209  Triangular Vertices
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Problem 209  Input Examples
Hy!
Can anyone help me? I need some critical input examples for problem 209.
Thanks,
Charlla
Can anyone help me? I need some critical input examples for problem 209.
Thanks,
Charlla
209 WA
Why does this code get WA?
My program analises: 2 different triangle types, 3 different paralelogram types; 1 hexagon type;
[cpp]
#include<stdio.h>
#include<iostream>
#include<sstream>
using namespace std;
struct Coord {
int x;
int y;
};
int triangle(Coord *p, long *d);
int parallelogram(Coord *p,long *d);
int hexagon(Coord *p,long *d);
int sort(int *p,int size) {
int i,j;
int k;
for(i=0;i<size1;i++) {
if (p>p[i+1]) {
j=i+1;
while ((j>0) && (p[j1]>p[j])) {
//cout << "j=" <<j << " ";
k=p[j1];
p[j1]=p[j];
p[j]=k;
j;
}
}
}
return 0;
}
int readline(const char *str,int *p)
{
int i=0;
istringstream ist(str);
while (ist >> p) {
printf("%d ",p);
if (i<=6)
i++;
}
return i;
}
int main() {
int p[8]; //points;
int i,count;
char line[5000];
Coord c[6];
long d[6];
while (!cin.eof()) {
cin.getline(line,5000); //read 1 input line
count=readline(line,p);
if(count>6  count<3  count==5) {
printf("are not the vertices of an acceptable figure\n");
}
else {
sort(p,count);
d[0]=0;
//convert p to (x,y)
for(i=0;i<count;i++)
{
if (i>0)
d=d[i1];
while((d+1)*(d+2)/2 < p) d++;
c.x=(d[i]+1)*(d[i]+2)/2p[i];
c[i].y=d[i]c[i].x;
}
switch(count) {
case 3:if (triangle(c,d))
printf("are the vertices of a triangle\n");
else printf("are not the vertices of an acceptable figure\n");
break;
case 4:if (parallelogram(c,d))
printf("are the vertices of a parallelogram\n");
else printf("are not the vertices of an acceptable figure\n");
break;
case 6:if (hexagon(c,d))
printf("are the vertices of a hexagon\n");
else printf("are not the vertices of an acceptable figure\n");
break;
}
}
}
return 0;
}
int triangle(Coord* p,long *d) {
if ((d[0]==d[1]) && (p[0].x!=p[1].x) &&
(p[0].x==p[2].x) && (p[1].y==p[2].y))
return 1;
else if ((p[1].x+p[1].y==p[2].x+p[2].y) && (p[1].x!=p[2].x) &&
(p[0].x==p[2].x) && (p[0].y==p[1].y))
return 1;
else return 0;
}
int parallelogram(Coord *p,long *d) {
if ((d[0]==d[1]) && (p[0].x!=p[1].x) &&
(d[2]==d[3]) && (p[2].x!=p[3].x) && (d[0]<d[2])) {
if ((p[0].y==p[2].y) && (p[1].y==p[3].y) && (p[0].x==p[3].x))
return 1;
else if ((p[0].x==p[2].x) && (p[1].x==p[3].x) && (p[1].y==p[2].y))
return 1;
else return 0;
}
if ((d[1]==d[2]) && (d[0]<d[1]) && (d[1]<d[3]) &&
(p[0].x==p[2].x) && (p[0].y==p[1].y) &&
(p[1].x==p[3].x) && (p[2].y==p[3].y))
return 1;
return 0;
}
int hexagon(Coord *p,long *d) {
if ((d[0]==d[1]) && (d[2]==d[3]) && (d[4]==d[5]) && (p[0].x!=p[1].x) &&
(p[0].y==p[2].y) && (p[1].y==p[4].y) && (p[3].y==p[5].y) &&
(p[0].x==p[5].x) && (p[1].x==p[3].x) && (p[2].x==p[4].x))
return 1;
else return 0;
}
[/cpp]
My program analises: 2 different triangle types, 3 different paralelogram types; 1 hexagon type;
[cpp]
#include<stdio.h>
#include<iostream>
#include<sstream>
using namespace std;
struct Coord {
int x;
int y;
};
int triangle(Coord *p, long *d);
int parallelogram(Coord *p,long *d);
int hexagon(Coord *p,long *d);
int sort(int *p,int size) {
int i,j;
int k;
for(i=0;i<size1;i++) {
if (p>p[i+1]) {
j=i+1;
while ((j>0) && (p[j1]>p[j])) {
//cout << "j=" <<j << " ";
k=p[j1];
p[j1]=p[j];
p[j]=k;
j;
}
}
}
return 0;
}
int readline(const char *str,int *p)
{
int i=0;
istringstream ist(str);
