All categories

2 years ago

Edhesive: Assignment 4: Evens and Odds

Computers and Technology

1 answer:

solniwko [45]2 years ago

6 0

Answer:

Written in Python

n = int(input("How many numbers do you need to check? "))

odd = 0

even = 0

for i in range(1,n+1):

num = int(input("Enter Number: "))

if num%2 == 0:

print(str(num)+" is an even number")

even = even + 1

else:

print(str(num)+" is an odd number")

odd = odd + 1

print("You entered "+str(even)+" even number(s).")

print("You entered "+str(odd)+" odd number(s).")

Explanation:

<em>I've added the full source code as an attachment where I use comments (#) as explanation</em>

Download txt

You might be interested in

Lucas put a lot of thought into the design for his company's new white paper. He made sure to include repeating design elements

Alenkasestr [34]

C. provide consistency

8 0

2 years ago

Read 2 more answers

Mitchell has noticed that his co-workers are unable to open attachments in the emails he sends. What is one possible reason for

hammer [34]

Maybe it could be the website or the email it self
<span />

8 0

2 years ago

Create an array w with values 0, 0.1, 0.2, . . . , 3. Write out w[:], w[:-2], w[::5], w[2:-2:6]. Convince yourself in each case

larisa86 [58]

Answer:

w = [i/10.0 for i in range(0,31,1)]

w[:] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]

w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]

w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]

w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]

Explanation:

List slicing (as it is called in python programming language ) is the creation of list by defining start, stop, and step parameters.

w = [i/10.0 for i in range(0,31,1)]

The line of code above create the list w with values 0, 0.1, 0.2, . . . , 3.

w[:] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]

since start, stop, and step parameters are not defined, the output returns all the list elements.

w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]

w[:-2] since stop is -2, the output returns all the list elements from the beginning to just before the second to the last element.

w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]

w[::5] since step is -2, the output returns the list elements at index 0, 5, 10, 15, 20, 25,30

w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]

the output returns the list elements from the element at index 2 to just before the second to the last element, using step size of 6.

6 0

2 years ago

Suppose that each row of an n×n array A consists of 1’s and 0’s such that, in any row i of A, all the 1’s come before any 0’s in

Anna71 [15]

Answer:

Check the explanation

Explanation:

Each row of nxn array A consists of 1’s and 0’s such that , in any row of A, all the 1’s come before any 0’s in that row.
Use binary search algorithm to find the index of the last 1 in a row.
Perform this process for each row.
Now, searching for last occurrence of 1 in a row will take O (log n) time.
There are n such rows, therefore total time will be O (n log n).

Complexity analysis:

The method would be to use binary search for each row to find the first zero starting with index of A[i][n/2+1].

Let’s say j=n/2.

The number of 1’s in a row would be j+1.

This would take O (log n).

An algorithm that divides by 2 until the number gets sufficiently small then it terminates in O (log n) steps.

As there are n rows the complexity would be O (n log n).

Pseudo-code:

A = [[1,0,0,0],[0,0,0,0],[1,1,1,1],[1,1,0,0]]

n=4

c=0

for i in range(n): # Loop in rows

j = n/2 # Search from middle index

while j>0: # Loop in column

if(A[i][j]==0): # search for first zero

if(A[i][j-1]==1): # confirm first zero

c = c+j # add 1's count to c

break

else: # reduce index by 1 or j/2

if(j/2 == 0):

j = j-1

else:

j = j - j/2

else: # increase index by 1 or j/2

if(j/2 == 0):

j = j+1

else:

j = j + j/2

if(j==n): # For all 1's

c = c+n

break

print c

8 0

2 years ago

Data is: a. Information endowed with relevance and purpose b. Set of specific objective facts or observations c. Some informatio

Maru [420]

Answer: b) Set of specific objective facts or observations

Explanation: Data is referred as the information that consumes the facts and figures . These facts and figure information is usually viewed or observed in the form of images, text, facts, graphs,reports etc.Data of any sort has to be stored for any use or analyzing in future so it is generally stored in the computer and other devices.Therefore,option(b) is the correct option.

4 0

2 years ago