10215  The Largest/Smallest Box ...
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 little joey
 Guru
 Posts: 1080
 Joined: Thu Dec 19, 2002 7:37 pm
Did you look at the problemstats of this problem? It's impossible to get Accepted using Pascal. I wrote two identical programs, one in Pascal (my 'natural' language) and one in C, after many hours of frustration. The C program got accepted at once.
I guess it's all about implementation specific rounding errors and a lazy problemsetter... (Mark my words: I don't say lousy, but lazy. I consider Manzoor to be one of the most original and productive problemsetters, but IMHO he makes some slips in the rounding error department once in a while, see 10402 for which nobody to date reproduced his (seemingly) erronious result (124 WA and 0 AC point in that direction)).
My Pascal code is (WA):[pascal]program p10215;
var
w,l,x,x1,x2,vol1,vol2:real;
begin
while not eof do begin
readln(w,l);
if (w>l) then begin
x:=w;
w:=l;
l:=x;
end;
x1:=((l+w)sqrt(l*l+w*wl*w))/6.0;
x2:=((l+w)+sqrt(l*l+w*wl*w))/6.0;
vol1:=x1*(w2.0*x1)*(l2.0*x1);
vol2:=x2*(w2.0*x2)*(l2.0*x2);
if (vol2<vol1) then x:=x1 else x:=x2;
writeln(x:0:3,' 0.000 ',(w/2.0):0:3);
end;
end.[/pascal]
And my C code is (AC):[c]#include <stdio.h>
#include <math.h>
int main(void){
double w,l,x,x1,x2,vol1,vol2;
while (scanf("%lf %lf",&w,&l)==2) {
if (w>l) {
x=w;
w=l;
l=x;
}
x1=((l+w)sqrt(l*l+w*wl*w))/6.0;
x2=((l+w)+sqrt(l*l+w*wl*w))/6.0;
vol1=x1*(w2.0*x1)*(l2.0*x1);
vol2=x2*(w2.0*x2)*(l2.0*x2);
if (vol2<vol1) x=x1; else x=x2;
printf("%.3lf %.3lf %.3lf\n",x,0.0,w/2.0);
}
return 0;
}[/c]
Looks the same, feels the same, tastes the same. Please don't steal it...
I guess it's all about implementation specific rounding errors and a lazy problemsetter... (Mark my words: I don't say lousy, but lazy. I consider Manzoor to be one of the most original and productive problemsetters, but IMHO he makes some slips in the rounding error department once in a while, see 10402 for which nobody to date reproduced his (seemingly) erronious result (124 WA and 0 AC point in that direction)).
My Pascal code is (WA):[pascal]program p10215;
var
w,l,x,x1,x2,vol1,vol2:real;
begin
while not eof do begin
readln(w,l);
if (w>l) then begin
x:=w;
w:=l;
l:=x;
end;
x1:=((l+w)sqrt(l*l+w*wl*w))/6.0;
x2:=((l+w)+sqrt(l*l+w*wl*w))/6.0;
vol1:=x1*(w2.0*x1)*(l2.0*x1);
vol2:=x2*(w2.0*x2)*(l2.0*x2);
if (vol2<vol1) then x:=x1 else x:=x2;
writeln(x:0:3,' 0.000 ',(w/2.0):0:3);
end;
end.[/pascal]
And my C code is (AC):[c]#include <stdio.h>
#include <math.h>
int main(void){
double w,l,x,x1,x2,vol1,vol2;
while (scanf("%lf %lf",&w,&l)==2) {
if (w>l) {
x=w;
w=l;
l=x;
}
x1=((l+w)sqrt(l*l+w*wl*w))/6.0;
x2=((l+w)+sqrt(l*l+w*wl*w))/6.0;
vol1=x1*(w2.0*x1)*(l2.0*x1);
vol2=x2*(w2.0*x2)*(l2.0*x2);
if (vol2<vol1) x=x1; else x=x2;
printf("%.3lf %.3lf %.3lf\n",x,0.0,w/2.0);
}
return 0;
}[/c]
Looks the same, feels the same, tastes the same. Please don't steal it...
Hm
I switched float to double and got AC.
The problem says:
so are we getting the length and width of the card (as in input specification) or the width and height of the box???Given the width and height of the box, you will have to find the value of x for which the box has maximum and minimum volume.
Input
The input file contains several lines of input. Each line contains two positive floatingpoint numbers L(0<L<10000) and W(0<W<10000); which indicate the length and width of the card respectively.
10215 WA
Here is my code:
[c]
#include <stdio.h>
int main()
{
double l,w;
double res1,res2,a,b,c;
while(scanf("%lf %lf",&l,&w)!=EOF)
{
a=12;
b=4*(w+l);
c=w*l;
res1=(b + sqrt(b*b  4*a*c))/(2*a);
res2=(b  sqrt(b*b  4*a*c))/(2*a);
if (res1>res2)
printf("%.3lf 0.000 %.3lf\n",res2,res1);
else
printf("%.3lf 0.000 %.3lf\n",res1,res2);
}
return 0;
}
[/c]
It seems to be ok....
plz if someone knows help me!!