while (ist >> p) {
printf("%d ",p);
if (i<=6)
i++;
}
return i;
}
int main() {
int p[8]; //points;
int i,count;
char line[5000];
Coord c[6];
long d[6];
while (!cin.eof()) {
cin.getline(line,5000); //read 1 input line
count=readline(line,p);
if(count>6  count<3  count==5) {
printf("are not the vertices of an acceptable figure\n");
}
else {
sort(p,count);
d[0]=0;
//convert p to (x,y)
for(i=0;i<count;i++)
{
if (i>0)
d=d[i1];
while((d+1)*(d+2)/2 < p) d++;
c.x=(d[i]+1)*(d[i]+2)/2p[i];
c[i].y=d[i]c[i].x;
}
switch(count) {
case 3:if (triangle(c,d))
printf("are the vertices of a triangle\n");
else printf("are not the vertices of an acceptable figure\n");
break;
case 4:if (parallelogram(c,d))
printf("are the vertices of a parallelogram\n");
else printf("are not the vertices of an acceptable figure\n");
break;
case 6:if (hexagon(c,d))
printf("are the vertices of a hexagon\n");
else printf("are not the vertices of an acceptable figure\n");
break;
}
}
}
return 0;
}
int triangle(Coord* p,long *d) {
if ((d[0]==d[1]) && (p[0].x!=p[1].x) &&
(p[0].x==p[2].x) && (p[1].y==p[2].y))
return 1;
else if ((p[1].x+p[1].y==p[2].x+p[2].y) && (p[1].x!=p[2].x) &&
(p[0].x==p[2].x) && (p[0].y==p[1].y))
return 1;
else return 0;
}
int parallelogram(Coord *p,long *d) {
if ((d[0]==d[1]) && (p[0].x!=p[1].x) &&
(d[2]==d[3]) && (p[2].x!=p[3].x) && (d[0]<d[2])) {
if ((p[0].y==p[2].y) && (p[1].y==p[3].y) && (p[0].x==p[3].x))
return 1;
else if ((p[0].x==p[2].x) && (p[1].x==p[3].x) && (p[1].y==p[2].y))
return 1;
else return 0;
}
if ((d[1]==d[2]) && (d[0]<d[1]) && (d[1]<d[3]) &&
(p[0].x==p[2].x) && (p[0].y==p[1].y) &&
(p[1].x==p[3].x) && (p[2].y==p[3].y))
return 1;
return 0;
}
int hexagon(Coord *p,long *d) {
if ((d[0]==d[1]) && (d[2]==d[3]) && (d[4]==d[5]) && (p[0].x!=p[1].x) &&
(p[0].y==p[2].y) && (p[1].y==p[4].y) && (p[3].y==p[5].y) &&
(p[0].x==p[5].x) && (p[1].x==p[3].x) && (p[2].x==p[4].x))
return 1;
else return 0;
}
[/cpp]
209  Triangular Vertices
Hi!
I've got a question about problem 209. A polygon with side length 0 is permitted? I mean, is 1 1 1 a valid triangle?
I'm getting WA again and again and I don't know any other special case...
Thank you very much!!
Bernat
I've got a question about problem 209. A polygon with side length 0 is permitted? I mean, is 1 1 1 a valid triangle?
I'm getting WA again and again and I don't know any other special case...
Thank you very much!!
Bernat
 GreenPenInc
 Learning poster
 Posts: 53
 Joined: Sat May 01, 2004 9:31 pm
 Contact:
reading input whose end is not signalled
OK, I'm officially sick of reading in input whose end isn't signalled by a zero or something. Example: problem 209. I've had the logic down for a while now, and it runs like a dream on my own computer, but I alternate between RTE and TLE when I submit. What's a good way to handle cases where you keep on reading until EOF? I'd love to hear multiple people's thoughts on this. Thanks!
_(GPI)_
"Finally I have freed myself from the clutches of the garbage fairy!"
"Finally I have freed myself from the clutches of the garbage fairy!"
 GreenPenInc
 Learning poster
 Posts: 53
 Joined: Sat May 01, 2004 9:31 pm
 Contact:
Hmm... thanks, that's what I have been doing. I just hate doing it
What REALLY burns me up is the following. I've been writing a function to take a string and return an int. It all compiles fine on my machine. The following also compiles on Valladolid.
