you're right..shadow wrote:The problem hasnt precisely defined the answer when the dog and the gopher both reach a hole at the same time. I think the answer in this case is "can reach".
10310  Dog and Gopher
Moderator: Board moderators
Re: 10310: vague problem statement

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 Posts: 3
 Joined: Tue Mar 13, 2007 11:04 am
 Contact:
Re: math is always important...
And don't forget them outputs, oh boy do that thing blows up if you don't make it a 3 decimal pecision exact.stcheung wrote:People, perhaps this may be the most useful tip on getting your 10310 program to AC. First of all, you can use doubles (no need to use long long and multiply input by 1000 blahblah). Secondly, when the dog & gopher reach the hole at same time, it's still considered a successful escape. Lastly, and definitely NOT the least, if you are lazy like me and didn't take sqrt when calculating distances, then when you compare the dog & gopher distances, remember to multiply gopher distance by 4 (or divide dog distance by 4). Multiplying/dividing by 2 doesn't work if you HAVE NOT taken square root (and that's how I keep getting WA at first).
~Sunny

 New poster
 Posts: 10
 Joined: Mon May 07, 2007 4:45 am
Hi I'm trying to do this problem since the algorithm is straight forward but I can't AC for some reason so I was wondering if someone could kindly help me find the bugs in code =)
Here it is..hope someone can tell me what's wrong with it
Here it is..hope someone can tell me what's wrong with it
Code: Select all
#include<iostream>
#include<stdio.h>
using namespace std;
typedef struct
{
double x, y;
}point;
inline double sqr(double n) { return n*n; }
int main()
{
while( !cin.eof() )
{
point dog, gopher;
int nholes, i, flag=0;
point hole, escape;
cin>>nholes;
cin>>gopher.x>>gopher.y>>dog.x>>dog.y;
for( i=0; i<nholes; i++ )
{
cin>>hole.x>>hole.y;
if( flag==1 ) continue;
if( (sqr(hole.xdog.x)+sqr(hole.ydog.y))>=4.0*(sqr(hole.xgopher.x)+sqr(hole.ygopher.y)) )
{
escape.x=hole.x; escape.y=hole.y;
flag=1;
}
}
if( flag==0 )
printf("The gopher cannot escape.\n");
else
printf("The gopher can escape through the hole at (%.3lf,%.3lf).\n", escape.x, escape.y);
}
return 0;
}
#include<stdio.h>
#include<math.h>
int main()
{
int n,k,temp;
double x1,y1,x2,y2;
double holeX[10000],holeY[10000];
double d1,d2;
while(scanf("%d%lf%lf%lf%lf",&n,&x1,&y1,&x2,&y2)==5)
{
temp=0;
for(k=0;k<n;k++)
{
scanf("%lf%lf",&holeX[k],&holeY[k]);
d1=sqrt((holeX[k]x1)*(holeX[k]x1)+(holeY[k]y1)*(holeY[k]y1));
d2=sqrt((holeX[k]x2)*(holeX[k]x2)+(holeY[k]y2)*(holeY[k]y2));
if(2*d1<=d2)
{
printf("The gopher can escape through the hole at (%.3lf,%.3lf).\n",holeX[k],holeY[k]);
temp=1;
break;
}
}
if(temp==0)printf("The gopher cannot escape.\n") ;
}
return 0;
}
why i got wa......?
#include<math.h>
int main()
{
int n,k,temp;
double x1,y1,x2,y2;
double holeX[10000],holeY[10000];
double d1,d2;
while(scanf("%d%lf%lf%lf%lf",&n,&x1,&y1,&x2,&y2)==5)
{
temp=0;
for(k=0;k<n;k++)
{
scanf("%lf%lf",&holeX[k],&holeY[k]);
d1=sqrt((holeX[k]x1)*(holeX[k]x1)+(holeY[k]y1)*(holeY[k]y1));
d2=sqrt((holeX[k]x2)*(holeX[k]x2)+(holeY[k]y2)*(holeY[k]y2));
if(2*d1<=d2)
{
printf("The gopher can escape through the hole at (%.3lf,%.3lf).\n",holeX[k],holeY[k]);
temp=1;
break;
}
}
if(temp==0)printf("The gopher cannot escape.\n") ;
}
return 0;
}
why i got wa......?
Try this:

Rio
Code: Select all
3 2.000 2.000 1.000 1.000
2.500 2.500
1.500 1.500
0.000 0.000
Rio

 New poster
 Posts: 4
 Joined: Mon Dec 08, 2008 11:03 am
I got AC in UVa judge, but i got WA in Programmingchallenge
Please Help..
I got AC in UVa judge, but why i got WA in http://www.programmingchallenges.com judge? I don't understand..
I'm sorry posting my AC code here , but, for me it's still WA , coz it's WA in programmingchallenges, please help..thanks in advance..
Here is my code, for anyone who got AC, please try at http://www.programmingchallenges.com and please tell me why i got WA there?
I got AC in UVa judge, but why i got WA in http://www.programmingchallenges.com judge? I don't understand..
I'm sorry posting my AC code here , but, for me it's still WA , coz it's WA in programmingchallenges, please help..thanks in advance..
Here is my code, for anyone who got AC, please try at http://www.programmingchallenges.com and please tell me why i got WA there?
Code: Select all
#include<iostream>
#include<iomanip>
#include<cmath>
#define eps 1e7
using namespace std;
double panjang(double x1, double y1, double x2, double y2);
int main()
{
int n;
double G1, G2, D1, D2;
double H[1000][2];
int i, check;
cout<<fixed<<showpoint<<setprecision(3);
while(cin>>n>>G1>>G2>>D1>>D2)
{
check=0;
for(i=0; i<n; i++)
cin>>H[i][0]>>H[i][1];
for(i=0; i<n; i++)
{
if(panjang(D1,D2,H[i][0],H[i][1])>2*panjang(G1,G2,H[i][0],H[i][1]))
{
cout<<"The gopher can escape through the hole at ("<<H[i][0]+eps<<","<<H[i][1]+eps<<")."<<endl;
check++;
break;
}
}
if(check==0)
cout<<"The gopher cannot escape."<<endl;
}
return 0;
}
double panjang(double x1, double y1, double x2, double y2)
{
return sqrt((x1x2)*(x1x2)+(y1y2)*(y1y2));
}

