must be integer,
so r1=k^2*r2(k is rational number)
if not, t will be irrational number,
so 4*k^2*r2^2=(t/2)^2
then we can got r2=t/4k,r1=tk/4
so the big circle area:
(t/4k + tk/4)^2*pi=(t^2/16k^2+t^2*k^2/16+t^2/8)*pi
the gray area still (t^2/8)*pi....

so the two pictures that post before maybe correct,
but the area of gray part maybe have precision errors,
or it will be the same.

This is my first post here,if I make a mistake,plz let me know.
Hope this will be a little help...

[pavelph]

As I understand if in input "t" then write “Impossible.” else write area of grey part. Am I right???

[shamim]

Well not quite.
I will give you one hint, the part about "Impossible" here is only for confusing. My AC program does not consider cases where output will be "Impossible".

[pavelph]

I think that with <const t> and different <Radius of big circle> we will have different answers

[Eduard]

Please help i don't know were is an error i thing my formula is right.
[pascal]var t,i,j,k,n:longint;
r:array[1..3] of longint;
rr:real;
begin
readln(n);
for j:=1 to n do
begin
k:=0;
while not eoln do
begin
inc(k);
read(r[k]);
if k=2 then break;
end;
if j<>n then readln;
if k=2 then rr:=2*pi*r[1]*r[2]
else rr:=(pi*r[1])/8;
writeln(rr:0:4);
end;
end.[/pascal]

[Eduard]

I put this question in Volume CV, but did not get answer.
I put it here because I think my problem is with input. Plase help me. What is wrong with my program why I'm getting WA.
[pascal]var t,i,j,k,n:longint;
r:array[1..3] of longint;
rr:double;
begin
readln(n);
for j:=1 to n do
begin
k:=0;
while not eoln do
begin
inc(k);
read(r[k]);
if k=2 then break;
end;
if j<>n then readln;
if k=2 then rr:=2*pi*r[1]*r[2]
else rr:=(pi*r[1])/8;
writeln(rr:0:4);
end;
end. [/pascal]

[Observer]

No. Your formula for the "one integer" case is wrong. Check that again.

[Eduard]

Thank you i did not sea that write the second formula wrong.
I get AC.

[MarC]

there have been a lot of discussions about this problem...
but, in my opinion, the clue is that in the problem's text, they say "each set contains two integer. If it contains one integer..."
but in the judge's files, there is one test with double data!!!

that's why i got a thousand times WA!

I expect they will fix this error!!!

[UFP2161]

No, it says "Each set either contains one or two integer."

Thus, a line like "5 1.1161" contains only one integer, and is technically valid.

[ansiemens]

HI! I always got WA. I don't know whats wrong with it.
can anyone tell me?

Code: Select all

#include<iostream>
#include<sstream>
#include<string>
using namespace std;

const double M_PI=3.14159265358979323846;

int main() {
    int n,r1,r2,t;
    string input;
    cin >> n; cin.ignore();
    cout.setf(ios::fixed);
    cout.precision(4);
    while(n--) {
        getline(cin,input);
        if(input[1]==0 || input[2]==0) {
            istringstream(input) >> t;
            cout << t*t*(M_PI/8.0) << endl;
        }
        else {
            istringstream(input) >> r1 >> r2;
            cout << r1*r2*(2.0*M_PI) << endl;
        }
    }
    return 0;
}

[Sedefcho]

The formulas you use are obviously OK.
Try to simplify the way you read the input. That should
be your problem. You may want to use simpler functions
( pure C functions ) like gets() and sscanf().

[kbr_iut]

code deleted.Now AC,thanx obaida

[Obaida]

If only t is given, you can assume that r1 = r2, t become the diameter of the outer circle Thus the area of the outer circle becomes PI*(t/2)*(t/2) and area of the two identical inner circles become 2*PI*(t/4)*(t/4), thus the area of the gray part = (PI*t*t)/4 - (PI*t*t)/8 = (PI*t*t)/8.
If there are r1 and r2, then gray part =2*PI*r1*r2.
And there is no impossible case.