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Andrej [43]
3 years ago
7

F. Practical Questions

Computers and Technology
1 answer:
irinina [24]3 years ago
6 0

Answer:

Explanation:

public class rectangle{

public static void main(String args[])

{

int width=5;

int height=10;

int area=width*height;

System.out.println("Area of rectangle="+area);

}

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In theory, a(n) _____ can be an independent centralized database management system with proper interfaces to support remote acce
Vlad [161]

Answer:

Transaction Manger

Explanation:

Transaction Manger

Transaction Manager is responsible for transaction between resources that are available. It works as interface to support remote access form other systems in Data base management system.

8 0
3 years ago
How do keygens work?
Artist 52 [7]
What are keygens please

3 0
4 years ago
What is a benefit of using a combination of symmetrical and asymmetrical cryptography when transferring data across the internet
Luden [163]

Answer:

Explanation:

Asymmetric cryptography, is a cryptographic system that uses two different types of keys both public and private keys. Symmetric cryptography on the other hand uses the same type of key for both encryption. The benefit of using a combination when transferring data across the internet is that Asymmetrical cryptography is necessary to establish a secure connection while symmetrical cryptography can enhance the speed of the data transmission after connection has been established, thus both provide a unique benefit.

7 0
4 years ago
Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the middl
Stels [109]

Answer:

Check the explanation

Explanation:

#include<iostream.h>

#include<algorithm.h>

#include<climits.h>

#include<bits/stdc++.h>

#include<cstring.h>

using namespace std;

int partition(int arr[], int l, int r, int k);

int kthSmallest(int arr[], int l, int r, int k);

void quickSort(int arr[], int l, int h)

{

if (l < h)

{

// Find size of current subarray

int n = h-l+1;

 

// Find median of arr[].

int med = kthSmallest(arr, l, h, n/2);

 

// Partition the array around median

int p = partition(arr, l, h, med);

 

// Recur for left and right of partition

quickSort(arr, l, p - 1);

quickSort(arr, p + 1, h);

}

int findMedian(int arr[], int n)

{

sort(arr, arr+n); // Sort the array

return arr[n/2]; // Return middle element

}

int kthSmallest(int arr[], int l, int r, int k)

{

// If k is smaller than number of elements in array

if (k > 0 && k <= r - l + 1)

{

int n = r-l+1; // Number of elements in arr[l..r]

 

// Divide arr[] in groups of size 5, calculate median

// of every group and store it in median[] array.

int i, median[(n+4)/5]; // There will be floor((n+4)/5) groups;

for (i=0; i<n/5; i++)

median[i] = findMedian(arr+l+i*5, 5);

if (i*5 < n) //For last group with less than 5 elements

{

median[i] = findMedian(arr+l+i*5, n%5);

i++;

}

int medOfMed = (i == 1)? median[i-1]:

kthSmallest(median, 0, i-1, i/2);

int pos = partition(arr, l, r, medOfMed);

if (pos-l == k-1)

return arr[pos];

if (pos-l > k-1) // If position is more, recur for left

return kthSmallest(arr, l, pos-1, k);

return kthSmallest(arr, pos+1, r, k-pos+l-1);

}

return INT_MAX;

}

void swap(int *a, int *b)

{

int temp = *a;

*a = *b;

*b = temp;

}

int partition(int arr[], int l, int r, int x)

{

// Search for x in arr[l..r] and move it to end

int i;

for (i=l; i<r; i++)

if (arr[i] == x)

break;

swap(&arr[i], &arr[r]);

 

// Standard partition algorithm

i = l;

for (int j = l; j <= r - 1; j++)

{

if (arr[j] <= x)

{

swap(&arr[i], &arr[j]);

i++;

}

}

swap(&arr[i], &arr[r]);

return i;

}

 

/* Function to print an array */

void printArray(int arr[], int size)

{

int i;

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

cout << arr[i] << " ";

cout << endl;

}

 

// Driver program to test above functions

int main()

{

float a;

clock_t time_req;

int arr[] = {1000, 10, 7, 8, 9, 30, 900, 1, 5, 6, 20};

int n = sizeof(arr)/sizeof(arr[0]);

quickSort(arr, 0, n-1);

cout << "Sorted array is\n";

printArray(arr, n);

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(i*i*i*i);

}

time_req = clock()- time_req;

cout << "Processor time taken for multiplication: "

<< (float)time_req/CLOCKS_PER_SEC << " seconds" << endl;

 

// Using pow function

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(pow(i, 4));

}

time_req = clock() - time_req;

cout << "Processor time taken in pow function: "

<< (float)time_req/CLOCKS_PER_S

return 0;

}

..................................................................................................................................................................................................................................................................................................................................

OR

.......................

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

 

// Swap utility

void swap(long int* a, long int* b)

{

int tmp = *a;

*a = *b;

*b = tmp;

}

 

// Bubble sort

void bubbleSort(long int a[], long int n)

{

for (long int i = 0; i < n - 1; i++) {

for (long int j = 0; j < n - 1 - i; j++) {

if (a[j] > a[j + 1]) {

swap(&a[j], &a[j + 1]);

}

}

}

}

 

// Insertion sort

void insertionSort(long int arr[], long int n)

{

long int i, key, j;

for (i = 1; i < n; i++) {

key = arr[i];

j = i - 1;

 

// Move elements of arr[0..i-1], that are

// greater than key, to one position ahead

// of their current position

while (j >= 0 && arr[j] > key) {

arr[j + 1] = arr[j];

j = j - 1;

}

arr[j + 1] = key;

}

}

 

// Selection sort

void selectionSort(long int arr[], long int n)

{

long int i, j, midx;

 

for (i = 0; i < n - 1; i++) {

 

// Find the minimum element in unsorted array

midx = i;

 

for (j = i + 1; j < n; j++)

if (arr[j] < arr[min_idx])

midx = j;

 

// for plotting graph with integer values

printf("%li, %li, %li, %li\n",

n,

(long int)tim1[it],

(long int)tim2[it],

(long int)tim3[it]);

 

// increases the size of array by 10000

n += 10000;

}

 

return 0;

}

8 0
4 years ago
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You gotta do the paragraph group its that one
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3 years ago
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