 Experienced poster
 Posts: 136
 Joined: Sat Nov 29, 2008 8:01 am
 Location: narayangong,bangladesh.
 Contact:
10310  Dog and Gopher
whats wrong with my code??
Code: Select all
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define eps 1e7
double h[1001][1001];
int main()
{
int nhole,i,flag;
double x1,x2,y1,y2,dis_gof,dis_dog,h_x,h_y;
// freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
while(scanf("%d",&nhole)==1)
{
//sscanf(str,"%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
scanf("%lf %lf %lf %lf\n",&x1,&y1,&x2,&y2);
for(i=0;i<nhole;i++)
{
scanf("%lf %lf",&h_x,&h_y);
//h_x=atof(str1);
//h_y=atof(str2);
dis_gof=(((x1h_x)*(x1h_x))+((y1h_y)*(y1h_y)));
printf("%lf\n",dis_gof);
dis_dog=((x2h_x)*(x2h_x))+((y2h_y)*(y2h_y));
printf("%lf\n",dis_dog);
if(dis_dog4*dis_gof>=eps)
{
flag=1;
break;
}
else
{
flag=0;
}
scanf("\n");
}
//printf("flag=%d\n",flag);
if(flag==1)
printf("The gopher can escape through the hole at (%.3lf,%.3lf).\n",h_x,h_y);
else
printf("The gopher cannot escape.\n");
}
return 0;
}
Life is more complicated than algorithm.
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http://felixhalim.net/uva/hunting.php?id=32359
For Hints: http://salimsazzad.wordpress.com
Re: 10310  Dog and Gopher
@sazzadcsedu
you are breaking the "for(i=0;i<nhole;i++)" loop if you find a suitable hole for the gopher. this loop happens to do the scanning as well "scanf("%lf %lf",&h_x,&h_y);". now tell me what will happen if you find a suitable hole before scanning all the hole coordinates? I mean say nhole is 10 and the 2nd hole is a escape hole for the gopher..
you are breaking the "for(i=0;i<nhole;i++)" loop if you find a suitable hole for the gopher. this loop happens to do the scanning as well "scanf("%lf %lf",&h_x,&h_y);". now tell me what will happen if you find a suitable hole before scanning all the hole coordinates? I mean say nhole is 10 and the 2nd hole is a escape hole for the gopher..
hmm..

 Experienced poster
 Posts: 136
 Joined: Sat Nov 29, 2008 8:01 am
 Location: narayangong,bangladesh.
 Contact:
Re: 10310  Dog and Gopher
NEWKID,
IF NOT BREAK THE LOOP
THEN IT PRINT THE LAST ESCAPABLE HOLE.
IF MORE THEN 1 ESCAPABLE HOLE THEN WHAT SHOULD BE ANSWER????
IF NOT BREAK THE LOOP
THEN IT PRINT THE LAST ESCAPABLE HOLE.
IF MORE THEN 1 ESCAPABLE HOLE THEN WHAT SHOULD BE ANSWER????
Life is more complicated than algorithm.
http://felixhalim.net/uva/hunting.php?id=32359
For Hints: http://salimsazzad.wordpress.com
http://felixhalim.net/uva/hunting.php?id=32359
For Hints: http://salimsazzad.wordpress.com
Re: 10310  Dog and Gopher
oh.. thats the easy part.. you can do like this..
Code: Select all
typedef struct Hole {
double x, y;
}
Hole hole[MAXHOLES];
for (int i = 0; i < nholes; i++) {
scanf("%lf %lf", &hole[i].x, &hole[i].y);
}
for (int i = 0; i < nholes; i++) {
// if feasible with this hole then set ur flag and break..
}
hmm..
Re: 10310  Dog and Gopher
Can some one help me. I got TLE..
Code: Select all
removed
Last edited by Obaida on Tue Feb 03, 2009 9:13 am, edited 1 time in total.
try_try_try_try_&&&_try@try.com
This may be the address of success.
This may be the address of success.
Re: 10310  Dog and Gopher
whats the logic behind this line?
just comment this line and you will get accepted..
Code: Select all
if(((dog_x>=gophar_x)&&(gophar_x>=hole_x[i]))((dog_x<=gophar_x)&&(gophar_x<=hole_x[i]))((dog_xgophar_x==0)&&(((dog_y>=gophar_y)&&(gophar_y>=hole_y[i]))((dog_y<=gophar_y)&&(gophar_y<=hole_y[i])))))
hmm..
Re: 10310  Dog and Gopher
Sorry i was kidding.
try_try_try_try_&&&_try@try.com
This may be the address of success.
This may be the address of success.