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Marianna [84]
3 years ago
9

Nuclear batteries use devices called thermocouples, which convert the ____ of a nuclear reaction into electricity.

Computers and Technology
1 answer:
pshichka [43]3 years ago
8 0
The answer would be c
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Read the following characteristic:
Ivanshal [37]

Answer:

A goal of procedural programming

An advantage of object oriented programming

A disadvantage of object oriented programming

A drawback of procedural programming

3 0
2 years ago
What are some short term goals for getting a bachelors degree
snow_tiger [21]
1. stay determined and motivated
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3 0
3 years ago
Int [] val = { 3, 10, 44 };
Shtirlitz [24]

Answer:

4) 3 11 44

Explanation:

Given data

int [] val = { 3, 10, 44 };

The total number of parameters of given array are 3, so total length of array is also 3.

The indexing of array starts with '0', Therefore the indexes of array with length zero are: {0,1,2}

The value of array at index 0 is = 3

similarly,

value at index 1 = 10

value at index 2 = 44

Here, Int i = 1 is storing the value '1' in integer variable i.

In addition to that, any value of index 'i' of an array is selected using array[i].

Therefore,

val[i] is selecting the value of array located at index '1' because i = 1.

val[i] = val[1] = 10

val[i]+1 is selecting the value of array located at index 'i' that is (1) and adding 1 to it

=> val[i] = 10

=> val[i]+1 = 10+1 = 11

Finally,

val[i] = val[i]+1; is copying the val[i]+1 = 11 to value placed at index 1 (10). Hence, the output would be {3 11 44}. So 4th option is correct.

6 0
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.

8 0
3 years ago
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