# STRTLN Editorial

**URL:** <https://discuss.codechef.com/t/strtln-editorial/51476>\
**Category:** editorial\
**Tags:** help, coordinate, editorial, maths, standard-template-library\
**Created:** [January 27, 2020, 7:08am UTC](https://discuss.codechef.com/t/strtln-editorial/51476 "2020-01-27T07:08:14Z")\
**Posts on this page:** 1\
**Page:** 1

<div class="post-metadata">

**Author:** ![xenon\_shivam](https://discuss.codechef.com/user_avatar/discuss.codechef.com/xenon_shivam/32/10089_2.png) [@xenon\_shivam](https://discuss.codechef.com/u/xenon_shivam)\
**Post date:** [January 27, 2020, 7:08am UTC](https://discuss.codechef.com/t/strtln-editorial/51476/1 "2020-01-27T07:08:14Z")

</div>

# PROBLEM LINK:

[Practice](https://www.codechef.com/problems/STRTLN)  
[Contest](https://www.codechef.com/AABH2020/problems/STRTLN)

# SETTER:

[xenon\_shivam](https://www.codechef.com/users/xenon_shivam)

# TESTER:

[subash23](https://www.codechef.com/users/subash23)

# DIFFICULTY:

EASY

# PREREQUISITES:

Math, STL

# PROBLEM:

Given N points find how many straight line we can make such that non of them will neither parallel to X-axis nor parallel to Y-axis.

# QUICK EXPLANATION:

As if n points are given so number of straight lines we can form is \binom {n}{2}.  
Now we have to remove those lines which are parallel to X- axis or Y-axis.

# EXPLANATION:

Now removing the lines which are parallel to X- axis or Y-axis, we just have to count the number of points having X-coordinates same and number of points having Y-coordinates same (of course independently as all points are unique). Now just subtract the lines which can be formed using these X-coordinates and Y-coordinates by applying the same formula \binom {x}{2} and \binom {y}{2}..

_Note:_ Overlapping Straight lines are counted differently as they are formed using different set of (x,y) co-ordinates, since, we have asked about straight lines and not about the straight lines equation.

# SOLUTIONS:

> **Setter's Solution**
>
> #include\<bits/stdc++.h\>  
> using namespace std;  
> #define ll long long int
> 
> int main(){  
> ll t;t=1;  
> cin\>\>t;  
> while(t–){  
> ll n;  
> cin\>\>n;  
> map\<ll,ll\>mpx,mpy;  
> for(int i=0;i\<n;i++){  
> ll x,y;  
> cin\>\>x\>\>y;  
> mpx++;  
> mpy[y]++;  
> }  
> ll ans=(n\*(n-1))/2;  
> for(auto i:mpx)  
> ans-=(i.second\*(i.second-1))/2;
> 
> ```
> for(auto i:mpy)
> ans-=(i.second*(i.second-1))/2;
>     
> cout<<ans<<"\n";
> }
> 
> ```
> 
> return 0;  
> }