maroonrk's blog

AtCoder Regular Contest 145 Announcement
By maroonrk, history, 2 years ago, In English

We will hold AtCoder Regular Contest 145.

The point values will be 400-500-600-700-800-1200.

We are looking forward to your participation!

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2 years ago, # |
  Vote: I like it +32 Vote: I do not like it

Looking forward to ARC !!

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2 years ago, # |
  Vote: I like it +151 Vote: I do not like it

As the writer, good luck and enjoy yourself! :D

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2 years ago, # |
  Vote: I like it +1 Vote: I do not like it

What's the approximate rating of ARC's problems B & C according to CodeForces rating distribution?

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2 years ago, # |
  Vote: I like it +19 Vote: I do not like it

math and counting round

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2 years ago, # |
  Vote: I like it +6 Vote: I do not like it

I really suck at counting, but today I couldn't figure out what the 16 different permutations were in C, the most I found was 12 :P

Can anyone list those out please?

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    2 years ago, # ^ |
      Vote: I like it +22 Vote: I do not like it 
    1 2 3 4
2 1 3 4
1 2 4 3
2 1 4 3
3 4 1 2
4 3 1 2
3 4 2 1
4 3 2 1
1 3 2 4
2 3 1 4
1 4 2 3
2 4 1 3
3 1 4 2
4 1 3 2
4 2 3 1
3 2 4 1
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    2 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it
    Code that prints
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2 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Could someone point out what case am I missing in my submission for B?

Submission : https://atcoder.jp/contests/arc145/submissions/33632229

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

My first AtCoder contest. The problems were good, it was interesting for me. Thanks !

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2 years ago, # |
  Vote: I like it +65 Vote: I do not like it

D is just cantor set...

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2 years ago, # |
  Vote: I like it +10 Vote: I do not like it

As a contestant, I was not a big fan of A,B,D. Those problems' ideas are not hard to get, but have very annoying corner cases...

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem D is magic, but I want to know, why the set $$${1,2,4,7,11,...}$$$, which the differential sequence is $$${1,2,3,4,...}$$$ is not the optimal one.

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    2 years ago, # ^ |
      Vote: I like it +19 Vote: I do not like it

    The set $$$1,2,4,7,11,\dots$$$ has an $$$\mathrm{O}(n^2)$$$ value on the $$$n$$$-th element, while the optimal set has $$$\mathrm{O}(n^{\log_2 3}) \approx \mathrm{O}(n^{1.585})$$$ value on the $$$n$$$-th one ,which is enough to meet the constraint of $$$[-10^7,10^7]$$$ .

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      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      U R right.I tried that,and the number can be as large as 4e7

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    2 years ago, # ^ |
      Vote: I like it +30 Vote: I do not like it

    That doesn't even work, $$$1+7=2\cdot 4$$$

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2 years ago, # |
  Vote: I like it +16 Vote: I do not like it

Sequence of pC can be found here.

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2 years ago, # |
  Vote: I like it +24 Vote: I do not like it

My friend told me the problem C is in the OEIS, what do you think of it?

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

void jaiishriRam(){ ll n; cin>>n; string s; cin>>s; if(n==2) { if(s=="BA" || s=="AB") { cout<<"NO"<<endl; return; } } if(s[0]=='A' && s[n-1]=='B') { cout<<"NO"<<endl; return; }

cout<<"YES"<<endl;

} could anyone help me to know what thing I did wrong in A above is the code

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    2 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    You have to print "Yes", not "YES". Same for "No" and "NO"

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2 years ago, # |
  Vote: I like it +16 Vote: I do not like it

Can anyone plz tell me whats wrong here in my submission for problem B

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    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    Consider the test case :- 6 5 4 ,
    Your O/P — 4
    Correct O/P — 2
    for value (5 and 6 only Alice will win )

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      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Corrected that mistake but still getting WA on 3 test cases ;-; Here is my submission

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        2 years ago, # ^ |
        ← Rev. 2   Vote: I like it +1 Vote: I do not like it

        I'm not sure if binary search is a good idea here. You can try 30 10 3 correct ans: 7 your code gives: 9

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        2 years ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        Consider this test case :- 8 4 3
        Your O/P — 6
        Correct O/P — 4
        Aice wins game (4,5,6 and 8)
        Error is in line 58 loop that you are multiplying . They will depend on remainder of (n%a).

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    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    your code doesn't handle the cases for $$$n$$$ in the range $$$a \le n < a+b$$$

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2 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Can anyone please tell me whats wrong here in my submission for problem D?

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    2 years ago, # ^ |
      Vote: I like it +16 Vote: I do not like it

    I don't know how your code works, but it fails at the input 4 2

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      2 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      thx.

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      2 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      It works in such way:

      $$$\pi(n)\sim \frac n{\ln n}$$$

      So the dfs is expected to stop in very low depth.

      p.s. What about this one? It could pass 4 2.

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        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        It fails at 10000 0.

