AtCoder Grand Contest 020

6 лет назад, # |

← Rev. 2 →

+71

Previous AGC set by tourist was 4,5 months ago, it's kinda hard to believe but these two AGCs are two AGCs in a row xD. Glad to see AGCs coming back (next one is scheduled three weeks after this).

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6 лет назад, # ^ |

+20

Isn't the next one three weeks after this and not two?

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6 лет назад, # ^ |

+14

Indeed, thanks for nice catch, I will correct it

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6 лет назад, # ^ |

Pls kill me, I forgot about it and slept too long >_> Moreover before changing timezone AGC was always at 2PM, but now it is at 1PM, so it was 13min versus 1h13min being late in my case xd

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6 лет назад, # ^ |

Hm so the takeaway here for me is that you sleep past 1pm. Impressive.

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6 лет назад, # ^ |

+28

1pm? These are rookie numbers

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rng_58

6 лет назад, # ^ |

+100

Judging from your personality, I thought you were trying F.

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HOMIARA_RUBY

6 лет назад, # |

Score for D: 1100(500) means? is 500 partial score like in Codechef?

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6 лет назад, # ^ |

+40

1100 (500) means that the full score of problem D is 1100, but it's possible to get 500 points for solving the problem for partial constraints mentioned in the problem statement.

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VladaMG98

6 лет назад, # |

It is unfortunate, I wasn't able to submit in last 2 minutes, page just didn't want to load :( You maybe should have extended the contest for 5/10 minutes because of such issues.

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matthew99

6 лет назад, # |

+18

What is the solution to problem F?

I found the solution to the case where all L's are equal here, but the obvious generalization to the case where L's can be different was incorrect.

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rng_58

6 лет назад, # ^ |

Uploaded the editorial.

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matthew99

6 лет назад, # ^ |

Thanks. I've already seen it. By the way, why is there no subtask for solving the case where all L's are equal?

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6 лет назад, # ^ |

+107

Well, the fact that it's searchable on the Internet is a good counterargument.

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ikatanic

6 лет назад, # ^ |

+26

It seems that we can speed up editorial solution a little by using one more bitmask in the dp state, instead of generating all (N - 1)! permutations..

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6 лет назад, # ^ |

+21

Nice, you are right, we didn't notice that. In fact, 5! < 2⁵·5, but the complexity becomes O(2^2N - 2·N³·C²).

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mitsuki

6 лет назад, # |

Solution for C?

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square1001

6 лет назад, # ^ |

I solved by using bitset. O(n² × a) DP is obvious but if I use bitset seems like we can achieve O(n² × a / 64). Is it true that bitset is always very fast in use of bitwise operation?

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6 лет назад, # ^ |

+13

Something else that passes is probabilistic solution of shuffling array and then assuming that optimal solution likely never has delta at any time exceeding 50000 (that is when we are solving problem of partitioning array into two halves as close in sum as possible, in our dp we can limit current delta between halves to [-50000, 50000]).

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zscoder

6 лет назад, # ^ |

The delta between halves is at most 2000 because if it exceeds, we can move one element from the larger side to the smaller side and get a better solution.

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6 лет назад, # ^ |

So this makes sense for the final delta, however how does this help us during our dp? As in although final delta is <= 2000, the delta at when we are say midway through our dp can be greater, or is there some easy way to avoid this?

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zscoder

6 лет назад, # ^ |

Right, I didn't get exactly what you were trying to say. In that case the delta can indeed be larger.

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6 лет назад, # ^ |

+23

The final delta is indeed at most 2000, but if we add elements to halves in random order, intermediate deltas may be larger. I assume they are bounded by something like $\text{[math]}$ with high probability, since the process looks like a random walk.

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xennygrimmato

6 лет назад, # ^ |

Can we binary search on the answer (say X) and see if exactly (2^N - 1) / 2 subsequences have sum less than X? Would that work within TL?

We can use DP to check that. f(i, j) represents no. of subsequences using first i numbers with sum < j. Maintain only 2 rows in the DP table to avoid exceeding memory limit.

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6 лет назад, # ^ |

Numbers can get very large, on order of 2^2000, what is your plan of avoiding this? Or if you do have such large numbers, I imagine TLE.

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xennygrimmato

6 лет назад, # ^ |

Right, I did not consider that. Thanks.

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E869120

6 лет назад, # |

+26

What a coincidence... Getting AC after fraction of a second...

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mnaeraxr

6 лет назад, # ^ |

+36

You were 8ms late

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square1001

6 лет назад, # ^ |

Interesting, how did you get this information?

