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3 years ago

9

Design the logic for a program that allows a usher to continuously enter numbers until the usher enters 0. Display the sum of th

e numbers entered.

Computers and Technology

1 answer:

uysha [10]3 years ago

5 0

int sum = 0, n;

do {cin>>n; sum+=n;}while (n!=0);

cout<<sum;

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Tower of Hanoi algorithm 4 disks

Sophie [7]

????is this a question

4 0

3 years ago

Read 2 more answers

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

4 0

2 years ago

Randomly choose rock, paper, or scissors for the computer

Pepsi [2]

import random

computer_choice = random.choice(["rock", "paper", "scissors"])

user_choice = input("rock, paper, or scissors? ")

if computer_choice == user_choice:

print("It's a draw!")

elif user_choice == "rock" and computer_choice == "scissors":

print("You win!")

elif user_choice == "paper" and computer_choice == "rock":

print("You win!")

elif user_choice == "scissors" and computer_choice == "paper":

print("You win!")

else:

print("The computer wins!")

I wrote my code in python 3.8. I hope this helps.

8 0

2 years ago

Suppose you wish to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm. Your algo

Kazeer [188]

Answer:

The number of subproblems are given as $T(n)=a*T(n/8)+\theta(n^2)$ while the value of T(n) to be less than S(n) is for 342.

Explanation:

The number of subproblems are given as

$T(n)=a*T(n/8)+\theta(n^2)$

Asymptotic running time for Strassen’s algorithm is $S(n)=\theta(n^{log(7)})$

Now, when a increases, number of subproblems determines the asymptotic running time of the problem and case 1 of master theorem applies. So, in worst case, asymptotic running time of the algorithm will be

$T(n)=\theta(n^{logb(a)})=\theta(n^{log8(a)})=\theta(n^{log_{2}(a^{1/3})})$

Now, for T(n) to be smaller than S(n)

$n^{loga^{1/3}}$

So,

$log(a^{(1/3)})$

So,

a=342

5 0

3 years ago

Two hundred workstations and four servers on a single LAN are connected by a number of switches. You’re seeing an excessive numb

quester [9]

Answer:

Add more servers or decrease the amount of workstations

5 0

2 years ago