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First unrated educational round of 2022! :D

There will be at most five hundred interactive problems in the round, which means all problems might be interactive. That's interesting!

On BlueSmokeCodeforces Round #641, 6 years ago
0

5 1 1 1 6 6 1

-> 5 1 1 6 6 6 1 // operate [4, 6]

-> 5 5 5 5 5 5 1 // operate [1, 6]

-> 5 5 5 5 5 5 5 // operate [1, 7]

On BlueSmokeCodeforces Round #641, 6 years ago
+15

Try if this test is yes:

1
7 5
5 1 1 1 6 6 1

Of course, we can't make sure we'll always find such $$$a$$$ and $$$b$$$ that $$$a + b \geq x + y + 1$$$. And at that time, the answer will be increased by 1, because whichever we select the $$$b$$$, we can't make $$$a + b \gt = x + y + 1$$$, in other world, the person who get the $$$a$$$ place in round 1 will finally placed before Nikolay.

The answer should in range [1,n].

+24
  • For the minimum possible overall place:

For a number $$$a$$$ in round 1 and another number $$$b$$$ in round 2, we should make sure $$$a + b \geq x + y + 1$$$. $$$1$$$ should find $$$x + y$$$, $$$2$$$ should find $$$x + y - 1$$$... And if $$$x + y \gt n$$$, $$$1$$$ should find the smallest number in round 2 which can be selected.

So, we can see that the final answer will be $$$x + y - n + 1$$$, cause we can not arrange number $$$[1, x + y - n]$$$ to make them greater than $$$x + y$$$, and don't forget the ans must in $$$[1,n]$$$.

  • For the maximum possible overall place:

In round 1, for each number $$$a$$$ between $$$1$$$ and $$$x-1$$$(it can be empty), we can find a number $$$b$$$ in round 2, and make $$$a + b \leq x + y$$$.

In round 2, for each number $$$b$$$ between $$$1$$$ and $$$y-1$$$(it can be empty), we can find a number $$$a$$$ in round 1, and make $$$a + b \leq x + y$$$.

We cannot find any other pairs to make $$$a + b \leq x + y$$$. Thus, the answer is $$$x - 1 + y - 1 + 1$$$, and don't forget the ans must in $$$[1,n]$$$.

Let $$$a = x - 1, b = y + 1$$$, while $$$a$$$ is decreased by $$$1$$$, we can add $$$1$$$ for $$$b$$$. If $$$b$$$ is greater than $$$n$$$, we can select any number remaining, cause at this time, $$$a + b \lt x + y$$$. To proof the second one, we can just swap the round 1 and round 2.

It seems like it's a kind of greedy, but the editorial can proof the answer strictly. And also sorry for my poor English. Wish it can help you. :D

UPD: swap((a,b),(x,y)), sorry for the mistake.

How to register as a team member...

I can't see it...

UPD: Fixed now, and wish everyone will enjoy the contest, GL & HF.

On crausCodeforces Simulator, 8 years ago
0
On crausCodeforces Simulator, 8 years ago
0

That's amazing!

And how about to add hack to the game? :)

The time I showed was UTC+8, so I didn't notice the difference in the start time.

The time is right for Chinese students! That's great!

UPD: The Problem was fixed, and wish everyone will increase your rating.

+8

Chinese users will like it because of the time of the contest~