Dijkstra.'s blog

By Dijkstra., history, 8 years ago, In English

i have been on cf for about 7 weeks most of the problems i solve are problems A division 2 and some times i cant solve them now i am at a stage where i am feeling that i cant solve any problem can any one give me some tips to improve myself as i dont know what to do and sorry for poor english

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8 years ago, # |
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8 years ago, # |
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C'mon man! It's only been 7 weeks. Take a look at my graph in comparison to yours!!

If you say you feel you can never improve, take me at least as an example! I have been coding for 6 months roughly. You just have to motivate yourself. As for improving, I would say A2OJ Ladders are a good start. I have improved a LOT because of it. Last thing would be to keep believing in yourself and keep practicing and participating. I don't know if I am eligible to give someone advice but this would still be my say.

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    8 years ago, # ^ |
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    How to merge graphs of two users?

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    8 years ago, # ^ |
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    Which specific ladder in A2OJ...The ones divided on basis of ratings or the ones divided on basis of division(Div2A,Div2B....)?

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      8 years ago, # ^ |
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      As long as you are solving problems where you need to think for some time and as long as you are not skipping problems that you can't solve it really doesn't matter. It's not like one of those ladders is magic or something. You just need to solve problems that are hard for you.

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      8 years ago, # ^ |
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      I normally do the ones divided on the basis of division. I have problems with solving C but I can solve almost any Div2 A and B, so I work for C right now. I did the same thing for practicing with A and B too.

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8 years ago, # |
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I have only three advices: work,work and work!

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7 years ago, # |
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There is no shortest path to succes in CP , Dijkstra.. You have to practice a lot :)

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    7 years ago, # ^ |
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    There is a shortest path to success in CP. Just that it is NP-hard to compute. So the graph is just not suitable for applying Kidd's O((V+E)log(V)) single-source shortest path algorithm.