11068 - An Easy Task
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11068 - An Easy Task
you might try to mirror the plane first at one line, then at the second line. What does that statement mean . i don't understand the problem clearly. Can anyone help
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I keep getting WAs on this problem. Can anyone help? 
This is my code.

This is my code.
Code: Select all
removed after AC
Last edited by cytmike on Tue Aug 15, 2006 11:23 pm, edited 1 time in total.
Impossible is Nothing.
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Re: 11068
In my opinion, the question is asking for the intersection point of the lines (if any), is that the whole thing this question is asking?temper_3243 wrote:you might try to mirror the plane first at one line, then at the second line. What does that statement mean . i don't understand the problem clearly. Can anyone help
Impossible is Nothing.
Re: 11068
Well, the question doesn't ask that directly, although it ends up being just that. The question is asking whether the composition of two "mirrorings" (that is, a reflection over one of the lines followed by another reflection over the other line) has a fixed point (a point which is left at the same place afther these two steps).cytmike wrote: In my opinion, the question is asking for the intersection point of the lines (if any), is that the whole thing this question is asking?
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Try this one:cytmike wrote:I keep getting WAs on this problem. Can anyone help?
Code: Select all
-19 100 43
19 -53 55
0 0 0
0 0 0
Code: Select all
The fixed point is at 8.71 2.09.
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Here's a couple of tests:
The correct output is:
Code: Select all
1 2 3 2 4 7
0 1 0 1 2 0
0 0 0 0 0 0
Code: Select all
No fixed point exists.
The fixed point is at 0.00 0.00.
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11068 An Easy Task (WA) Why? Please, help
/* Removed */
Last edited by uvasarker on Sat Jan 28, 2012 7:33 pm, edited 1 time in total.
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Re: 11068 An Easy Task (WA) Why? Please, help
You should post in the existing thread.
Check input and AC output for thousands of problems on uDebug!
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Re: 11068 - An Easy Task
I check previous posted of this topic and try all test cases and my code gives correct answer, but when I submit my code to the judge they reply to verdict WA WA WA.
Why? Please, give me some information on my code or critical input-output. Please...............
Why? Please, give me some information on my code or critical input-output. Please...............
Code: Select all
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<iostream>
#include<fstream>
#include<numeric>
#include<string>
#include<vector>
#include<cstring>
#include<map>
#include<iterator>
using namespace std;
int main()
{
float a1,a2,b1,b2,c1,c2;
while(scanf("%f %f %f %f %f %f",&a1,&b1,&c1,&a2,&b2,&c2)==6)
{
if(a1==0 && a2==0 && b1==0 && b2==0 && c1==0 && c2==0)
break;
float x1=0.0,y1=0.0;
float difa=0.0,difb=0.0,minx=0.0;
if(a1==a2 && b1==b2)
printf("No fixed point exists.\n");
else
{
if(a1>a2)
{
if(a1<0)
{
a1=a1*(-1.0);
b1=b1*(-1.0);
c1=c1*(-1.0);
}
if(a2<0)
{
a2=a2*(-1.0);
b2=b2*(-1.0);
c2=c2*(-1.0);
}
if(a2==0) difa=a1;
else
difa=a1/a2;
}
else
{
if(a1<0)
{
a1=a1*(-1.0);
b1=b1*(-1.0);
c1=c1*(-1.0);
}
if(a2<0)
{
a2=a2*(-1.0);
b2=b2*(-1.0);
c2=c2*(-1.0);
}
if(a1==0) difa=a2;
else
difa=a2/a1;
}
if(b1>b2)
{
if(b1<0)
{
b1=b1*(-1.0);
a1=a1*(-1.0);
c1=c1*(-1.0);
}
if(b2<0)
{
b2=b2*(-1.0);
a2=a2*(-1.0);
c2=c2*(-1.0);
}
if(b2==0) difb=b1;
else
difb=b1/b2;
}
else
{
if(b1<0)
{
b1=b1*(-1.0);
a1=a1*(-1.0);
c1=c1*(-1.0);
}
if(b2<0)
{
b2=b2*(-1.0);
a2=a2*(-1.0);
c2=c2*(-1.0);
}
if(b1==0) difb=b2;
else
difb=b2/b1;
}
minx=min(difa,difb);
if( (minx==difa && a1<a2 && difa>1) || (minx==difb && b1<b2 && difb>1) )
{
a1=a1*minx;
b1=b1*minx;
c1=c1*minx;
}
else if( (minx==difa && a2<a1 && difa>1) || (minx==difb && b2<b1 && difb>1) )
{
a2=a2*minx;
b2=b2*minx;
c2=c2*minx;
}
if(a1==a2 && b1==b2)
printf("No fixed point exists.\n");
else
{
if(b1==b2)
{
x1=(c1-c2)/(a1-a2);
y1=(c1-(a1*x1))/b1;
}
else if(a1==a2)
{
y1=(c1-c2)/(b1-b2);
x1=(c1-(b1*y1))/a1;
}
printf("The fixed point is at %.2f %.2f.\n",x1,y1);
}
}
}
return 0;
}
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Re: 11068 - An Easy Task
Input:
Output from my AC code:
Code: Select all
1 0 1
2 0 2
0 0 1
1 2 3
1 2 3
0 0 1
1 2 4
1 0 1
1 0 1
1 2 5
1 2 3
1 2 4
6 5 11
2 43 -23
0 0 0
0 0 0
Code: Select all
No fixed point exists.