        The output has these three integers:

        -9999991 -9999937 -9999883
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          2 years ago, # ^ |
            Vote: I like it +8 Vote: I do not like it

          Sorry, but it got 9999991 -9999991 9999973 -9999973 9999971 ... here.

          Maybe there's some Undefined Behavior in my code?

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            2 years ago, # ^ |
              Vote: I like it +8 Vote: I do not like it

            It is not a UB, because the output is a set, and these three numbers are in it, which is against the constraint.

            Actually, they are not the first three numbers of the output on my computer as well.

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        2 years ago, # ^ |
          Vote: I like it +27 Vote: I do not like it

        Are you assuming primes don't have arithmetic sequences of length 3? Because that is not true, for example: $$$101, 107, 113$$$.

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          2 years ago, # ^ |
            Vote: I like it +8 Vote: I do not like it

          Oh, that's it.

          Sincerely thanks for your help!

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2 years ago, # |
  Vote: I like it +40 Vote: I do not like it

C could be solved by writing a brute force and searching the sequence on OEIS. Maybe those problems should be avoided by for example having problems parametrized by some more values, or making sure it isn't easy to find on the internet. It makes the contest experience less fun, if the answer is easily found on the internet. Problems A and B were fine. Problem D also suffered from this issue.

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2 years ago, # |
  Vote: I like it +18 Vote: I do not like it

C is actually A152029(OEIS).

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2 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

Why does this code give WA for problem A?

lng dp[3][SZ],n;string s;
lng solve(lng i,lng x)
{
 if(i==n-i){return 1;}
 if(i>n-i){return (x==0);}
 if(dp[x][i]!=-1){return dp[x][i];}
 lng j=(n-1)-i;
 char a=s[i],b=s[j];
 if(x==1){a='B';}
 if(x==2){b='A';}
 if(a==b)
 {
  lng z=solve(i+1,0);
  if(a=='A'){z=max(solve(i+1,1),z);}
  else{z=max(solve(i+1,2),z);}
  return dp[x][i]=z;
 }
 if(j-i==1){return 0;}
 if(a=='A'){return 0;}
 return dp[x][i]=max(solve(i+1,1),solve(i+1,2));
}
int main()
{
 ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
 cin >> n >> s;
 for(lng i=0;i<3;i++){for(lng j=0;j<SZ;j++){dp[i][j]=-1;}}
 if(solve(0,0)==1){cout << "Yes";rtr;}
 cout << "No";rtr;
 return 0;
}

Here lng=long long, rtr=return 0, SZ=2e5+7

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

This was my first contest on AtCoder, Can someone tell me, when do the editorials be out ?

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2 years ago, # |
  Vote: I like it +8 Vote: I do not like it

nok0 My submission for A get AC but can be hacked with n=4 AAAA My submission

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2 years ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

In problem A:

Why is "AB" a valid palindrome or can be changed to a palindrome according to editorial and tests?

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    AB doesn't satisfy the first condition (s[0] == "B" or s[-1] == "A") in the editorial.

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2 years ago, # |
  Vote: I like it +8 Vote: I do not like it

for D you can just use brute force to get a set meet the last restriction,and change the largest/smallest number,then add x to all numbers to get sum=M

base 3 solution is beautiful but not necessary

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2 years ago, # |
  Vote: I like it +117 Vote: I do not like it

I have different approaches for problem D and E:

D:

Ingore the limit of sum for now. Let $$$f(n)$$$ be the set of integers, one can find that

$$$ f(1)=\{0\} \\ f(2n)= f(n) \cup (f(n)+2\max f(n)+ {\color{red}1}) $$$

is good. Now consider the limit of sum. One can find that if we change the $$$\color{red}1$$$ into $$$2$$$, and add $$$1$$$ to some greatest elements in $$$f(n)$$$, it is still good.

E:

We can change $$$A_j$$$ to $$$A_i\oplus A_j$$$ ($$$i<j$$$) by performing operations $$$j,i+2,i+3,\ldots,j$$$ in order.

Now we change $$$A_i$$$ into $$$B_i$$$ from right to left, taking the lexicological largest basis each time. It turns out that the number of operations is OK. (It can take up to $$$N-1$$$ operations to change a single number, but each operation will move some basis vectors to the right)

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    2 years ago, # ^ |
      Vote: I like it +100 Vote: I do not like it

    Your code on E submitted during contest can be hacked using the data generated by the code below.

    #include<bits/stdc++.h>
#define rep(i,a,n) for (int i=(a);i<=(n);i++)

int main() {
	int n = 1000;
	printf("%d\n", n);
	rep(i, 0, 59) printf("%lld ", 1ll << i);
	rep(i, 1, n - 60) printf("%lld ", (1ll << 60) - 1);
	printf("\n");
	rep(i, 0, 59) printf("%lld ", 1ll << i);
	rep(i, 1, n - 60) printf("%lld ", (1ll << 60) - 1 - (i & 1));
	return 0;
}
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      2 years ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      Thanks for pointing this out! I tried to add some randomization and it worked pretty good.

