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vova2212 [387]
3 years ago
13

What is a typical use for a MAN?

Computers and Technology
1 answer:
Leto [7]3 years ago
6 0
A.to connect devices in five offices adjacent to each other
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I think it's Graph Search because there isn't any graphs in Facebook. That is my opinion. I don't use social media.
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3 years ago
You are asked to simulate a binary search algorithm on an array of random values.An array is the list of similar type of element
Alex Ar [27]

Answer:

Explanation:

Problem statement:

to simulate a binary search algorithm on an array of random values.

Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.

Input/output description

Input:

Size of array: 4

Enter array:10  20 30 40

Enter element to be searched:40

The Output will look like this:

Element is present at index 3

Algorithm and Flowchart:

We basically ignore half of the elements just after one comparison.

Compare x with the middle element.

If x matches with middle element, we return the mid index.

Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.

Else (x is smaller) recur for the left half.

The Flowchart can be seen in the first attached image below:

Program listing:

// C++ program to implement recursive Binary Search

#include <bits/stdc++.h>

using namespace std;

// A recursive binary search function. It returns

// location of x in given array arr[l..r] is present,

// otherwise -1

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

{

   if (r >= l) {

       int mid = l + (r - l) / 2;

       // If the element is present at the middle

       // itself

       if (arr[mid] == x)

           return mid;

       // If element is smaller than mid, then

       // it can only be present in left subarray

       if (arr[mid] > x)

           return binarySearch(arr, l, mid - 1, x);

       // Else the element can only be present

       // in right subarray

       return binarySearch(arr, mid + 1, r, x);

   }

   // We reach here when element is not

   // present in array

   return -1;

}

int main(void)

{ int n,x;

cout<<"Size of array:\n";

cin >> n;

int arr[n];

cout<<"Enter array:\n";

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

{ cin >> arr[i]; }

cout<<"Enter element to be searched:\n";

cin>>x;

int result = binarySearch(arr, 0, n - 1, x);

   (result == -1) ? cout << "Element is not present in array"

                  : cout << "Element is present at index " << result;

   return 0;

}

The Sample test run of the program can be seen in the second attached image below.

Time(sec) :

0

Memory(MB) :

3.3752604177856

The Output:

Size of array:4

Enter array:10  20 30 40

Enter element to be searched:40

Element is present at index 3

Conclusions:

Time Complexity:

The time complexity of Binary Search can be written as

T(n) = T(n/2) + c  

The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is Theta(Logn).

Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.

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its a  return address

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How to unblock a website on a school computer? ASAP who's ever works will get Brainliest and will get a follow.
Angelina_Jolie [31]

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You can't unblock them without admin access to the router. The only way to visit such sites is to find a VPN that is not also blocked by the school.

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. The variable numbers is a 2-dimensional array of type int; output all of its elements that are strictly greater than 10
lana66690 [7]

Answer:

// here is code in c++.

#include <bits/stdc++.h>

using namespace std;

int main()

{

   // variable to read row and column of 2-d array

  int r,c;

  cout<<"enter the number of row:";

  // read number of row

  cin>>r;

  cout<<"enter the number of column:";

  // read number of column

  cin>>c;

     // create an array of size rxc

  int arr[r][c];

  cout<<"enter the elements of the array:"<<endl;

     // read the elements of 2-d array

  for(int i=0;i<r;i++)

  {

      for(int j=0;j<c;j++)

      {

          cin>>arr[i][j];

      }

  }

  cout<<"numbers which are greater than 10 in 2-d array"<<endl;

  for(int i=0;i<r;i++)

  {

      for(int j=0;j<c;j++)

      {

          // if element is greater than 10, print it

          if(arr[i][j]>10)

          cout<<arr[i][j]<<" ";

      }

      cout<<endl;

  }

  return 0;

}

Explanation:

Read the size of 2-d array i.e row and column.Create a 2-d array of size rxc of integer type.Read the elements of the array. Then traverse the array and if an element if greater than 10 then print that element.

Output:

enter the number of row:3                                                                                                        

enter the number of column:3                                                                                                    

enter the elements of the array:                                                                                                

23 45 53 78 1 9 6 8 77                                                                                                                                

numbers which are greater than 10 in 2-d array                                                                                  

23 45 53                                                                                                                        

78                                                                                                                              

77        

3 0
3 years ago
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