WA in SICRANO http://www.spoj.pl/problems/SICRANO/

here is my code sir......

#include<stdio.h>
int main()
{
    int nt=0,i=0,n=0,m=0,j=0,x1=0,x2=0,y1=0,y2=0,xt=0,yt=0;
    scanf("%d",&nt);
    for(i=0;i<nt;i++)
    {
                     scanf("%d %d",&n,&m);
                     int x[n],y[n];int out[m],k=0,l=0;
                     for(j=0;j<n;j++)
                     {
                                     scanf("%d %d",&x[j],&y[j]);
                                     }
                     for(j=0;j<m;j++)
                     {
                                     scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
                                     int lhs=0,rhs=0;
                                     lhs=(xt-x1)*(y2-y1);
                                     rhs=(yt-y1)*(x2-x1);
                                     int cnt=0;
                                     for(l=0;l<n;l++)
                                     {
                                                     if(x[l]>=x1 && y[l]>=y1 && x[l]<=x2 && y[l]<=y2)
                                                     {
                                                     if(((x[l]-x1)*(y2-y1))==((y[l]-y1)*(x2-x1)))
                                                                 cnt++;
                                                                 }
                                                                 }
                                     out[j]=cnt;
                                     }
                     for(j=0;j<m;j++)
                     {
                                     printf("%d\n",out[j]);
                                     }
                                    }
          return 0;
          }

Your issue lies mainly in this line of code:
if(x[l]>=x1 && y[l]>=y1 && x[l]<=x2 && y[l]<=y2)

sir i am getting correct aswer for test case and the case given in forum but WA when i submit it at spoj.
dont know why.
[

Your issue is almost 100% identical to the poster above...
if(a[j]>=x1[i]&&a[j]<=x2[i])

if((a[j]>=x1[i]&&a[j]<=x2[i])||(a[j]<=x1[i]&&a[j]>=x2[i]))

this is also giving wrong answer

but why?
coordinates are positive only not negative so i shouold not consider the negative case

You are making an assumption about the input data.

ok i read it again .
coordinated of segment can be -ve bt of points are +ve.i am right?
i changed code but getting wrong only.

if 2nd x coordinate is -ve i have interchanged both points of segment

u want to say that 5 5 0 0 can be input so

if(1>=5 && 1<=0)

will be false.
i have considered this case also in below code.

Your algorithm is not accurate. You have not considered all possibilities.

code removed got ac

the ques. is very easy and shorter in code length.

kkkkk......i wil code with a new approach but.....i just want 2 know what is wrong with this approach......inspite it is lengthy i want 2 know where do i fail......???

The biggest issue is that you are assuming that the slope will be an integer value. As mentioned before you should abandon this solution and simplify it greatly. Work with a pen and paper for a while and find a better algorithm.

i dont have patience to read this much length of code like leppy sir
but what i caught in 3 seconds was

m1=(points[j][1]-start[i][1])/(points[j][0]-start[i][0]);

if points[j][0]=0 and start[i][0]=0;
x/0 underministic
and rest follow what sir say

@Leppy-----i worked on some more furious test cases and debugged my code .............CODE REMOVED

Please fix your indentation. Please explain your algorithm. Please explain why you think this algorithm is accurate.

I have some trouble with that problem though it looks like an easy one. I'm not sure if it's a matter of wrong logic, I/O problem (with Python) or just misunderstanding the problem description. "Segment" is a rectangle, right?

For this testcase from a post above:

1
4 5
0 3
2 2
3 3
4 4
0 0 5 5
3 3 4 4
0 0 1 0
0 0 0 1
0 0 0 5

The output should be:

4
2
0
0
1

Right?

m.cliffsnotes.com/study_guide/Se ... 18771.html2

Thanks a lot, leppy! Neither my printed dictionary nor the google translator did offer the correct meaning.