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elena-s [515]
3 years ago
14

Suppose that a server sends four packets of audio data, in 15 seconds each. The first is sent by the server exactly at noon (12:

00:00), while each of the three successive packets is sent immediately after the completion of its predecessor. The client, on the other hand, receives the four packets beginning at 12:00:03, 12:00:20, 12:00:41, 12:00:59 respectively. What is the smallest delayed wait time (buffer) in seconds that would ensure the client had seamless playback (i.e. each packet plays immediately after the previous one ends)?
Computers and Technology
1 answer:
Romashka [77]3 years ago
3 0

Answer:

15

Explanation:

for seamless playback each packet should be played before their predecessor so

4th packet will start at 12:01:03

3rd packet will start at 12:00:48

2nd packet will start at 12:00:33

1st packet will start at 12:00:18

as first packet arrives at 12:00:03 so wait time is 15 seconds for seamless playback

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Answer:

<em>The program written in Python is as follows:</em>

def solution(N):

     concat = ""

     for i in range(1,N+1):

           if not(i%2 == 0 or i%3 ==0 or i%5 == 0):

                 print(str(i))

           else:

                 if i%2 == 0:

                       concat= concat+"Codility"

                 if i%3 == 0:

                       concat= concat+"Testers"

                 if i%5 == 0:

                       concat= concat+"Coders"

                 print(concat)

                 concat = ""

N = int(input("Enter a positive integer: "))

solution(N)

Explanation:

This line declares the function

def solution(N):

This line initializes a variable named concat to an empty string

     concat = ""

This line iterates from 1 to the input integer

     for i in range(1,N+1):

<em>This line checks if the current number of iteration is divisible by 2,3 or 5, if no, the number is printed</em>

           if not(i%2 == 0 or i%3 ==0 or i%5 == 0):

                 print(str(i))

<em>If otherwise</em>

           else:

<em>This lines checks if current number is divisible by 2; if yes the string "Codility" is concatenated to string concat</em>

                 if i%2 == 0:

                       concat= concat+"Codility"

<em>This lines checks if current number is divisible by 3; if yes the string "Testers" is concatenated to string concat</em>

<em>                 </em> if i%3 == 0:

                       concat= concat+"Testers"

<em>This lines checks if current number is divisible by 2; if yes the string "Coders" is concatenated to string concat</em>

                 if i%5 == 0:

                       concat= concat+"Coders"

<em>The concatenated string is printed using this line</em>

                 print(concat)

This variable concat is intialized back to an empty string

                 concat = ""

The main method starts here

N = int(input("Enter a positive integer: "))

This line calls the defined function solution

solution(N)

6 0
3 years ago
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