[c]
#include <stdio.h>
int main()
{
double l,w;
double res1,res2,a,b,c;
while(scanf("%lf %lf",&l,&w)!=EOF)
{
a=12;
b=4*(w+l);
c=w*l;
res1=(b + sqrt(b*b  4*a*c))/(2*a);
res2=(b  sqrt(b*b  4*a*c))/(2*a);
if (res1>res2)
printf("%.3lf 0.000 %.3lf\n",res2,res1);
else
printf("%.3lf 0.000 %.3lf\n",res1,res2);
}
return 0;
}
[/c]
It seems to be ok....
plz if someone knows help me!!
Can anybody see what's wrong with this code?[cpp]#include <iostream>
using namespace std;
long double a, b;
inline long double f(const long double &x){
return (a  2*x)*(b  2*x)*x;
}
int main(){
while(cin >> a >> b){
long double mn = 0;
long double mx = a > b ? b/2 : a/2;
while(mx  mn > 0.0000000001)
if(f(mn + (mx  mn)/3.0) > f(mn + 2.0*(mx  mn)/3.0)) mx = mn + 2.0*(mx  mn)/3.0;
else mn = mn + (mx  mn)/3.0;
cout.setf(ios::fixed);
cout.precision(3);
cout << mn << " " << 0.0 << " " << (a > b ? b/2 : a/2) << endl;
}
return 0;
}
[/cpp]
using namespace std;
long double a, b;
inline long double f(const long double &x){
return (a  2*x)*(b  2*x)*x;
}
int main(){
while(cin >> a >> b){
long double mn = 0;
long double mx = a > b ? b/2 : a/2;
while(mx  mn > 0.0000000001)
if(f(mn + (mx  mn)/3.0) > f(mn + 2.0*(mx  mn)/3.0)) mx = mn + 2.0*(mx  mn)/3.0;
else mn = mn + (mx  mn)/3.0;
cout.setf(ios::fixed);
cout.precision(3);
cout << mn << " " << 0.0 << " " << (a > b ? b/2 : a/2) << endl;
}
return 0;
}
[/cpp]
 Abednego
 A great helper
 Posts: 281
 Joined: Tue Sep 10, 2002 5:14 am
 Location: Mountain View, CA, USA
 Contact:
minskcity, you are using long double. marian said his code got AC when he switched from long double to double. This is my code, and it works:
Code: Select all
int main()
{
double L, W;
while( cin >> L >> W )
printf( "%.3f %.3f %.3f\n",
( L + W  sqrt( ( double )( L*L  L*W + W*W ) ) ) / 6.0,
0.0, ( W <? L ) * 0.5 );
return 0;
}
If only I had as much free time as I did in college...

 New poster
 Posts: 32
 Joined: Fri Oct 01, 2004 10:53 pm
someone wanna check my code???
btw... formatting decimals in java is a pain in the ass
[java]import java.io.IOException;
import java.util.StringTokenizer;
import java.text.NumberFormat;
class Main
{
static String ReadLn(int maxLg)
{
byte lin[]=new byte[maxLg];
int lg=0,car=1;
try
{
while(lg<maxLg)
{
car=System.in.read();
if ((car<0)(car=='\n'))
break;
lin[lg++]+=car;
}
}
catch(IOException e)
{
return(null);
}
if ((car<0)&&(lg==0))
return(null);
return (new String (lin, 0, lg));
}
public static void main(String[] args)
{
Main newMain = new Main();
newMain.start();
}
void start()
{
String input;
StringTokenizer tokenizer;
double w,l,m;
while ((input=Main.ReadLn(20))!=null)
{
tokenizer = new StringTokenizer(input);
l = new Double(tokenizer.nextToken()).doubleValue();
w = new Double(tokenizer.nextToken()).doubleValue();
m = Math.min(l,w);
System.out.print(FormatDecimal((((w+l)Math.sqrt((w*w+l*lw*l)))/6.0),3)+" 0.000 ");
System.out.println(FormatDecimal(m/2,3));
}
}
String FormatDecimal(double inp,int dec)
{
String in = new String(new Double(inp).toString());
NumberFormat nf = NumberFormat.getInstance();
nf.setGroupingUsed(false);
nf.setMaximumFractionDigits(dec);
in = nf.format(new Double(in).doubleValue());
int k=0;
if(in.charAt(0)=='.')
in = "0"+in;
while((in.charAt(k)!='.')&&!(k>=in.length()1))
{
if(k<in.length())
{
if(in.charAt(k)=='.')
break;
else k++;
}
else break;
}
int diff = in.length()k1;
if(in.charAt(k)!='.')