[cpp]
int make_int(string s)
{
int result = 0, i, temp;
for (i = 0; i < s.length(); i++)
{
//result = result * 10;
temp = s.at(i)  48;
//result += temp;
//result = result + (s.at(i)  '0');
}
return result;
/*
for (int i = 0; i < s.length(); i++)
{
result *= 10;
result += (int)(s.at(i)  '0');
}
return result;
*/
}
[/cpp]
[cpp]
The following does NOT.int make_int(string s)
{
int result = 0, i, temp;
for (i = 0; i < s.length(); i++)
{
//result = result * 10;
temp = s.at(i)  48;
result += temp;
//result = result + (s.at(i)  '0');
}
return result;
/*
for (int i = 0; i < s.length(); i++)
{
result *= 10;
result += (int)(s.at(i)  '0');
}
return result;
*/
}
[/cpp]
Now, why the hell would result += temp, when they're both integers, cause me to get TLE? I have isolated it (through scores of intentionally wrong submissions) to this very line of code. What the hell is going on?
What REALLY burns me up is the following. I've been writing a function to take a string and return an int. It all compiles fine on my machine. The following also compiles on Valladolid.
[cpp]
int make_int(string s)
{
int result = 0, i, temp;
for (i = 0; i < s.length(); i++)
{
//result = result * 10;
temp = s.at(i)  48;
//result += temp;
//result = result + (s.at(i)  '0');
}
return result;
/*
for (int i = 0; i < s.length(); i++)
{
result *= 10;
result += (int)(s.at(i)  '0');
}
return result;
*/
}
[/cpp]
[cpp]
The following does NOT.int make_int(string s)
{
int result = 0, i, temp;
for (i = 0; i < s.length(); i++)
{
//result = result * 10;
temp = s.at(i)  48;
result += temp;
//result = result + (s.at(i)  '0');
}
return result;
/*
for (int i = 0; i < s.length(); i++)
{
result *= 10;
result += (int)(s.at(i)  '0');
}
return result;
*/
}
[/cpp]
Now, why the hell would result += temp, when they're both integers, cause me to get TLE? I have isolated it (through scores of intentionally wrong submissions) to this very line of code. What the hell is going on?
_(GPI)_
"Finally I have freed myself from the clutches of the garbage fairy!"
"Finally I have freed myself from the clutches of the garbage fairy!"
 Krzysztof Duleba
 Guru
 Posts: 584
 Joined: Thu Jun 19, 2003 3:48 am
 Location: Sanok, Poland
 Contact:
All my programs have the following shape:
[cpp]while(true){
Typename t;
if(cin >> t)break;
}[/cpp]
or
[cpp]while(true){
Typename t;
cin >> t;
if(t == EndOfInputCharacter)break;
}[/cpp]
It is also very easy to use scanf or fgets instead of cin here. In the first case, just look at the value returned by reading function (0 means EOF), in second  compare t with EndOfInputCharacter. So generally everything stays the same, and if the first input value is the number of cases, I simply ignore it.
[cpp]while(true){
Typename t;
if(cin >> t)break;
}[/cpp]
or
[cpp]while(true){
Typename t;
cin >> t;
if(t == EndOfInputCharacter)break;
}[/cpp]
It is also very easy to use scanf or fgets instead of cin here. In the first case, just look at the value returned by reading function (0 means EOF), in second  compare t with EndOfInputCharacter. So generally everything stays the same, and if the first input value is the number of cases, I simply ignore it.
 GreenPenInc
 Learning poster
 Posts: 53
 Joined: Sat May 01, 2004 9:31 pm
 Contact:
Okay, after asking around in #c++ (and getting badgered for the bad habits I developed by doing competitions ) I found the following scheme. As always, it works great on my own comp, but I'm getting a SIGSEGV on the Online Judge. Help?!
[cpp]
int main()
{
int nums[6], length, i;
char * line;
string msg;
while (cin.getline(line, 256))
{
length = 0;
istringstream iss(line);
while (iss >> nums[length++]);
length;
msg = "are not the vertices of an acceptable figure";
switch(length)
{
case 3:
if (check_triangle(nums))
msg = "are the vertices of a triangle";
break;
case 4:
if (check_parallelogram(nums))
msg = "are the vertices of a parallelogram";
break;
case 6:
if (check_hexagon(nums))
msg = "are the vertices of a hexagon";
break;
}
for (i = 0; i < length; i++)
cout << nums << " ";
cout << msg << endl;
}
return 0;
}
[/cpp]
[cpp]
int main()
{
int nums[6], length, i;
char * line;
string msg;
while (cin.getline(line, 256))
{
length = 0;
istringstream iss(line);
while (iss >> nums[length++]);
length;
msg = "are not the vertices of an acceptable figure";
switch(length)
{
case 3:
if (check_triangle(nums))
msg = "are the vertices of a triangle";
break;
case 4:
if (check_parallelogram(nums))
msg = "are the vertices of a parallelogram";
break;
case 6:
if (check_hexagon(nums))
msg = "are the vertices of a hexagon";
break;
}
for (i = 0; i < length; i++)
cout << nums << " ";
cout << msg << endl;
}
return 0;
}
[/cpp]
_(GPI)_
"Finally I have freed myself from the clutches of the garbage fairy!"
