I didn't get it in DIV 1 A it is important to delete the middle letters in the last two paragraphs to avoid appearing a new occurrence after a line is collapsed can anyone provide some testcase

My solution to Div.2 E :

1) Perform DFS starting at Vertex a. This DFS function returns when reached at Vertex b.

2) Perform DFS starting at Vertex b. This DFS function returns when reached at Vertex a.

3) When each function saves the visited vertex, there is the vertex that only the first function visited, the vertex that only the second function visited, and the vertex at which both functions visited.

4) Multiply the number of vertices only the first function visited by the number of vertices only the second function visited, and it's the answer.

For example, on this graph, the answer is 5 x 6 = 30.

Code : 67127301

I really liked this contest, I wish I had participated live.

Here's my solution for Div2-E using BFS. Uses very similar approach to what is mentioned in the editorial. 66995096

VERY nice editorial MikeMirzayanov

Div2 D

5th paragraph, 1st line "In fact, the set of words of kind n10 has no more than n10"

Shouldn't it be "In fact, the set of words of kind n01 has no more than n10"?

In the editorial for the problem G:

• In the recursive equation for dp[v][0], the first product should be "j = 1 to i — 1", not "j = 1 to i = 1".
• In the recursive equation for dp[v][1], the first product should be "j = 1 to d", not "j = 1 to i = d".
• In the recursive equation for dp[v][2], the first product should be "j = 1 to i — 1", not "j = 1 to i = 1".

In the problem 1277D - Let's Play the Words?, how come the solution to the last sample testcase is given as 1 2

can anyone give me a testcase that broke my code?

I have a question about div1 E.

How do you perform operations to make $$$\Delta$$$ decrease to $$$\frac{3}{4}\Delta$$$ for stones located at 0, 1, 99999, 100000 according to your alogrithm?

In the editorial of question D: Let's Play The Words? Why do we need to reverse $$$n_{01} - n_{10} - 1$$$ strings and not $$$(n_{01} - n_{10}) / 2$$$. Also, can anyone give hints on how to prove that the problem is equivalent to "the Euler traversal of a directed graph with 2 nodes"? Thanks!

can anyone help me in problem B I am getting WA on tc 2 https://mirror.codeforces.com/contest/1277/submission/81699789 thanks

In Two Fairs, I am not able to understand what exactly (alpha u ,beta u) represents.Can someone clarify? Any help will be appreciated thanks :)

For those confused with the editorial for div2 E, I think there is a much easier approach with a similar idea.

1. build an adjacency list that removes the vertex A from the graph(i.e., don't include edges to/from A) and then calculate the number of vertices connected to B(by performing dfs, starting from B). Let the number be x.

2. similarly, build an adj list without B and count the number of vertices connected to A. Let the number be y.

3. Now, the answer is simply (n-x-1)*(n-y-1). This is because (n-x-1) is the number of vertices that cant reach B if A is not present(we subtracted 1 to exclude A). This means that to connect these vertices to B, a path should exist only via A.

A similar conclusion can be made for the (n-y-1) vertices. Thus answer will be (n-x-1)*(n-y-1).

For those confused with the editorial for div2 E, I think there is a much easier approach with a similar idea.

1. build an adjacency list that removes the vertex A from the graph(i.e., don't include edges to/from A) and then calculate the number of vertices connected to B(by performing dfs, starting from B). Let the number be x.

2. similarly, build an adj list without B and count the number of vertices connected to A. Let the number be y.

3. Now, the answer is simply (n-x-1)*(n-y-1). This is because (n-x-1) is the number of vertices that cant reach B if A is not present(we subtracted 1 to exclude A). This means that to connect these vertices to B, a path should exist only via A.

A similar conclusion can be made for the (n-y-1) vertices. Thus answer will be (n-x-1)*(n-y-1).

Here is my submission.