AtCoder Regular Contest 145 Announcement

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!

• 5 weeks ago, # ^ | me too

• 5 weeks ago, # ^ | As a contestant, I want to be blue :)

• 5 weeks ago, # ^ | Can I do up-hack to my AC submission? This submission can be hacked by n=4 AAAA

• 5 weeks ago, # ^ | Code that prints

• 5 weeks ago, # ^ | I believe you miss the cases where .

  • 5 weeks ago, # ^ | Thanks for pointing that out.

• 5 weeks ago, # ^ | The set has an value on the -th element, while the optimal set has value on the -th one ,which is enough to meet the constraint of .

  • 5 weeks ago, # ^ | U R right.I tried that,and the number can be as large as 4e7

• 5 weeks ago, # ^ | That doesn't even work,

• 5 weeks ago, # ^ | You have to print "Yes", not "YES". Same for "No" and "NO"

• 5 weeks ago, # ^ | ← Rev. 2 →   +1 Consider the test case :- 6 5 4 , Your O/P — 4 Correct O/P — 2 for value (5 and 6 only Alice will win )

  • 5 weeks ago, # ^ | Corrected that mistake but still getting WA on 3 test cases ;-; Here is my submission

    • 5 weeks ago, # ^ | ← Rev. 2 →   +1 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

    • 5 weeks ago, # ^ | 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).

      • 5 weeks ago, # ^ | Thank you guys so much, my code got AC , there was implementation mistake at the end . In case anyone wants to see binary search solution — Here

• 5 weeks ago, # ^ | ← Rev. 2 →   0 your code doesn't handle the cases for in the range

• 5 weeks ago, # ^ | I don't know how your code works, but it fails at the input 4 2

  • 5 weeks ago, # ^ |

  • 5 weeks ago, # ^ | It works in such way: So the dfs is expected to stop in very low depth. p.s. What about this one? It could pass 4 2.

    • 5 weeks ago, # ^ | It fails at 10000 0. The output has these three integers: -9999991 -9999937 -9999883

      • 5 weeks ago, # ^ | Sorry, but it got 9999991 -9999991 9999973 -9999973 9999971 ... here. Maybe there's some Undefined Behavior in my code?

        • 5 weeks ago, # ^ | 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.

    • 5 weeks ago, # ^ | Are you assuming primes don't have arithmetic sequences of length 3? Because that is not true, for example: .

      • 5 weeks ago, # ^ | Oh, that's it. Sincerely thanks for your help!

• 5 weeks ago, # ^ | Whoa, i didn't brute force on the actual answer tho, I bruteforced on a subproblem of it, The Catalan Numbers

  • 5 weeks ago, # ^ | Catalan numbers are well known.

• 5 weeks ago, # ^ | AB doesn't satisfy the first condition (s[0] == "B" or s[-1] == "A") in the editorial.

• 5 weeks ago, # ^ | Can you share your submission?

• 5 weeks ago, # ^ | 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; }

  • 4 weeks ago, # ^ | Thanks for pointing this out! I tried to add some randomization and it worked pretty good.

• 5 weeks ago, # ^ | 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.

  • 4 weeks ago, # ^ | ← Rev. 3 →   0 Let we can found that . This is my submission for D.