パナソニックグループプログラミングコンテスト2023（AtCoder Beginner Contest 301）...

We will hold パナソニックグループプログラミングコンテスト2023（AtCoder Beginner Contest 301）.

The point values will be 100-200-300-400-475-500-600-625. (Since this ABC, we adjust the point values depending on problems.) We are looking forward to your participation!

• 3 days ago, # ^ | Moral of this contest: Never use alts to try to consume all the penalties for your main account. (even with unrated register) If so, both accounts will be deleted automatically.

• 3 days ago, # ^ | Digit DP

  • 3 days ago, # ^ | it's not digit DP

    • 3 days ago, # ^ | Can you point out the mistake in this approach? We first replace all '?' with '0' and store indices of all '?' in a vector. Then we iterate on every element x of the vector and check if setting xth bit to 1 would give a number smaller than or equal to N. If you get a valid number, change '0' to '1' at that index. Code

      • 3 days ago, # ^ | Never mind, I got it. It should be 1LL << i instead of 1 << i. Got WA x 2 due to this small mistake. :(

    • 3 days ago, # ^ | I didn't know that there exists an easier solution, I just said how I solved it.

  • 3 days ago, # ^ | ← Rev. 2 →   +29 You guys did DP? I did it by greedy

    • 3 days ago, # ^ | There is no username that goes by "perfect".

• 3 days ago, # ^ | Implementation :( I unwanted memoized it with dp my submission. DDOS didn't get chill with me.. when I play good round gets unrated and when bad gets rated :(

• 3 days ago, # ^ | ← Rev. 3 →   +1 Solution of D

• 3 days ago, # ^ | ← Rev. 5 →   0 I've solved it with a greedy technique. My Algorithm is: 1- Compute the current value of string S in decimal representation and neglect all '?' symbols in it 2- Iterate over string S from MSB (most significant bit) and check if the current symbol is '?' and the (value of current value of S + pow (2, size(s)-idx-1)) is less than or equal to N, then I'll flip this symbol to 1 and add pow (2, size(s)-idx-1) to current value of s, else I will flip it to 0. My Source code

• 3 days ago, # ^ | I tried solving it using DP. Here is my solution (in case, you are interested).

• 3 days ago, # ^ | I also was getting WA on that test case. I had a typo in my solution. While checking the distance from the starting point to the goal, I was doing instead of

• 2 days ago, # ^ | I faced the same problem, I thought he had to stop once he reached the Goal square (meaning he can't visit the goal square multiple times). This was giving WA. Maybe the wording of the question could have been better :(

  • 2 days ago, # ^ | I thought exactly the same lol, could not figure out what the mistake was for 1 hour.. Thanks!

  • 43 hours ago, # ^ | Explanation for sample 1 shows that you can move through the goal square

• 3 days ago, # ^ | ← Rev. 2 →   0 Counter: Input: ttttac @@@tac Output: No Correct Output: Yes You should reduce firstone[i] by 1 when firstone[i] > secondone[i] similar for secondone[i].

• 3 days ago, # ^ | you just re-invented the traveling salesman problem. Congrats!

• 2 days ago, # ^ | Can't we use recursion on all the candies, because the maximum count is 18. I followed your first and second as it is but my solution got TLE.

• 3 days ago, # ^ | ← Rev. 3 →   +30 Please run the following snippet in AtCoder's custom test page: Spoiler

• 2 days ago, # ^ | That's because in the first case, "atcoder" is considered as a character array and not as a string.