No fixed point exists.
No fixed point exists.
The fixed point is at 1.00 1.50.
The fixed point is at 1.00 2.00.
No fixed point exists.
The fixed point is at 2.37 -0.65.
Check input and AC output for thousands of problems on uDebug!
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Re: 11068 - An Easy Task
Hi
Boss
I am frustrated.............Still WA. Here is my code:
Boss
I am frustrated.............Still WA. Here is my code:
Code: Select all
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<iostream>
#include<fstream>
#include<numeric>
#include<string>
#include<vector>
#include<cstring>
#include<map>
#include<iterator>
using namespace std;
int main()
{
double a1,a2,b1,b2,c1,c2;
while(scanf("%lf %lf %lf %lf %lf %lf",&a1,&b1,&c1,&a2,&b2,&c2)==6)
{
if(a1==0 && a2==0 && b1==0 && b2==0 && c1==0 && c2==0)
break;
double x1=0.0,y1=0.0;
double difa=0.0,difb=0.0,minx=0.0;
if( (a1==a2 && b1==b2) || (a1==0.0 && b1==0.0) || (a2==0.0 && b2==0.0) || (a1==c1 && a2==c2 && b1==b2))
printf("No fixed point exists.\n");
else
{
if(a1>a2)
{
if(a1<0)
{
a1=a1*(-1.0);
b1=b1*(-1.0);
c1=c1*(-1.0);
}
if(a2<0)
{
a2=a2*(-1.0);
b2=b2*(-1.0);
c2=c2*(-1.0);
}
if(a2==0) difa=a1;
else
difa=a1/a2;
}
else
{
if(a1<0)
{
a1=a1*(-1.0);
b1=b1*(-1.0);
c1=c1*(-1.0);
}
if(a2<0)
{
a2=a2*(-1.0);
b2=b2*(-1.0);
c2=c2*(-1.0);
}
if(a1==0) difa=a2;
else
difa=a2/a1;
}
if(b1>b2)
{
if(b1<0)
{
b1=b1*(-1.0);
a1=a1*(-1.0);
c1=c1*(-1.0);
}
if(b2<0)
{
b2=b2*(-1.0);
a2=a2*(-1.0);
c2=c2*(-1.0);
}
if(b2==0) difb=b1;
else
difb=b1/b2;
}
else
{
if(b1<0)
{
b1=b1*(-1.0);
a1=a1*(-1.0);
c1=c1*(-1.0);
}
if(b2<0)
{
b2=b2*(-1.0);
a2=a2*(-1.0);
c2=c2*(-1.0);
}
if(b1==0) difb=b2;
else
difb=b2/b1;
}
minx=min(difa,difb);
if( (minx==difa && a1<a2 && difa>1) || (minx==difb && b1<b2 && difb>1) )
{
a1=a1*minx;
b1=b1*minx;
c1=c1*minx;
}
else if( (minx==difa && a2<a1 && difa>1) || (minx==difb && b2<b1 && difb>1) )
{
a2=a2*minx;
b2=b2*minx;
c2=c2*minx;
}
if(a1==a2 && b1==b2)
printf("No fixed point exists.\n");
else
{
if(b1==b2)
{
x1=(c1-c2)/(a1-a2);
y1=(c1-(a1*x1))/b1;
}
else if(a1==a2)
{
y1=(c1-c2)/(b1-b2);
x1=(c1-(b1*y1))/a1;
}
printf("The fixed point is at %.2lf %.2lf.\n",x1,y1);
}
}
}
return 0;
}
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Re: 11068 - An Easy Task
Input
My AC output
Code: Select all
24 -18 67
73 -81 -33
0 0 0
0 0 0
Code: Select all
The fixed point is at 9.56 9.02.
Check input and AC output for thousands of problems on uDebug!