Please, guys, don't always compare him with tourist. He is best in his own way. Here he makes tutorial videos that help many people around the world. According to me he is best and I don't care about, he wins the competition or not but he is only one who helps us, tourist not.
great gesture of helping others, it helps a lot please continue these streaming as much as possible ,like me lots of people will love to see your streams very very useful
sir please make more videos on this Type of contest problems of codeforces they teach us alot..your efforts are awesome ...and are helping the students very much...
sir please make more videos on the algorithms TOPICS and concepts and please tell us about the books our resources to boost our CP skills. Your videos are Great...
This is helping me a lot because this kind of problem almost always have simple and/or smart tricks involved. Thanks for the time invested and hardwork! =D
Codeforces Round #517 - B: Minimum path in the first part of the problem instead of DP, we can use Dijkstra`s algorithm to get the same matrix as of the DP matrix.
If you want to split A numbers into two groups, one group must have at least A/2 numbers. Here we have A=k*(k+1)/2. k is the number of remaining numbers on the left.
ok now i know that if we want the remaining space between 2 sets k*(k+1)/4 but how we make sure that there's a space left so using the first while loop we make sure that the highest k here equal to i -1 can be splitted to 2 groups am i getting it right ?
I proved why taking from right to left worked. I also said some intuition why the opposite doesn't work: when you decide about some first numbers, you are left with big numbers that might not fit in either set/bag.
2:29 Codeforces Round #517 - A: Cram Time
29:48 Codeforces Round #517 - B: Minimum path
1:30:30 Codeforces Round #124 - B: Infinite Maze
2:25:33 Technocup 2019 - Elimination Round 1 - E: Vasya and Good Sequences
3:31:56 Codeforces Round #272 (Div. 2) - D: Dreamoon and Sets
Extra:
3:07:34 Improving suboptimal solution for 517B: Infinite Maze
3:26:47 Explanation for Educational Codeforces Round 3 - E: Minimum spanning tree for each edge
Thank you a lot!
Please, guys, don't always compare him with tourist. He is best in his own way. Here he makes tutorial videos that help many people around the world. According to me he is best and I don't care about, he wins the competition or not but he is only one who helps us, tourist not.
👍
Appreciate your effort in making this video :)
Cool, I'm glad you liked it :)
I always thot D.E and F were very tough, but you actually made me believe that they are not. Thank you.
F is very tough
great gesture of helping others, it helps a lot please continue these streaming as much as possible ,like me lots of people will love to see your streams very very useful
you are a genius! I wonder how that feels. Although I know not a stitch of what you do, I find it very interesting.
sir please make more videos on this Type of contest problems of codeforces they teach us alot..your efforts are awesome ...and are helping the students very much...
sir please make more videos on the algorithms TOPICS and concepts and please tell us about the books our resources to boost our CP skills. Your
videos are Great...
Thankyou so much Erichto! Your explanations were really helpful. I hope you keep making such videos and your channel keeps growing!
This is helping me a lot because this kind of problem almost always have simple and/or smart tricks involved. Thanks for the time invested and hardwork! =D
Make more videos like this which have more explaination
Love to watch your videos Big fan errichto
thanks a lot. the video is great and very educational. please keep making as many of these as possible.
thanks for making this kind of videos, very appreciated
Codeforces Round #517 - B: Minimum path
in the first part of the problem instead of DP, we can use Dijkstra`s algorithm to get the same matrix as of the DP matrix.
YOU GET TO READ MORE NOTES IF THE NOTES TAKE LESS TIME TO READ. but u read the same amount of words...
Wow..! Its really a informative video
It's really helpful for me. Though best results at 1.25x.
very intersting stuff
WOW! Its cool maaaan
the videos are really good ,thank you
You are a god !
why at 16:00 at least one set has (k*(k+1))/4 ??
and what is k ? the number of remaining numbers from right?
If you want to split A numbers into two groups, one group must have at least A/2 numbers. Here we have A=k*(k+1)/2.
k is the number of remaining numbers on the left.
ok now i know that if we want the remaining space between 2 sets k*(k+1)/4 but how we make sure that there's a space left
so using the first while loop we make sure that the highest k here equal to i -1 can be splitted to 2 groups
am i getting it right ?
You're getting it right. And the proof was explained in the video.
@@kamildebowski7003 thanks kamil your work is really appreciated looking forward for new live streams
greetings from Egypt
for the first problem how could we prove that you need to select numbers from right to left instead of left to right?
I proved why taking from right to left worked. I also said some intuition why the opposite doesn't work: when you decide about some first numbers, you are left with big numbers that might not fit in either set/bag.
is this cutoff ?
UA-cam takes a few hours to process a stream, and then it's available in full length, including chat on the side.
okay, thank you.
its full now
ดี
Good
nostalgia
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