Contest Talk ------------------ Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)

Highlights : Did the 1st (A) question and Got AC in the first go, struggled till the end to get the 2nd question (B) right.

Analysis :

A. Arpa and the Mexican Wave The first question seemed difficult at first , so I took a lot of time thinking about all the scenarios that could be possible, and finally typed in the solution ( all in all taking 42 minutes , in contrast, the person who submitted it the first who took less than a minute :O )

And my code follows :

     int main(){

     	ll n,k,t; // Unnecessarily taken long long ( int would have done )
     	cin>>n>>k>>t;
     	if(t>(n+k))t=(t%(n+k)); // check out this UNNECESSARY STEP !! :-/

     	if(t<=k){
     		cout<<t;
     	}
     	else {
     		if(t<=n)cout<<k;
     		else cout<<k-(t-n);
     	}
     	return 0;
     }

Should have really given attention to this constraint :

1 ≤ t < n + k

All one had to do keeping the constraints in mind was this :

 int main(){

     	int n,k,t;
     	cin>>n>>k>>t;
     	cout<<min(t,min(k,n+k-t)); 
     	return 0;
     }

In retrospect, this could very well have been done in under a minute if one understood the problem statement fast enough, as the implementation is easy and short !

B. Arpa and an exam about geometry

If you've read my previous blog post , you know about my love for geometry problems and the deep rooted connection they have with Star Trek . This question is a perfect example of how I link those 2 things together .

The question talks about a rotation of the triangle about an axis, so as to find out if the previous points once rotated would move to the subsequent points or not. The first thing that I imagined was an equilateral triangle ( As I would later find out this to be a case of not reading the question thoroughly enough ). I submitted the solution right away and failed the pretest :-| . Then I went on to again read the question slowly and carefully this time. So, we only had to ensure that 2 sides were of the same length ( that the only way you could the point from one point to the next ( of 2 points ) ), and not just any 2 sided needed to be same ( the side B and C were the ones that needed to have equal length ).

This is the image that came to my mind right away :

I finally got it !! ( or so I thought ) I used the distance formula and took all the values as doubles to compare the 2 sides that mattered, got it Wrong on the pretests. (NOTE TO SELF: Avoid floating points like the PLAGUE! )

As I would later find out , almost every one who eventually got this problem right , got it wrong the first time ( because they too used floating points, but later realised that it was not needed and the length could be compared directly (in integer form ) as the square-roots of the integers could be cancelled on both sides ( and that would lead to integer numbers for comparison ) ).

I was still unable to get the problem accepted ( as I did not check for collinearity of all points ). Later I did it by using the point slope method ( which does not require floating points and gets an integer result ).

Code follow :

#define sqr(x) (x)*(x)
typedef long long ll;

   ll   length(ll  ax,ll  ay,ll  bx,ll  by){

     	return (sqr(bx-ax)+sqr(by-ay));

     }

     int main(){

     	ll  ax,ay,bx,by,cx,cy;

     	cin>>ax>>ay>>bx>>by>>cx>>cy;

     	ll  l1=length(ax,ay,bx,by);

     	ll  l3=length(bx,by,cx,cy);


     	if( (by-cy)*(bx-ax)==(bx-cx)*(by-ay)  ){
     		cout<<"No";

     	}
     	else if(l1==l3){
     		cout<<"Yes";
     	}
     	else cout<<"No";
     	return 0;
     }

....................

All questions after this remain unsolvable by me because of my lack of enough algo-ds knowledge ...

Effect on my progress graph: My graph dipped lower! :-|