Блог пользователя memset0

Автор memset0, история, 10 лет назад, По-английски

Hi All,

Topcoder SRM 700 will be held 13 October 07:00 AM BDT

Let's discuss the problems after the contest.

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10 лет назад, скрыть # |
 
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An important piece of information omitted from the blog post but present in match reminder: Booz Allen Hamilton will be sponsoring the SRM and giving out t-shirts to the winner of each room.

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Hello sir, I have two questions:

Is it rated? Can a former employee of Booz Allen take part on it?

Have a good day,

Ed.

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I just created an account on TC. When can i register for the contest ?

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Do I know you?

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Can someone please guide through the approach for Div2 C?

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    10 лет назад, скрыть # ^ |
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    Dp[i][j]= How many ways do you arrange the contestants , skipping at most j contestants. This DP assumes that the the first contestant of group 1 has a smaller rank that the first contest of group 2 and so on... Then you have to multiply the answer by fat[group] to cover all possibilities: DP[i][j]=DP[i+1][j+size-1]+DP[i][j-1]; sorry for my bad english ;/

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      dp[i][j] = dp[i + 1][j + size — 1] + dp[i][j — 1], in the second dp (dp[i][j — 1]) why do we have to go back to when j > 0 ?

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      What do you mean by skipping? Could you please elaborate it a bit more, specially the transition and the base case? Thanks

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        10 лет назад, скрыть # ^ |
         
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        Suppose that we have the rank of the first place of each group. R1 is the greatest rank of group 1 , R2 is the greatest rank of group 2 and so on. Now, we will suppose that R1<R2<R3<R4 and so on. Now, suppose that we have M contestants on each room. We can mantain a DP , that the state is I (the player we are) and J (how players we can skip). The dp count HOW many manners we can arrange the players in each room such that the rank of the first contestant on room 1 is smaller than the rank of the first contestant on room 2 and so on.. On state DP[I][J] -> WE HAVE TO POSSIBILITIES FOR THE PLAYER I: HE IS THE FIRST OF HIS ROOM , and he is not the first of his room; WHEN j is 0 , the i-th contestant MAY be the first of his room. Then: DP[I][J]=DP[I+1][J+SIZE-1] ( THE player I is the first of his room) + DP[i][j+1] ( THE PLAYER I IS the first of his room) ; The base case is: dp[1][0] = > the first player may be the first of his room. DP[lastplayer][anything] = 1; HOW WE assume that R1<R2<R3<R4<Rgroup. we may multiply this answer by fatorial[group].

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Can someone please tell from where can I get editorial of this round ???

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How to form dp in Div1 medium?

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Can anyone please explain the approach of div2 450?

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    you just have to check all subset of the '?' in the board. You can do this as, at first check all them as '.'. Then for each subset you say that this will include 'O'. Then check that by including this as 'O' .After making as 'O' ,you now check what is the current min of row or column and max of row or column . Then simply add to variable (max row-min row+1)*(max column -min column +1).

    You can check all subset by bitmask or recursion.

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I was hacked in the Div2 450 problem. I submit the same code after the SRM and got "Successful for XXX.XX points". Is there any way to see the test case that hacked my code? First round there, thanks.

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I have trouble understanding dev1 B. Some help would be great!

What's the Z(f) for the below graph?

1->2
2->1
3->4
4->3
5->6
6->5
7->8
8->7

What's the relevance of the condition Z(f) will be the minimum of S(f,w), taken over all nonnegative w. I think S(f,0) would give the minimum value we are looking. Is there a case where S(f,1) is lesser then S(f,0)?

Edit: link to problem statement for quick access: https://community.topcoder.com/stat?c=problem_statement&pm=14266