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

ABC345? Increasing sequence?

contest questions are in japanese language?

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

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.

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)

How C

Problems were more difficult than usual today.

Top 200 EZ

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

How to prune D?

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

F was easier than D :)

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;
     }
}

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].

Why A is giving WA?

bool a = false, b = false, c = false;
for(auto &ch : s)
{
    if(ch == '<')
    {
        a = true;
        b = false;
        c = false;
    }
    if(ch == '=')
    {
        b = true;
        c = false;
    }
    if(ch == '>')
    {
        c = true;
    }
    if(a && b && c)
    {
        cout<<"Yes"<<endl;
        return;
    }
}
cout<<"No"<<endl;

Atcoder, where are the testcases of abc344 & abc345?

where are the test cases atcoder_official