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Monoxer Programming Contest 2024（AtCoder Beginner Contest 345） Announcement
By atcoder_official, history, 2 years ago, In English

We will hold Monoxer Programming Contest 2024（AtCoder Beginner Contest 345）.

We are looking forward to your participation!

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

ABC 345... Time for me to practice arithmetic series to prepare for this contest.

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

ABC345? Increasing sequence?

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

contest questions are in japanese language?

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

Does the contest provide problem statements in English or only in Japanese?

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

Please tell us if you only provide Japanese problem statements.

Although I'm a Chinese, I know that there're some differences between Chinese and Japanese, so it may cause ambiguity.

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2 years ago, hide # |
← Rev. 2  
Vote: I like it +26 Vote: I do not like it

To those who are confused

All previous rated ABC have Japanese and English statements. I don't know this time, but I guess it's the same since no special announcement was made.(I'm not an official)

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

How C

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    2 years ago, hide # ^ |
    ← Rev. 2  
    Vote: I like it 0 Vote: I do not like it
    hint
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2 years ago, hide # |
 
Vote: I like it +11 Vote: I do not like it

Problems were more difficult than usual today.

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2 years ago, hide # |
 
Vote: I like it -22 Vote: I do not like it

Top 200 EZ

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

Don't know if it was intended or not, but I guess the time limit is too loose for G. My $$$O(n^2/k)$$$ solution with a fast mod managed to squeeze into the time limit. XD

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

How to prune D?

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    2 years ago, hide # ^ |
    ← Rev. 2  
    Vote: I like it -6 Vote: I do not like it

    with ‌Backtracking.

    You can first solve this problem for learning backtrack. Eight queens puzzle

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

      Well it turned out I shouldn't sort the tiles by increasing area. I should sort them in decrease

      Never mind

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    2 years ago, hide # ^ |
    ← Rev. 5  
    Vote: I like it +9 Vote: I do not like it

    You don't really need to, there is a $$$O(n! \cdot 2^n \cdot H \cdot W)$$$ brute which fits into the time limit pretty easily.

    Basically, for all possible orders of the tiles, try all combinations of both possible orientations (initial or rotate 90) among all tiles and for each of them try to fit the grid using the following pseudocode:

    for i in [1, H]:
    for j in [1, W]:
        if (i, j) is not covered:
            try to place the next tile with its top left corner at (i, j)

    Code — 51308848

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2 years ago, hide # |
 
Vote: I like it +9 Vote: I do not like it

Toughest ABC I've seen in a while, problem E feels tougher to me than most CF Div2E problems @_@.

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

F was easier than D :)

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2 years ago, hide # |
← Rev. 5  
Vote: I like it 0 Vote: I do not like it

if you get wrong in c this because if there is a repetition of the letter, when you make a switch between the same letter, the resulting string calculates a new string just once. aab -> aab , baa , aba so the answer is 3 not 2 just do it before your code

for(int i = 0;i < s.size();i++){
    if(mp[s[i]] > 1){
        sum++;
        break;
     }
}
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2 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Problem E is fantastic, and I spend about two hours and finally get it accepted. My idea is to use dpmax[j][0] = pair<color0, value0>, and dpmax[j][1] = pair<color1, value1> to denote that, until now, if we remove j balls, the optimal two values with color0 and color1 as the rightmost ball. Besides, we use dp[i][j] to denote the optimal value that we can get, if we consider the first i balls and remove j of them. dp[i][j] can be computed simply by using dpmax[j][0] and dpmax[j][1], and then we should update dpmax[j][0] and dpmax[j][1].

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

Why A is giving WA?

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

Atcoder, where are the testcases of abc344 & abc345?

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

where are the test cases atcoder_official

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