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P___

6 лет назад, # ^ |

+13

You should ask organizers to accept this. The only thing that matters is submission time.

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EnumerativeCombinatorics

6 лет назад, # |

+64

What makes you to make it required to use bitset in problem C? Is there any trouble if n<500?

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6 лет назад, # ^ |

← Rev. 3 →

We can calculate Dp[i][j][t] (1 ≤ i ≤ N, 1 ≤ j ≤ N·Max, 1 ≤ t ≤ C) the number of ways to have sum j using the first i elements modulo P[t] in O(N²·Max·C). We can choose P[t] (1 ≤ t ≤ C) to be a prime number ~2⁶². Now we can find for sure the (2^N - 1)^th element if we choose C = 8 primes.

I'm not sure that's the reason though...

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6 лет назад, # ^ |

+11

Right, we didn't want straightforward counting solutions to get accepted.

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EnumerativeCombinatorics

6 лет назад, # ^ |

And you didn't think about slow languages such as Java or C#? (I know Python is too slow for programming contests though)

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6 лет назад, # ^ |

+21

We did. Our Java solution using built-in java.util.BitSet gets accepted in 619 ms, which is less than 1 / 3 of TL.

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xennygrimmato

6 лет назад, # ^ |

Can you also tell why we'll be able to surely find (2^N - 1)^th element using C = 8 primes? That doesn't seem very intuitive to me.

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6 лет назад, # ^ |

← Rev. 2 →

Suppose you have that x ≡ 2^N (mod P_t) for every t (1 ≤ t ≤ C) then by Chinese Reminder Theorem $\text{[math]}$ . Because we choose 2^N to be smaller than $\text{[math]}$ the solution is unique in our case.

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ikatanic

6 лет назад, # ^ |

Why not just use big integers to compute dp?

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6 лет назад, # ^ |

← Rev. 2 →

wrong

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ikatanic

6 лет назад, # ^ |

How do you do it with fewer primes?

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6 лет назад, # ^ |

Oh, yeah, you're right, I feel stupid now. It's just the first that came to my mind...

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GeniusCoder

6 лет назад, # |

← Rev. 4 →

Atcoder is another judge?

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t1016d

6 лет назад, # ^ |

+10

It's the contest of another online judge, Atcoder.

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GeniusCoder

6 лет назад, # ^ |

Thanks

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JayZhan

6 лет назад, # ^ |

-15

Also there is another online judge Csacademy

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Zain

6 лет назад, # |

What could be the worst case for problem D subask.1 ? I try q=1000 with all a=b=1000, and c=1, d=100. and it runs very fast on my computer, but I get TLE on it when submitting it.

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teja349

6 лет назад, # ^ |

It gives verdict on all cases combined.So if you attempt for partial you will get TLE. But the score reflects on standings page.

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rng_58

6 лет назад, # |

+21

AtCoder Race Ranking

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xuanquang1999

6 лет назад, # |

In problem E, I don't understand the explanation of the "Programmer's way" to figure out that the number of DP state is not large.

Is each element of V a mask of bit or a single bit?
The editorial said "V_i means that character i is logical AND of the corresponding characters in the original string". In the DP recurrence, different value of K and |A| will give different set of "corresponding characters". So what exactly is "corresponding characters" here?
The editorial said "Now, call f({{1}, {2}, ..., {N}}) for N = 100". What does {1}, {2}, ..., {N} here mean?

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6 лет назад, # ^ |

+14

Suppose we are given string S as input. Then, we call f(S). Let T be any string passed as a parameter to f in a later call. It turns out that for every i, $\text{[math]}$ for some sets $\text{[math]}$ .

Now, instead of accepting S as input, accept only its length |S| and assume the general case. Initially, $\text{[math]}$ for all i. For subsequent calls of f, these sets can be formed in the same way as DP is calculated.

Thus:

Every element of V is a bitmask.
I'm not sure I understand this question correctly, but the "corresponding characters" are meant to be "in the original string". Just like in DP calculation, every pair of K and |A| results in a different call to f.
I abuse notation of sets and bitmasks and consider them equivalent. $\text{[math]}$ is a set containing one element i, or, equivalently, a bitmask with a single set bit i.

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xuanquang1999

6 лет назад, # ^ |

I think I got it. Thanks for the clarification.

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cheater2k

6 лет назад, # ^ |

Just a curious question... Did you intend that many people would come up with the "Programmer's way" and "Mathematician's way" during the contest, tourist?

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