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AtCoder Beginner Contest 363 Announcement
By atcoder_official, history, 21 month(s) ago, In English

We will hold AtCoder Beginner Contest 363.

We are looking forward to your participation!

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21 month(s) ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

GL & HF!

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21 month(s) ago, hide # |
 
Vote: I like it +22 Vote: I do not like it

I have a rating of 1598 and I'm sure I'm going to reach 2 Kyu this time.

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21 month(s) ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

I hope I can get a good grade.

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21 month(s) ago, hide # |
 
Vote: I like it +17 Vote: I do not like it

675 pts G! Wow!

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21 month(s) ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Good Luck!

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21 month(s) ago, hide # |
 
Vote: I like it +43 Vote: I do not like it

Palindromeforces

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21 month(s) ago, hide # |
 
Vote: I like it +5 Vote: I do not like it

Felt E was easier than D, spent too much time in C thinking generating all permutations would TLE :/

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21 month(s) ago, hide # |
 
Vote: I like it +30 Vote: I do not like it

First time among the top 5 on an Atcoder Contest! I just can't express the thrill I'm feeling when passing G!

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21 month(s) ago, hide # |
 
Vote: I like it +7 Vote: I do not like it

Am I just a scrub, or is Python doomed to TLE for problem C?

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21 month(s) ago, hide # |
← Rev. 2  
Vote: I like it -7 Vote: I do not like it

You are right, but G is an easier version of this problem (that this problem is on a tree). I think many Chinese participants have seen this problem. (Though few people solved it) Atcoder, it's a good job and don't set problems like this again.

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    21 month(s) ago, hide # ^ |
     
    Vote: I like it +20 Vote: I do not like it

    Disagree with this opinion: don't set problems like this again.

    This problem('s statement) is much more beautiful than that (even if i am a Chinese, i still think the problem you mentioned is too complex for me to appreciate).

    And most of all, this is an atcoder BEGINNER contest (I do agree with that on a ARC or AGC original-idea-problem is important), reusing some beautiful ideas should be encouraged: If evan ABC could not introduce this one, how people around the world could know it.

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

getting 1 testcase wrong in D please help to make correction

submission:code

thanks

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21 month(s) ago, hide # |
← Rev. 2  
Vote: I like it 0 Vote: I do not like it

I solved ABCE

my approach for C is as follows

  • try all the permutations using next_permutation
  • find the palindromes' sizes using manacher's algorithm for each string in the permutations
  • find if there are any sub-palindrome of size K or no, if you didn't find one, increase the result

are there any easier solutions ?

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Why wrong D. I've checked it for one hour.

include<bits/stdc++.h>

using namespace std; typedef long long ll; ll n,cnt,l; int main(){ ios::sync_with_stdio(false); cin.tie(0),cout.tie(0); cin>>n; while(true){ l++; if(l==1){ if(n<=10){cout<<n-1; return 0;} cnt=10; }else{ ll dgt=(l+1)/2; ll sum=9*pow(10,dgt-1); if(cnt+sum<n) cnt+=sum; else{ ll nd=n-cnt; nd--; ll to=nd+pow(10,dgt-1); nd=to; string s=""; for(ll i=1;i<=dgt;i++){ if(nd%10==0) s="0"+s; else if(nd%10==1) s="1"+s; else if(nd%10==2) s="2"+s; else if(nd%10==3) s="3"+s; else if(nd%10==4) s="4"+s; else if(nd%10==5) s="5"+s; else if(nd%10==6) s="6"+s; else if(nd%10==7) s="7"+s; else if(nd%10==8) s="8"+s; else if(nd%10==9) s="9"+s; nd/=10; } cout<<s; if(l%2==1) for(ll i=s.size()-2;i>=0;i--) cout<<s[i]; else for(ll i=s.size()-1;i>=0;i--) cout<<s[i]; return 0; } } } return 0; }

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    21 month(s) ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    Put your code like this:

    ``` cpp
#include<bits/stdc++.h>
(more codes)
```

    or this:

    ~~~~~
#include<bits/sdtc++.h>
~~~~~

    When you are using the wave line version, please make sure that there is at least one blanket line before and after it. No one likes reading a code like this.

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Usually, how long does AtCoder take to updating ratings? New user here

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    21 month(s) ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    I bet It takes about 1-2 hours.

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      21 month(s) ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      You're right, that's fast.

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        21 month(s) ago, hide # ^ |
         
        Vote: I like it 0 Vote: I do not like it

        I like Codeforces and Atcoder very much,because the ratings is updated very fast.But Luogu is very famous for its slow calculation of ratings.

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          21 month(s) ago, hide # ^ |
           
          Vote: I like it 0 Vote: I do not like it

          However, the rating system of Luogu is still relatively new. But I agree that it's truly slow.

          As we can see, Luogu is not specifically designed for contests. I think I don't have any objections if the rating of this round is updated just before the next Rated round.

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

How to solve F?

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    21 month(s) ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    Probably there was an easier solution, but I wrote a dp(i, val) = can we have a string of length i with product = val? The transition is just to try using a divisor and it 'reverses' when you have (val / divisor) % reverse(divisor) == 0

    In my code, I also have to fix the middle element before calling the dp

    You can check out my code here

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

For problen F, I almost have the same idea as the editorial, but couldn't figure out the complexity, and thus missed the dfs+memorizarion method.

Learned a lot from this problem, thank you so much, atcoder team!

By the way, I think the complexity could be reduced from O(sqrt(N) * number of divisors of N), to O(number of divisors of N * number of divisors of N), because it is not necessary to check every integer from 2 to sqrt(n) when calling dfs. Instead, we can just find all the divisors of n at first, and then check these divisors every time when dfs is called.

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

How to solve D ?

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    21 month(s) ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    dont think about palindrome just you have any number let say 235 then you can make two type of palindrome number from this 235532 and 23532 just this for every number

    code:code

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

https://atcoder.jp/contests/abc363/submissions/55829821 where i am doing wrong pls help problem f

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    21 month(s) ago, hide # ^ |
    ← Rev. 2  
    Vote: I like it 0 Vote: I do not like it

    Same as mine (earlier submission).

    Of course, in the samples, $$$363=11\times3\times11$$$, but that's a misdirection. $$$363=363$$$ is also correct. The problem didn't force you to use asterisks.

    I found that problem during contest, and my friends have found this typical after contests: $$$114411$$$, it's just $$$114411$$$ and can't divide anymore (in the condition of the problem), but $$$114411$$$ itself is a possible string. Your program prints $$$-1$$$.

    Most of the F (WA*8) could be hacked on this number: $$$114411$$$, the correct output is $$$114411$$$.

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21 month(s) ago, hide # |
 
Vote: I like it +5 Vote: I do not like it

is D solvable by digit dp ?

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21 month(s) ago, hide # |
← Rev. 2  
Vote: I like it 0 Vote: I do not like it

[NEVER MIND]

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21 month(s) ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Problem E, I am getting runtime error on just 1 testcase on this submission. Can anyone figure out the bug?

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