Codeforces should be about the player's ability to code, data Structures and algorithms, not the ability to play the genshin's mini-game.

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As you can see, today's Div2 Rank1 syf2008 is a **cheater**, He and his friend Tx_Lcy cheats together. They have cheated at least twice between them.

For the first time, they received a cheating check message from Codeforces, as they wrote in their comments.

The second cheating occurred in Codeforces Round 857 (Div. 1) Problem C. They actually wrote the same code and made a lot of equivalent replacements, but due to FST, they were not caught by the system.

In this Div2, he submitted all the questions in an **extremely abnormal order**. Therefore, I reasonably suspect that they committed the crime in a team. (Upd: in comments)

I no longer want to care about any large-scale cheating incidents, but I want to say, if Rank1 in a contest can be obtained through a cheater, then the ranking of Codeforces becomes a joker.

rui_er At least do something, thx.

2 . Right click on any Latex and it will provide a settings interface.

3 . Enter Math settings → Math renderer option

5 . Try each option until it displays normally.

7 . If it still doesn't work, try close Accessibility → Assistive MathML

11 . Don't forget to thank me :P

https://codeforces.com/ratings/country/North%20Korea/city/Pyongyang

People like RGB_ICPC use a whole $$$\Huge\text{ten}$$$ alt to take away ratings that belong to others, which is the real cause of rating deflation.

Everyone knows that the priority queue for C++ is log n, but this is very slow, so we hope to optimize it.

Some people believe that using vjudgian theorem can help you optimize to $$$O(-1)$$$. But this is wrong. The definition of complexity is $$$O(f(x))$$$ representation, and you can find an $$$f(x)$$$ and two cons

tants $$$c_1,c_2$$$ with true run time $$$g(x)$$$ such that $$$\text{lim}_{x→\infty}\text{ }c_1f(x)\le g(x)\le c_2f(x)$$$. So essentially $$$O(-1)=O(1)$$$. The Vjudgian theorem is very difficult and cannot be run on Codeforces testers, so I won't introduce this.

Let me present a method that can be run in Codeforecs and that can be easily coded by tourist and benq. First, go offline for the first time. Offline means that you can know in advance what all operations are and give the results of all queries at the last time. For priority queues, we believe that there are two operations: (inserting a number) and (accessing and popping a minimum).

For these two operations, we will convert them into a sequence, use — 1 for pop-up operations, and use the inserted integer for insert operations. For example, for $$$[11,12,-1,3,-1]$$$, we insert $$$11$$$, insert $$$12$$$, pop up a number, insert $$$3$$$, and pop up a number. Because this is offl

ine, we can handle this in any way we like. First, we will access all non $$$-1$$$ numbers from small to large. Any smart person knows how to do this within $$$O(n)$$$ using cardinality sorting.

Subsequently, for each accessed number, we find the $$$-1$$$ closest to it and on the right, and delete this $$$-1$$$. We can easily do this by using and searching the set.

So I took the dsu operation offline again. Create a graph that treats each dsu merge operation as an edge, and the edge weight is the order in which the operation is executed. Then perform a minimum spanning tree on the graph (Klein P N, Tarjan R E. A Randomized Linear Time Algorithm for Finding Minimum Sp

anning Trees [M]. Brown University, 1993).

Then only the edges on the minimum spanning tree is valuable, and then you discover that you can solve it using four Russian algorithms.

https://www.spoj.com/problems/BALIFE/

Because I'm not smart enough, so I don't know how to use Google.

Give me all the code, just like feeding.

https://codeforces.com/blog/entry/113075

today i read this blog and found a very useful trick i will introduce it to such interval problems consider constructing a cartesian coordinate system for an interval $$$[x,y]$$$ and mapping it to the point $$$(x,y)$$$ for the query operation stated in the blog given $$$(l,r)$$$ translate to coordinate points is to ask how many points there are in a certain rectangle then the problem was transformed into a simple one then lets say mos algo put each $$$[l,r]$$$ on the axis then the problem becomes that you can go up down left right and ask for the minimum number of times to traverse all the points so that you can understand mo more intuitively lets take another example of a topic i encountered while talking with my friend my friend asked there is an array that supports swapping two numbers and querying the number of inverse pairs my friend wants to know if there is anything below $$$O(n\log^2n)$$$ just put $$$(i,a_i)$$$ to the coordinate points then the problem becomes to support modifying the coordinates of a point and querying the number of points inside a rectangle it is impossible to go below $$$O(n\log^2n)$$$ you can see that this trick is very helpful to think about the problem try using this method of thinking when you encounter problems with interval bounded comparisons

https://codeforces.com/contest/1785/status?order=BY_PROGRAM_LENGTH_ASC

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