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anzhelika [568]
2 years ago
14

Write short notes about monitor printer and speaker​

Computers and Technology
2 answers:
Bond [772]2 years ago
8 0

Answer:

<em>A computer monitor</em> is an electronic device that shows pictures for computers. Monitors often look similar to televisions. The main difference between a monitor and a television is that a monitor does not have a television tuner to change channels. Monitors often have higher display resolution than televisions.

<em>A printer</em> is a device that accepts text and graphic output from a computer and transfers the information to paper, usually to standard size sheets of paper. Printers vary in size, speed, sophistication, and cost. In general, more expensive printers are used for higher-resolution color printing.

<em>Speakers</em> receive audio input from the computer's sound card and produce audio output in the form of sound waves. Most computer speakers are active speakers, meaning they have an internal amplifier which allows you to increase the volume, or amplitude, of the sound.

andreev551 [17]2 years ago
3 0

Explanation:

Here is ur answer

HAVE A NICE DAY

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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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