Target audience: newbies and pupils (rating up to 1400).
Group link: https://codeforces.com/group/yg7WhsFsAp/contests (hit "join" on the right).
Hi. Enjoy a series of 8 problem lists for beginners. The example topics are strings, arrays, math, and binary search. You are allowed to discuss anything with others, or just look up solutions online. There are also 3 exams, each recommended for a 2-hour individual virtual participation. Use the displayed order, e.g. take exam 1 after day 3. It all should take you 2 weeks of intense bootcamp-like work (or a few months if you take your time).
The problems were originally used two years ago in a Saudi Arabia camp. It's a mix of around 70 existing CF problems and 30 new original problems, mainly prepared by kostka, with some help from me and mustafabar. I asked them for permission to publish everything.
I will put hints to some problems in this blog (or in the group? not sure). Expect a few videos and/or streams for beginners too. You should also read two first chapters of Competitive Programmer's Handbook.
UPD: On Sunday evening I'm making a stream with explanations to a few hard problems: P8, P11, P18, P30, P31, P33. You can try it with hints first:
Thanks a lot :D I'll try to solve all of them in the upcoming days
It is really usefull
Can those not in the rating range specified above not submit ?
Edit : Just checked, they can
You are allowed, but the problems should be easy and boring for you.
you mind putting some more tough questions in there .
These are problems from a camp two years ago. The participants were beginners just after learning basic C++.
Can you prepare another one for 1400 to 1600? I will be very helpful! And thanks for preparing this one. I will submit these problems.
There is no problem numbered P17 in day 2
Oh, you are right. I see now that we removed it back then because it required an algorithm that participants didn't know yet. It was 1041B - Buying a TV Set.
I've renumerated problems.
Reminds me of whenever I say I should start over competitive programming from the beginning since I am not good at the basics.
Great , it will help us a lot.
I have just finished the basics of java and solved 100 URI basic problems and some hackerrank easy problems. need to read the first two chapters of "Competitive Programmer's Handbook". it will take some time. I will try it.
I have a question. if someone wants to try it after 3/4 months will these problems will be there?
Sure. I don't plan to ever hide these problems.
When I try to register it says "you are not allowed to enter", I am new to this platform. how to register for this.
Actually it says " You are not allowed to take part in the group contest "
Come on. Read the top of the blog. Or just scroll up by a few comments.
Any hints/tips on how to solve P48 The line problem
The hint is in the original question already. It gave an unnecessary/redundant information which could have been omitted.
hey there i am trying to solve Day 2 P13 lost rectangle problem for almost a day now and ain't able to come up with a possible solution nor could find anything on the interent to help, coud anyone please help my with some possible idea . I would be realy helpful.
https://codeforces.com/group/yg7WhsFsAp/contest/355493/problem/P13
Prime is a big hint in itself.
The Area A is made by two integer (as given) sides. So, using this fact you can determine that for any integer to be the height or breadth of rectangle, It should evenly divide the A.
Just find all the prime factors of A, and then say for any factor p, we can treat it as 1 of the side of rectangle. Therefore, the other side is A/p. Now, you have both the side. Find the Perimeter using the same idea for all the factors of A, and print minimum one to the result.
You use one word incorrectly:
"prime" -> "divisor"
If the area is 100, possible shorter side lengths are 1, 2, 4, 5, 10. These are divisors, not primes.
I don't know if it is right to discuss it here, but it is related to a question in this bootcamp. do let me know if I am not supposed to ask or post anything here. I am new to this platform.
In a particular problem if I use.
for(size_t i =0; i<=sqrt(n) ; i++) //The answer gets acceptedis there any difference(in performance)? if yes then an explanation will be helpful.
i * ioverflows. You need to use(long long) i * i.someone help me solving [problem:P48]
Please help in p54 https://codeforces.com/group/yg7WhsFsAp/contest/355504/problem/P54. Why XOR works??
XOR of two equal numbers is 0. So also XOR of any even number of equal numbers is 0.
Since someone necro'd this, I request 100 hard problems bootcamp!
Great
Hello, I am trying to solve the Lost Rectangle problem. In my first try, I used normal way to find out the factors which led me to failure because of time limit.
Later I tried to implement sieve for prime factors, and again I got stuck for the 7th test with the input 1000000000000. I took integer which will not support this long. But when I tried to use long long I started getting memeroy stack issues. Also, I am facing the array size limit here as well.
Here is my last submission code: https://github.com/ahsanaasim/problem-solving/blob/main/codeforces/355493/P13-Lost%20rectangle/main.cpp
Is my approach entirely wrong? Or is there a way to solve it the way I am trying
You don't need sieve. You can iterate over all factors in $$$O(\sqrt{A})$$$.
https://codeforces.com/group/yg7WhsFsAp/contest/355493/submission/200610343
https://codeforces.com/group/yg7WhsFsAp/contest/355496/problem/P41 any help with this problem , please;
can anyone please help me with this problem : (https://codeforces.com/group/yg7WhsFsAp/contest/419146/problem/P23)
I saw this interesting idea:
I think the natural instinct is to do the problem as in the statement. Find divisors of 1, then divisors of 2, divisors of 3, etc. up until n.
However, what if we did it the other way around? In other words, we found the number of numbers in (1, n) were divisible by 1, then the number that were divisible by 2, then divisible by 3, etc.
link of the solution stream??
[problem:P19] Can someone please explain this problem for me.