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topjm [15]
2 years ago
14

Write a function named firstLast2 that takes as input an array of integers and an integer that specifies how many entries are in

the array. The function should return true if the array starts or ends with the digit 2. Otherwise it should return false. Test your function with arrays of different length and with the digit 2 at the beginning of the array, end of the array, middle of the array, and missing from the array.
Computers and Technology
1 answer:
frosja888 [35]2 years ago
4 0

Answer:

<em>Written in C++</em>

#include<iostream>

using namespace std;

int firstLast2(int arr[], int n);

int main()

{

int n;

cout<<"Number of Elements: ";

cin>>n;  

int myarray[n];

for(int i = 0; i<n;i++) {

cin>>myarray[i];

}

firstLast2(myarray, n);

return 0;

}

int firstLast2(int arr[], int n){

if(arr[0] == 2 || arr[n - 1] == 2) {

 cout<<"True";

}

else {

cout<<"False";

}

}

Explanation:

<em>I've added the full source code as an attachment where I used comments to explain difficult lines</em>

Download cpp
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Answer:

Explanation:

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Answer:

.code

main proc

mov ecx,ebp

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push ecx

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invoke ExitProcess,0

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2 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

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