atcoder_official's blog

By atcoder_official, history, 2 years ago, In English

We will hold AtCoder Beginner Contest 332.

We are looking forward to your participation!

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2 years ago, hide # |
 
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E is worth 525 points!

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2 years ago, hide # |
 
Vote: I like it +11 Vote: I do not like it

gl & hf! (though unrated)

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2 years ago, hide # |
 
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hope you guys get uogrades and good luck ! :)

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2 years ago, hide # |
 
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glhf!

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2 years ago, hide # |
 
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Can indians participate in Atcoder contests?

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Good luck and have fun!

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2 years ago, hide # |
 
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Was it intended, that the main difficulty in problem $$$E$$$ is not to lose precision?

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2 years ago, hide # |
 
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can anybody tell why this backtracking logic for d is not working,

code
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    2 years ago, hide # ^ |
     
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    My guess is that when you marked a state as visited and would not visit it again, since you're using DFS, the first time a state is visited may not be the best answer.

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2 years ago, hide # |
 
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Looks like yet another segtree problem F.

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imho best ABC contest in a long time, the idea for EF more educational than the usual easy problems, thanks a lot :)

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2 years ago, hide # |
 
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Can anyone tell why this submission is getting WA? The logic should be correct after all.

Submission link

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2 years ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

How to avoid precision problem in E, I spent all my time to implement it but still get WA. :(

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2 years ago, hide # |
 
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On problem E, what makes

cout << (dp.back() * m - sum.back() * sum.back()) / m / m << "\n";

correct while

cout << dp.back() / m - sum.back() * sum.back() / m / m << "\n";

is wrong?

Submission link: AC, WA

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can anyone help me understand the problem D , i didn't got any clue regarding how to solve that , during the contest

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Again :(

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Wrong on three cases on problem F. Can someone find an error?

Submission

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2 years ago, hide # |
Rev. 4  
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In problem E, I believe my implementation has issues.I would like to understand where I made a mistake.

my code with memoization updated code

still showing WA on 6 testcases :(

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2 years ago, hide # |
Rev. 2  
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What is the purpose of problem E?

If I use this

long double ans=(long double)((__int128)D*dp[(1<<n)-1][D]-ave*ave)/(D*D);

That's OK.

But this

long double ans=(long double)dp[(1<<n)-1][D]/D-(long double)ave*ave/(D*D);

is wrong.

I know precision is important but this is too ...

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2 years ago, hide # |
Rev. 2  
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Problem E, I don't understand why the program that uses fewer double get 5 WAs but the program that uses more double get AC?

In the first program, because:

$$$ \begin{aligned} &\frac{1}{D}\sum_{i=1}^{D}\left(x_{i} - \bar x\right)^2 \\ = &\frac{1}{D}\sum_{i=1}^{D} x_{i}^2 + \bar x^2 - \frac{1}{D}\sum_{i=1}^{D} 2x_{i}\bar x \\ = &\color{red}{\frac{1}{D}\sum_{i=1}^{D} x_{i}^2} + \color{blue}{\bar x^2 - \frac{2 \bar x}{D}\sum_{i=1}^{N} w_{i}} \\ \end{aligned} $$$

I use a dp to calculate the minimal value of the red part. As all the x are integer, and I can avoid using double in dp, I believe it will be more precise. The second problem simply sums $$$\left(x_{i} - \bar x\right)^2$$$ and uses long double in dp. Why the second program AC but the first not?

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    2 years ago, hide # ^ |
    Rev. 2  
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    The two lines

    auto xbar = (long double)s[all] / d;
    long double ans = (f[all] + xbar * xbar * d - 2 * sa * xbar) / d;
    

    lost even more precision than dp with double. ($$$5$$$ double multiplications' precision loss in the worst case)

    Try to change the code to

    long double ans = (long double)1.0 * (d * f[all] - s[all] * s[all]) / (d * d);
    

    It will pass :)

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      2 years ago, hide # ^ |
      Rev. 2  
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      Thanks!

      I get it now. The first program uses a value many times larger than the answer, and the larger the value, the worse the precision. But in the second program, the value used in dp is always smaller than the answer, so there doesn't seem to be much precision lost.

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2 years ago, hide # |
 
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why E not output "variance * D * D" as the answer? just to edu us how to "avoid precision problem"?

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std:1224489579591846.408203

wolframalpha:1224489579591846.408163

So why does the question ask for an accuracy of 1e-6?

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Sorry, but my simple brute force (just do dfs) passed E... Submission.

Can anyone hack it?

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In problem D. How the number of operations required to obtain a desired sequence A(A1, A2, ... , An) starting from the sequence (1, 2 .... N) by repeatedly swapping adjacent two elements is given as the inversion number of the permutation A. Can anyone explain that!?

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2 years ago, hide # |
 
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Hi hope you are well!

Could you please check my submissions for problem E, submissions 48412388 and 48412226 should give the same answer mathematically. It does seem like this problem would be better if it requires an output in p/q format rather (maybe mod 1e9+7 or something) rather than output as a fixed point number. Thanks!

Regards Yuhao

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In problem F, how to do the replacement which is mentioned in editorial with lazy segtree.

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    2 years ago, hide # ^ |
    Rev. 3  
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    Based on the editorial, each update that happens on an interval changes each of its values from $$$x$$$ to $$$\alpha x + \beta$$$ for some $$$\alpha,\beta$$$ related to the update.

    Two different updates described by $$$(\alpha_{\ell},\beta_{\ell}),(\alpha_{k},\beta_{k})$$$ can be merge to $$$\left(\alpha_{k}\alpha_{\ell},\alpha_{k}\beta_{\ell}+\beta_{k}\right)$$$ because

    $$$ \alpha_{k}\left( \alpha_{\ell}x+\beta_{\ell} \right) +\beta_{k} = \alpha_{k}\alpha_{\ell}x + \alpha_{k}\beta_{\ell}+\beta_{k}. $$$
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Can anyone suggest any corner cases for this solution. I am confused.Submission Link

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There's a $$$O(lgN)$$$ solution for problem B:

let $$$x = GCD(G, M)$$$, $$$ G=ax; M=bx $$$, so there's a cycle with $$$2a+2b-1$$$ operations. After each cycle, both bottles become empty again. So first we mod $$$K$$$ with $$$2a+2b-1$$$ to make $$$0\lt K\lt 2a+2b-1$$$.

Because bottles only become both empty after each cycle, so in cycle it must formed like this:

  • operation type 1/2
  • operation type 3

So in each cycle, each operation with odd index is type 1/2 and each operation with even index is type 3.

If we know the total amount of two bottles, we can know how much in each before/after operation type 3:

  • Before operation type 3, mug holds as more water as possible.
  • After operation type 3, glass holds as more water as possible.

Now we should work out total amount of water.

We can see:

  • If total amount less than G, next operation should be type 2(fill water)
  • If total amount greator or equal than G, next operation should be type 1(empty water)

If there's i operations of type 2 and j operations of type 1, we got current water amount is $$$iM-jG$$$ . And we can find out:

$$$ 0\le iM - (k-i)G \lt G+M \\ 0\le i(G+M)-kG \lt G+M \\ kG \le i(G+M) \lt (k+1)G+M \\ i = \lceil\frac{kG}{G+M}\rceil $$$

So we can do all rest calculate in $$$O(1)$$$, the total time complexity will be $$$O(lg(min(G,M)))$$$ for "GCD" function.

submission link

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16 months ago, hide # |
 
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Seems a few greedy/wrong solutions passed in G.

(I know the contest is long over, just adding some cases for any upsolvers.)

A few extra cases
A wrong greedy idea that currently passes