"Finally I have freed myself from the clutches of the garbage fairy!"
 GreenPenInc
 Learning poster
 Posts: 53
 Joined: Sat May 01, 2004 9:31 pm
 Contact:
FOR ANYONE WHO GOT AC
I ONLY need to see your INPUT. I have tried everything and am at the end of my rope. Please, just show me how to input using cin. Here's my main method. Works GREAT on my computer.
[cpp]
int main()
{
int nums[6], length, i;
char line[257];
string msg;
while (!cin.getline(line, 256).eof())
{
length = 0;
istringstream iss(line);
while (iss >> nums[length++]);
length;
msg = "are not the vertices of an acceptable figure";
switch(length)
{
case 3:
if (check_triangle(nums))
msg = "are the vertices of a triangle";
break;
case 4:
if (check_parallelogram(nums))
msg = "are the vertices of a parallelogram";
break;
case 6:
if (check_hexagon(nums))
msg = "are the vertices of a hexagon";
break;
}
for (i = 0; i < length; i++)
cout << nums << " ";
cout << msg << endl;
}
return 0;
}
[/cpp]
This gives me TLE on the Judge; it doesn't know when to stop seeking input. I have over 50 submissions trying to debug this. Just help me. Somebody. Please.
I ONLY need to see your INPUT. I have tried everything and am at the end of my rope. Please, just show me how to input using cin. Here's my main method. Works GREAT on my computer.
[cpp]
int main()
{
int nums[6], length, i;
char line[257];
string msg;
while (!cin.getline(line, 256).eof())
{
length = 0;
istringstream iss(line);
while (iss >> nums[length++]);
length;
msg = "are not the vertices of an acceptable figure";
switch(length)
{
case 3:
if (check_triangle(nums))
msg = "are the vertices of a triangle";
break;
case 4:
if (check_parallelogram(nums))
msg = "are the vertices of a parallelogram";
break;
case 6:
if (check_hexagon(nums))
msg = "are the vertices of a hexagon";
break;
}
for (i = 0; i < length; i++)
cout << nums << " ";
cout << msg << endl;
}
return 0;
}
[/cpp]
This gives me TLE on the Judge; it doesn't know when to stop seeking input. I have over 50 submissions trying to debug this. Just help me. Somebody. Please.
_(GPI)_
"Finally I have freed myself from the clutches of the garbage fairy!"
"Finally I have freed myself from the clutches of the garbage fairy!"

 Guru
 Posts: 724
 Joined: Wed Dec 19, 2001 2:00 am
 Location: Germany
I am astonished that it works on your computer;
char * line;
This doesn't allocate any memory, so when you use
cin.getline(line,256);
where should the function getline write its result?
Just tried it on my computer, and it works, too. But I think that this is just luck, because line points to some address. Now if you write char *line = NULL you will see that it will not work.
So the thing you have to do is:
char line[258];
char * line;
This doesn't allocate any memory, so when you use
cin.getline(line,256);
where should the function getline write its result?
Just tried it on my computer, and it works, too. But I think that this is just luck, because line points to some address. Now if you write char *line = NULL you will see that it will not work.
So the thing you have to do is:
char line[258];
The online judge is very picky about memory addressing.Adrian Kuegel wrote:I am astonished that it works on your computer;
char * line;
This doesn't allocate any memory, so when you use
cin.getline(line,256);
where should the function getline write its result?
Just tried it on my computer, and it works, too. But I think that this is just luck, because line points to some address. Now if you write char *line = NULL you will see that it will not work.
So the thing you have to do is:
char line[258];
Most offbyone errors I made got unnoticed when compiled locally but got caught by the OJ.
Basically this happens because SIGSEGV signals are generated when accessing memory outside the memory segment assigned to the process. If you access unallocated memory that happens to be within a segment owned by your program you don't get a sigfault.
To catch those nasty mistakes you need extra checking enabled.
You may want to use stackguard to catch that nasty bugs...
Ciao!!!
Claudio