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astraxan [27]
3 years ago
6

Resize vector countDown to have newSize elements. Populate the vector with integers {newSize, newSize - 1, ..., 1}. Ex: If newSi

ze = 3, then countDown = {3, 2, 1}, and the sample program outputs:
#include
#include
using namespace std;
int main() {
vector countDown(0);
int newSize = 0;
int i = 0;
newSize = 3;
STUDENT CODE
for (i = 0; i < newSize; ++i) {
cout << countDown.at(i) << " ";
}
cout << "Go!" << endl;
return 0;
}
Computers and Technology
1 answer:
strojnjashka [21]3 years ago
4 0

Answer:

Following are the code to the given question:

#include <iostream>//defining header file

#include <vector>//defining header file

#include <numeric>//defining header file

using namespace std;

int main()//main method

{

vector<int> countDown(0);//defing an integer array variable countDown

int newSize = 0;//defing an integer variable newSize that holds a value 0

int i = 0;//defing an integer variable i that initialize with 0

newSize = 3;// //use newSize to hold a value

countDown.resize(newSize,0);// calling the resize method that accepts two parameters

for (i = 0; i < newSize; ++i)//defining a loop that reverse the integer value

{

countDown[i] = newSize -i;//reverse the value and store it into countDown array

cout << countDown.at(i) << " ";//print reverse array value

}

cout << "Go!" << endl;//print message

return 0;

}

Output:

3 2 1 Go!

Explanation:

In the given code inside the main method an integer array "countDown" and two integer variables "newSize and i" is declared that initializes a value that is 0.

In the next step, an array is used that calls the resize method that accepts two parameters, and define a for a loop.

In this loop, the array is used to reverse the value and print its value with the message.

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4 0
2 years ago
⦁ Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. ⦁ What is the maximum val
timurjin [86]

Answer:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

Explanation:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

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2 years ago
ven int variables k and total that have already been declared, use a while loop to compute the sum of the squares of the first 5
monitta

Answer:

Following are statement is given below

int  k=1,total=0; // variable declaration

while(k<50) // iterating the while loop

{

   total=total+k*k;//  calculating the square

   k=k+1; // increments the value of k by 1    

}

Explanation:

Following are the description of Statement.

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