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Monica [59]
3 years ago
12

Which of the following best describes Roblox?

Computers and Technology
2 answers:
liberstina [14]3 years ago
6 0

Answer:

B. An online platform for game creation!

Explanation:

Roblox is a massively multiplayer online and game creation system platform that allows users to design their own games and play a wide variety of different types of games created by other users.

--

Hope this helps! ✧.·:

Marta_Voda [28]3 years ago
5 0

I'd say B, An online platform for game creation. Millions of people can create their own game, clothes, and sometimes items on ROBLOX, for Millions of people to enjoy.

Honestly, ROBLOX isn't really like Minecraft, especially Survival. I'm not sure people always compare it.

And, the last 2 options C and D don't really fit ROBLOX. I wouldn't be able to guess you were talking about ROBLOX just hearing those choices.

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A contiguous subsequence of a list S is a subsequence made up of consecutive elements of S. For example, if S is 5,15,-30,10,-5,
Sindrei [870]

Answer:

We made use of the dynamic programming method to solve a linear-time algorithm which is given below in the explanation section.

Explanation:

Solution

(1) By making use of the  dynamic programming, we can solve the problem in linear time.

Now,

We consider a linear number of sub-problems, each of which can be solved using previously solved sub-problems in constant time, this giving a running time of O(n)

Let G[t] represent the sum of a maximum sum contiguous sub-sequence ending exactly at index t

Thus, given that:

G[t+1] = max{G[t] + A[t+1] ,A[t+1] } (for all   1<=t<= n-1)

Then,

G[0] = A[0].

Using the above recurrence relation, we can compute the sum of the optimal sub sequence for array A, which would just be the maximum over G[i] for 0 <= i<= n-1.

However, we are required to output the starting and ending indices of an optimal sub-sequence, we would use another array V where V[i] would store the starting index for a maximum sum contiguous sub sequence ending at index i.

Now the algorithm would be:

Create arrays G and V each of size n.

G[0] = A[0];

V[0] = 0;

max = G[0];

max_start = 0, max_end = 0;

For i going from 1 to n-1:

// We know that G[i] = max { G[i-1] + A[i], A[i] .

If ( G[i-1] > 0)

G[i] = G[i-1] + A[i];

V[i] = V[i-1];

Else

G[i] = A[i];

V[i] = i;

If ( G[i] > max)

max_start = V[i];

max_end = i;

max = G[i];

EndFor.

Output max_start and max_end.

The above algorithm takes O(n) time .

4 0
3 years ago
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