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I am so desperate that I risk no longer doing competitive programming because of my rating which didn’t even reach pupil after having solved 650 problems with different difficulties that qualify me to reach a better level.
i feel like i have lost 2 years…
any solutions ??
When you're down, the only way is up :)
I don't know why. The difficulty of the problems you've solved seems really good for a newbie, if you want you can DM me for doubts or doing some virtual contest, don't give up!
i think your solutions are too complicated, to solve easy problems fast, you need to think a simple solution, if the problem has low constraints sometime brute force works, othewise its greedy, prove it or guess that it works and then code. remember that if 5000 people solved the problem in 5 minutes then it must be very easy
let's see what you did in https://codeforces.net/contest/1620/problem/A that you struggled with
string s; cin>>s; int nb=0; if(s.size()==2){ if(s[0]!=s[1]) cout<<"NO"<<endl; else cout<<"YES"<<endl; } else { if(s[0]=='E'&&s[s.size()-1]=='E'){ for(int i=1;i<s.size()-1;i++) if(s[i]=='N') nb++; if(nb!=1) cout<<"YES"<<endl; else cout<<"NO"<<endl; } else{ for(int i=1;i<s.size()-1;i++) if(s[i]=='N') nb++; if(s[0]=='N'&&s[s.size()-1]=='N') cout<<"YES"<<endl; else if(nb>=1) cout<<"YES"<<endl; else cout<<"NO"<<endl; }my solution:
You can always put an equal or different number, unless the whole circle has equal numbers and there is a segment with different numbers, which is impossible and happens when there is only one "N"
in educational round, solutions for problem A should be very simple
string s; cin>>s; int c1=0; for(int i=0;i<s.size();i++){ if(s[i]=='N'){ c1++; } } if(c1==1){ cout<<"NO"<<"\n"; }else{ cout<<"YES"<<"\n"; } return;Your solution for problem C in last contest was extremely overcomplicated, and you should never use unordered_map or unordered_set unless you are LGM or something, because it isn't deterministic and probably that why it was hacked. also it is useless in div2 because the fastest problems can be solved in o(n log(n)), even reading the input is o(n log(n))
My solution was: sort the input in decreasing order and check numbers from n to 1 keeping a visited array, i used two loops in O(n^2) since the constraint is just 50. I thought of using bitsets at first but since it was div3 problem C, i did that and it worked
I recommend you to read https://usaco.guide/
While I agree that in some situations you need to be careful with them to avoid possible collisions or nondeterministic iteration order, waiting until you're an LGM to use unordered sets and maps seems unnecessary, given how useful they are. Also, in this case, the hack resulted in a WA, so hash collisions aren't the culprit.
Write more rounds. I spent 1.5 years to get a specialist
The graph is beautiful! What's it drawn by?
Your text to link here...
Thanks~
well, if you do feel so, then I would like to stress that people change, and so do their hobbies and interests, and it is fine if you no longer want to do something / take a break from something.
It doesn't mean that you tried something, failed at it, and had to quit. It can also mean that you tried something, stopped enjoying it, and moved on to something better. Good luck!
Honestly, I don't know why the number of solved problem matters. It all because of our strategy in practicing.
And you must get out of the safe zone.
Don't be sad. Look at my rating, it is about 300 lower than yours.
Solving problems in practice is very good. But try to give virtual contests more n more. Also when u r unable to solve a problem, see its editorial but don't code that, code it after 3-4 days so that u yourself are able to apply the logic.
Recently, there are several blogs talking about this situation, and I strongly suggest you read them. One of them is this one, http://codeforces.net/blog/entry/98806, and hope that you could get some inspiration there. As for me, I practice by participating virtual contests from round 1, and when I feel it is time to check whether I have got improved or not, I would like to take a real contest then.