### Iron_man_3000love's blog

By Iron_man_3000love, 5 weeks ago,
We are sorry for our many mistakes.

There Have many Solution and approach of any problem . We will discuss here only one approach for every problem which we found easy way .

Previousely you could only virtually , now the contest is open for practice .

A. Square Field

Tutorial
Solution in c++

Idea & Tutorial & Solution : Iron_man_3000love

B. Win Or Lose

Tutorial
Solution in c++

Idea : Iron_man_3000love Tutorial & Solution : muhammad_saidul_islam

C. Delete and insert

Tutorial
Solution in c++

Idea : muhammad_saidul_islam Tutorial & Solution : Iron_man_3000love

muhammad_saidul_islam's solution with less complexity :

Solution

D. Triple Fighting

Tutorial
Solution in c++

Idea & Tutorial & Solution : Iron_man_3000love

E. Absolute value and divisors

Tutorial
Solution in c++

Idea : muhammad_saidul_islam Tutorial & Solution : Iron_man_3000love

muhammad_saidul_islam's solution with less complexity :

solution

F. Home Work

Tutorial
Solution in c++

Idea & Tutorial & Solution : Iron_man_3000love

• +39

 » 5 weeks ago, # |   +4 waiting for the next round and thanks for the editorial.
 » 5 weeks ago, # |   +4 Since these problems are unofficial and CF does not give them ratings, Can you mention what are them based on your experience?
 » 5 weeks ago, # |   +1 last problem can be solved using dp => dp[N][26] dp[i][j] => upto index i max sorted subsequence ending at char('a' + j) state dp[i][j] = dp[i - 1][j] dp[i][s[j] - 'a'] = 1 + max(dp[i - 1]['a' to s[j] - 'a'])  Code (C++)void solve(int &T) { string s; cin >> s; ll n = s.size(); s = "#" + s; vector> dp(n + 1, vector(26, 0)); loop(i, 1, n + 1) { ll mx = 0; loop(j, 26) { dp[i][j] = dp[i - 1][j]; if(j <= s[i] - 'a') { mx = max(mx, dp[i][j]); } } dp[i][s[i] - 'a'] = mx + 1; } ll len = 0; loop(j, 26) { len = max(len, dp[n][j]); } cout << len << endl; string ans; ll i = n, j = 26; while(len > 0) { loop(k, 0, j) { if(dp[i][k] == len and s[i] == 'a' + k) { j = k + 1; ans += ('a' + k); len--; break; } } i--; } reverse(all(ans)); cout << ans << endl; } 
 » 4 weeks ago, # | ← Rev. 2 →   0 In problem D : Let's take this input :21 3 3 4 6 6The editorial solution output is 1 team, but it can be 2 like this : 1 3 4 and 3 6 6Am I wrong ?
•  » » 4 weeks ago, # ^ |   +3 $3 \times 3 \nleq 6$
 » 4 weeks ago, # | ← Rev. 2 →   -9 *
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   0 Lol he edited his comment