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kow [346]
3 years ago
12

Match each example to the type of component, either Input, Process, or Output.

Computers and Technology
1 answer:
statuscvo [17]3 years ago
5 0

Answer:

Find answers below.

Explanation:

1. Input: it takes in data in its raw format or an item that receives data and transfers them to the process.

  • <em>Employees in a management team. </em>
  • <em>Program code for a software application. </em>
  • <em>Flour for making bread.</em>
  • <em>A wrench. </em>

2. Process: it converts the data from an input to a usable format. Also, it conveys the processed data (informations) to the output.

  • <em>Fermentation. </em>
  • <em>Harvesting a crop. </em>
  • <em>Dryer spinning at top speed.</em>

3. Output: it is the result produced by a process i.e the finished product.

  • <em>Newly painted structure. </em>
  • <em>Freshly mowed lawn. </em>
  • <em>Hot muffins. </em>
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Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi
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Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

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Hello ur answer should be A

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An asymmetric encryption system utilizes how many keys?
ivanzaharov [21]
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)In a vector implementation of a stack ADT, you add an entry to the top of a stack using which vector method?
kotykmax [81]

Answer:

B.add.

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boolean add(element)

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