{
in+=".";
for(int i=0;i<dec;i++) in+="0";
}
else if(dec>diff)
for(int i=0;i<(decdiff);i++)
in+="0";
return in;
}
}[/java]
btw... formatting decimals in java is a pain in the ass
[java]import java.io.IOException;
import java.util.StringTokenizer;
import java.text.NumberFormat;
class Main
{
static String ReadLn(int maxLg)
{
byte lin[]=new byte[maxLg];
int lg=0,car=1;
try
{
while(lg<maxLg)
{
car=System.in.read();
if ((car<0)(car=='\n'))
break;
lin[lg++]+=car;
}
}
catch(IOException e)
{
return(null);
}
if ((car<0)&&(lg==0))
return(null);
return (new String (lin, 0, lg));
}
public static void main(String[] args)
{
Main newMain = new Main();
newMain.start();
}
void start()
{
String input;
StringTokenizer tokenizer;
double w,l,m;
while ((input=Main.ReadLn(20))!=null)
{
tokenizer = new StringTokenizer(input);
l = new Double(tokenizer.nextToken()).doubleValue();
w = new Double(tokenizer.nextToken()).doubleValue();
m = Math.min(l,w);
System.out.print(FormatDecimal((((w+l)Math.sqrt((w*w+l*lw*l)))/6.0),3)+" 0.000 ");
System.out.println(FormatDecimal(m/2,3));
}
}
String FormatDecimal(double inp,int dec)
{
String in = new String(new Double(inp).toString());
NumberFormat nf = NumberFormat.getInstance();
nf.setGroupingUsed(false);
nf.setMaximumFractionDigits(dec);
in = nf.format(new Double(in).doubleValue());
int k=0;
if(in.charAt(0)=='.')
in = "0"+in;
while((in.charAt(k)!='.')&&!(k>=in.length()1))
{
if(k<in.length())
{
if(in.charAt(k)=='.')
break;
else k++;
}
else break;
}
int diff = in.length()k1;
if(in.charAt(k)!='.')
{
in+=".";
for(int i=0;i<dec;i++) in+="0";
}
else if(dec>diff)
for(int i=0;i<(decdiff);i++)
in+="0";
return in;
}
}[/java]
Last edited by randomtaiwanese on Thu Nov 25, 2004 10:51 pm, edited 1 time in total.
 little joey
 Guru
 Posts: 1080
 Joined: Thu Dec 19, 2002 7:37 pm
 Abednego
 A great helper
 Posts: 281
 Joined: Tue Sep 10, 2002 5:14 am
 Location: Mountain View, CA, USA
 Contact:
Holy mother of god!
I think your only mistake is choosing the wrong language for this problem. :)
Somebody correct me if I'm wrong, but Java's NumberFormat by default rounds 0.5 to the nearest even number (which is 0), and printf() rounds up (which is 1). Since changing double to long double gives Wrong Answer, this little detail could be what is wrong with your code.
I think your only mistake is choosing the wrong language for this problem. :)
Somebody correct me if I'm wrong, but Java's NumberFormat by default rounds 0.5 to the nearest even number (which is 0), and printf() rounds up (which is 1). Since changing double to long double gives Wrong Answer, this little detail could be what is wrong with your code.
If only I had as much free time as I did in college...

 New poster
 Posts: 32
 Joined: Fri Oct 01, 2004 10:53 pm
shame, becuase i have spent soo much time on writing a method on formatting a double value to a certain decimal place... ><little joey wrote:Did you read my previous posting? It's impossible to get accepted in Pascal, and my strong impression is that it's also impossible to get accepted in Java. To date there are 599 people that got accepted, non of them with Java or Pascal.
Rewrite your code in C or C++.

 New poster
 Posts: 32
 Joined: Fri Oct 01, 2004 10:53 pm
nah... java isnt this bad, it is only just about 35 times slower than other languages if both uses the best compiler, but because of the STONEAGE java compiler they have... java is like 10 times slower, a program with just ONE println() method takes 0.004 seconds to run on the OJAbednego wrote:Holy mother of god!
I think your only mistake is choosing the wrong language for this problem.
Somebody correct me if I'm wrong, but Java's NumberFormat by default rounds 0.5 to the nearest even number (which is 0), and printf() rounds up (which is 1). Since changing double to long double gives Wrong Answer, this little detail could be what is wrong with your code.
Sigh...
I couldn't find what was wrong with my code (Java), so I came here, read that thing, went and checked the stats  791 so far, C/C++ only.
That would explain some other WAs on similarly easy problems, I think (10137 (AC at PC), 10056...)
I guess I'll just skip any problems involving precision. Yes, I know I could just submit it in C and be done with it, but that's not the point.
Darko
P.S. Btw, people got AC by printing 3 0's? (I tried both 2 and 3) From the problem description one could say that if x_min = 0, you print only one, not two?
I couldn't find what was wrong with my code (Java), so I came here, read that thing, went and checked the stats  791 so far, C/C++ only.
That would explain some other WAs on similarly easy problems, I think (10137 (AC at PC), 10056...)
I guess I'll just skip any problems involving precision. Yes, I know I could just submit it in C and be done with it, but that's not the point.
Darko
P.S. Btw, people got AC by printing 3 0's? (I tried both 2 and 3) From the problem description one could say that if x_min = 0, you print only one, not two?