<span>A peer-to-peer network (or workgroup) consists of multiple windows computers that share information, but no computer on the network serves as an authoritative source of user information. domain-based peer-to-peer server-based centralized</span>
Answer:
True: In binary search algorithm, we follow the below steps sequentially:
Input: A sorted array B[1,2,...n] of n items and one item x to be searched.
Output: The index of x in B if exists in B, 0 otherwise.
- low=1
- high=n
- while( low < high )
- { mid=low + (high-low)/2
- if( B[mid]==x)
- {
- return(mid) //returns mid as the index of x
- }
- else
- {
- if( B[mid] < x) //takes only right half of the array
- {
- low=mid+1
- }
- else // takes only the left half of the array
- {
- high=mid-1
- }
- }
- }
- return( 0 )
Explanation:
For each iteration the line number 11 or line number 15 will be executed.
Both lines, cut the array size to half of it and takes as the input for next iteration.
Answer:
The answer to this question is given below in the explanation section.
Explanation:
The correct answer to this question is the planning stage. Because the planning stage represents the development of documents that provide the basis for acquiring the resources and for developing the requirement document. at this stage, you plan about what you are going to develop and how to develop it. At this stage, you come out mainly with two documents i.e. project proposal and requirement document.
Other options are not correct because:
In the project management, after planning, you will start designing the product, and after designing you start developing the product, and at the implementation stage, you implement or deploy the product to the customer or to the client. The requirement document that is developed at the planning stage can be used in the later stages of the project.
Answer:
The program in Python is as follows:
num1 = int(input())
num2 = int(input())
if num1 >=0 and num2 >= 0:
print(num1+num2)
elif num1 <0 and num2 < 0:
print(num1*num2)
else:
if num1>=0:
print(num1**2)
else:
print(num2**2)
Explanation:
This gets input for both numbers
num1 = int(input())
num2 = int(input())
If both are positive, the sum is calculated and printed
<em>if num1 >=0 and num2 >= 0:</em>
<em> print(num1+num2)</em>
If both are negative, the products is calculated and printed
<em>elif num1 <0 and num2 < 0:</em>
<em> print(num1*num2)</em>
If only one of them is positive
else:
Calculate and print the square of num1 if positive
<em> if num1>=0:</em>
<em> print(num1**2)</em>
Calculate and print the square of num2 if positive
<em> else:</em>
<em> print(num2**2)</em>
