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Kryger [21]
3 years ago
7

Write a program called interleave that accepts two ArrayLists of integers list1 and list2 as parameters and inserts the elements

of list2 into list1 at alternating indexes. If the lists are of unequal length, the remaining elements of the longer list are left at the end of list1.
Computers and Technology
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

Explanation code is given below along with step by step comments!

Explanation:

// first we create a function which accepts two ArrayList of integers named list1 and list2

public static void interleave(ArrayList<Integer> list1, ArrayList<Integer> list2)

{

// we compare the size of list1 and list2 to get the minimum of two and store it in variable n

   int n = Math.min(list1.size(), list2.size());

   int i;

// here we are getting the elements from list2 n times then we add those elements in the list1 alternating (2*i+1)

   for (i = 0; i < n; i++)

     {

       int x = list2.get(i);

       list1.add(2 * i + 1, x);

      }

// if the size of list1 and list2 is same then program stops here else we need to append extra elements at the end of list1

// then we check if the size of list2 is greater than list1 then simply add the remaining elements into list1

   if (i < list2.size())

{

       for (int j = i; j < list2.size(); j++)

           {

                list1.add(list2.get(j));

            }  

     }  

}

Sample Output:

list1=[1, 2, 3]

list2=[5, 6, 7, 8, 9]

list1=[1, 5, 2, 6, 3, 7, 8, 9]

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Write a function stats that takes an array and the number of elements in the array. Then, it computes and prints the minimum val
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Answer:

#include<iostream>

using namespace std;

void stats(double [],int);  

int main()

{

int totalElements,i;

 

cout<<"Enter total elements:"<<endl;

cin>>totalElements;

double array[totalElements];

cout<<"Enter the elements in array:"<<endl;

for(i=0;i<totalElements;i++)

cin>>array[i];

stats(array,totalElements);  

}

void stats(double array[],int totalElements)

{

int i;

double minimum,maximum;

double Sum=0.0,average=0.0;

minimum=array[0],maximum=array[0];

for(i=0;i<totalElements;i++)

{

if(array[i]>maximum)

maximum=array[i];

if(array[i]<minimum)

minimum=array[i];

Sum+=array[i];

}

average=Sum/totalElements;

cout<<"Test: ";

for(i=0;i<totalElements;i++)

cout << fixed << setprecision(2) <<array[i]<<" ";

cout<<endl;

cout <<"minimum:"<< fixed << setprecision(2) <<minimum<<endl;

cout <<"maximum:"<< fixed << setprecision(2) <<maximum<<endl;

cout <<"average:"<< fixed << setprecision(2) <<average<<endl;

 

}

Explanation:

  • Loop through the total elements to get the input from user and call the stats function.
  • In the stats function check whether a number is maximum, minimum or average.
  • Calculate the average by finding the sum of all the numbers in array  and dividing it by total numbers.
  • Finally display the results.

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3 years ago
Where is the insert function button found in microsoft excel?
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A nonpipelined system takes 300ns to process a task. The same task can be processed in a 5-segment pipeline with a clock cycle o
Vladimir79 [104]

Answer:

Pipelined architecture is 1.667 times more speedy than non pipelined architecture.

Explanation:

There are five stages to to complete an instruction to execute in pipeline architecture. as

Stage 1:  Instruction Fetch

Stage 2: Instruction Decode

Stage 3: Execute

Stage 4: Memory Access

Stage 5 : Write Back

The pipeline architecture processed instruction as given below. Each Stage take 1 clock cycle, which is represented as CC.

                Stage 1           Stage 2         Stage 3         Stage 4            Stage 5

I1:              1cc                  2cc                 3cc                4cc                   5cc

I2              2cc                 3cc                  4cc               5cc                   6cc

I3              3cc                  4cc                 5cc              6cc                    7cc

I4              4cc                 5cc                  6cc              7cc                    8cc

each four Instructions set takes 8 Clock Cycles.

I5              9cc                  10cc                11cc               12cc                 13cc

I6              10cc                 11cc                  12cc             13cc                 14cc

I7              11cc                  12cc                 13cc              14cc                 15cc

I8              12cc                 13cc                  14cc              15cc                16cc

.

.

.

I97:             193cc            194cc              195cc              196cc               197cc

I98             194cc             195cc             196cc               197cc               198cc

I99            195cc             196cc              197cc              198cc               199cc

I100         196cc                197cc             198cc            199cc                 200cc

<u>Pipelined architecture Time Calculation</u>

So in pipeline architecture 100 instructions takes 200 clock cycles to execute.

1 Clock Cycle time period is = 60ns.

200 Clock Cycle =  60ns * 200 = 12000 ns

<u>Non Pipelined architecture Time Calculation</u>

1 task Time period= 300ns

100 task time period = 100 * 300 =  30000 ns

The difference between pipelined and non pipelined architecture time period is = 30000 - 12000 = 18000 ns

<u>Speedup ratio</u>

<em>Speed up ratio= 30000/18000 =  1.667</em>

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Software on your computer is taking a long time to load. What could help solve this problem?
Andrei [34K]

Answer: close the program and reopen it.

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Reboot the computer.

Check for known issues on the web or updates to the software.

Use Task Manager (Ctrl+Alt+DEL) if program is non-responsive to "end task."

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KatRina [158]

Answer:

Written in C++

void number(int n){

if(n%2 == 0)

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else

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}

Explanation:

The programming language is not stated.

However, I answered using C++

This line defines the function as void

void number(int n){

This line checks if the number is even

if(n%2 == 0)

If yes, it doubles the number and prints the output

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If otherwise,

else

It multiplies the number by 5 and prints the output

cout<<5 * n;

}

To call the function from main, use:

number(n);

Note than n must be declared as integer

See attachment

Download cpp
6 0
3 years ago
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