How about implementing a more advanced problem filter? For example, I would like to solve Atcoder problems within the range of 1200-1400, but I don't want to see any problems without English problem statements (allowing users to manually mark which problems should not be displayed). The filter should also be able to select only the problems from contests with the name format "Atcoder Regular Contest XXX" (supporting regular expressions).

Hi you guys, I'd like to share my OfflineJudge with you. What you need to do is to set up your topcoder account and problem's statistics page, and just run it. feel free to change it to meet your own preferences.

ok, guys, I made a script to fetch test cases from topcoder, it requires your account info(yes password also) and someone's problem statistics page like this one, you can find these statistics on competitiveprogramming.

for now, I just throw away brackets, commas, etc , and treat it as plain text, making code get input from the file. when I got a better solution, I'll share my offline_judge with you guys.

I submit an issue on GitHub but got no response, maybe it's high time to crawl test cases from Topcoder, and build an offline judge to verdict my code. Should I do this?

now, I see why we had kenkooo for atcoder, cf tracker for codeforces, but nothing for topcoder.

Practise and have fun! Just solve it!

dang! I forgot that according to bubble sorting, if we can swap two adjacent element, then we can sort the whole segment! why I rushed to implement my bad fragil 'solution' without further thinking, so sad.

it means the triangle's three vertices won't be in a single line, that's why it's a triangle.

I participated this contest virtually. for problem D, my observation was that: 1. we can reverse a substring s[l:r] by inserting s[r] to s[l] to the front. 2. we never reverse two substring which one is contained in the other. So I tired to solve it with dp[i][j][0/1] which means for s[1:i], the last segment was s[i — j:j], 1 stand for it was reversed, and vice versa. after I calcatuled this dp array, I will construct the operation sequence. But I failed at dp processing, and wondering if this method is resonable.

thanks, I didn't realise it's buggy when swap element in array.

you need help of Santa Claus

How do you solve problem F?

thanks for your hint, I solved this problem. the interesting thing is this code will TLE on my computer, but got AC after submit it.

ll dfs(int k, int r, int c) {
  if (K == k) return r && c;
  ll& ans = dp[k][r][c];
  if (ans != -1) return ans;
  ans = 0;
  if (r && c) {
    ans = dfs(k + 1, 0, 1) * (n - 1) % mod + dfs(k + 1, 1, 0) * (m - 1) % mod;
  }
  if (r && !c) {
    ans = dfs(k + 1, 0, 0) * (n - 1) % mod + dfs(k + 1, 1, 0) * (m - 2) % mod +
          dfs(k + 1, 1, 1);
  }
  if (!r && c) {
    ans = dfs(k + 1, 1, 1) + dfs(k + 1, 0, 1) * (n - 2) % mod +
          dfs(k + 1, 0, 0) * (m - 1) % mod;
  }
  if (!r && !c) {
    ans = dfs(k + 1, 0, 0) * (n - 2 + m - 2) % mod + dfs(k + 1, 1, 0) +
          dfs(k + 1, 0, 1);
  }
  ans %= mod;
  return ans;
}

update: it's beacause of stack size, it works after i add --stack=536870912 to compiler flag.

now, he did it.

thanks for your advice

my code done right when k=0, but WA on test 5

I think sometimes the problem setter found that problem can be solved by another approach, so they wanna us know that.

914F - Substrings in a String supposed to be solved by string suffix structures, but solved by bitset.

the for-loop is m-size, so there may be more than n element in the vector, up to 2^n.

for m segements, [l, r], we can just think it as there is a x in the array, no matter the position.

thanks!

can't understand solution of E since "iterate the number of message".

absolutely, "st" is the intial of this "shuai tou" emoji.

in this problem, $$$y_i \gt y_{i-1}$$$, why it also can be calculate by this fomula?