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2 years ago, # |
← Rev. 2   Vote: I like it +39 Vote: I do not like it

My randomized solution for E:

If the answer is Yes, we first randomly manipulate the sequence for a small number of times, and then operate on index $$$n$$$ more than $$$n$$$ times repeatedly. Then we can (roughly) assume that $$$a_n \oplus b_n$$$ is a linear combination of $$$a_{n-r+1 \cdots a_{n-1}}$$$, where $$$t$$$ is about $$$60$$$.

We can then solve $$$a_n \oplus b_n = \bigoplus_{j \in S} a_j$$$ with Gaussian elimination. If $$$a_{n-1} \not \in S$$$, operate on $$$n$$$ so that $$$a_{n-1} \in S$$$.

Then we operate on the second minimum element $$$t$$$ in $$$S$$$. Then $$$\min S$$$ is removed from $$$S$$$ and all indices between $$$\min S$$$ and $$$t$$$ are added. Repeat this process until $$$S = {a_{n-1}}$$$ and operate on $$$n$$$. Do this for $$$n - 1 \dots 2$$$ and we are done. This solution uses ~60000 operations in randomly generated inputs.

implementation

UPD: More than 1000 random operations at first are needed to guarantee randomness.

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can somebody explain here problem D, how exactly are we constructing the required set after we have figured out the base 3 condition where each digit should be 0/1.

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2 years ago, # |
  Vote: I like it +10 Vote: I do not like it

F is some wild generalization of IMO 1995 P6. I like it

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2 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Such a magical problem D!

Base-3 representation is such a trick that could perfectly beat the rule of $$$2y \neq x + z$$$.

Next, by first keeping the rightmost digit of all n integers as zero and then change some of them to 1, so that $$$|m - sum| \% n = 0$$$. At the same time, all n integers are still distinct and satisfy the "good set" rule.

Finally, add $$$(m - sum) / n$$$ to each integer so that $$$sum = m$$$, and at the same time, adding the same integer will not lead to any $$$2y = x + z$$$, although they may not belong to the "good set" anymore.

Thanks to the problem writer for coming up with such a clever and unbelievable construction problem!

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

This round is good, arc should contain magical counting and constructing problems like this.

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2 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

A different way to think for D.

We can construct as follow (P is a set,initial $$$P={1},len=1$$$): $$$P\to P\cup (P+(2len-1)),len\to 3len-1$$$

$$$P+a$$$ means add $$$a$$$ to all elements in $$$P$$$.

And it can prove that $$$y−x\not=z−y$$$ for every triple $$$x,y,z (x<y<z)$$$ of distinct elements in $$$P$$$.

(Which is similar to the official editorial)

and we can construct the set with $$$n+2$$$ elements, and enumerate which two to delete. It can solve the problem.

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Could you please talk more about why your construction will not lead to 2y=x+z? And, how to construct these n+2 elements, and how to determine which two to delete?

    Thank you so much for providing such a clever idea.

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      2 years ago, # ^ |
      ← Rev. 3   Vote: I like it 0 Vote: I do not like it

      Let $$$P'=P+(2len-1),\forall x\in P,y\in P,z\in P',$$$ we can found that $$$|x-y|<len,|x-z|\ge len$$$.

      This is my submission for D.

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2 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Sorry for asking it late,

Can any one please the explain this line of editorial :

Here, if the elements in P are scanned from left to right, there must always be more left elements than right elements. The number of such arrangements corresponds to that of parenthesis sequences.

What I understood is that, lets say we have (1,2),(3,4),(5,6),(7,8) as corresponding (ai and bi).
Permutation : (1,2,4,3,5,6,7,8) have more right elements(2) than left elements(1) in first 3 elements (1,2,4) but still it can be divided it into two sequences A(1,3,5,7) and B(2,4,6,8), but in editorial it is told that it should not be possible ( as per I understood ) .
I know I have understood it wrong.
Anyone who did this problem or got the editorial please explain the line in blue

Editorial link : Link

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

sorry for asking late. i'm confused about problem C on how/why the legal arrangement corresponds to parenthesis sequences. (that seems strange to me)

i know we are arranging some pairs like (1,2)(3,4)... , but in my opinion and based on some experiments and some comments like this. i claim an arrangement is legal if and only if there is no pair contains another pair.

for example, (1,2,3,4) legal because pair(1,2) and (3,4) doesn't contains the other. (1,3,2,4) is legal because two pair intersect not contain. (1,3,4,2) not legal because of containing.

however, if you see the pair as parentheses, (1,3,4,2) is legal because they can be mapped to "(())". that's a contradiction (do i misunderstand something?)

however again, when we consider the number of it, these two somehow are equal. here i take some example on n=3 to illustrate.

you see, if we take the view of parentheses, their correspondence(left parentheses and right parentheses) is really strange (at least differ from matching them greedily)

in general, i'm not negate the editorial, but i think i need some more clear intuition to understand this transform

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