Which of these is not the correct method for moving text in a document in Word 2016?
1 answer:
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Answer:
Option A:
<input name="name" id="id" type="number" value="value" step="value" min="value" max="value" />
Explanation:
Spinner control is a graphical control element where user can adjust the value by pressing up or down arrow button. An example is given in the attached image.
In HTML, one of the key attributes we must use to create a spinner control is "step". The attribute "step" is required to specify the interval of the step value when user press the up or down arrow button.
If we set the attribute values as follows:
- type = "number"
- value = 2
- min = 0
- max = 10
The setting above will give a spinner control with a range of legal numbers between 0, 2, 4, 6, 8 and 10.
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.
Answer:
Following are the solution to this question:
Explanation:
In point A:
It applies the AND operation with the subnet-mask that is
This corresponds to a 128.96.39.0 subnet. This is why the router sent the packet to 0.
In point B:
It applies the AND operation with the subnet-mask that is
This suits the 128.96.40.0 subnet. The router then sent R2 to the packet. The packet.
In point C:
It applies the AND operation with the subnet-mask that is
It applies the AND operation with the subnet-mask that is
All of the matches of its sublinks above.
Its default R4 router was therefore sent to the router by router.
In point D:
It applies the AND operation with the subnet-mask that is
It matches the 192.4.153.0 subnet. It router has sent packet to R3 then.
In point E:
It applies the AND operation with the subnet-mask that is
There, neither of the subnet inputs correlate with the ip address with the same subnet mask.
It applies the AND operation with the subnet-mask that is
It does not correlate to the subnet entries for same subnet mask, and therefore neither of the entries matches their corresponding ip address entries.
Mask of the subnet. That packet would therefore be sent to the default R4 router by the router.