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1 year ago

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Without using any additional variables, and without changing the values of ndays or the elements of the parkingTickets array, wr

ite some code that results in mostTickets containing the largest value found in parkingTickets.

Computers and Technology

1 answer:

dusya [7]1 year ago

7 0

Answer:

Explanation:

mostTickets=0;

for (k=0; k< ndays; k++){

if (parkingTickets[k]>mostTickets) mostTickets=parkingTickets[k];

}

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You have been tasked with building a URL file validator for a web crawler. A web crawler is an application that fetches a web pa

mixas84 [53]

Answer:

Python Code:

def validate_url(url):

#Creating the list of valid protocols and file name extensions

valid_protocols = ['http', 'https', 'ftp']

valid_fileinfo = ['.html', '.csv', '.docx']

#splitting the url into two parts

url_split = url.split('://')

isProtocolValid = False

isFileValid = False

#iterating over the valid protocols and file names for validity

for x in valid_protocols:

if x in url_split[0]:

isProtocolValid = True

break

for x in valid_fileinfo:

if x in url_split[1]:

isFileValid = True

break

#Returning the result if the URL has both valid protocol and file extension

return (isProtocolValid and isFileValid)

url = input("Enter an URL: ")

print(validate_url(url))

Explanation:

The image of the output code is attached. Hope it helps.

4 0

2 years ago

A contact list is a place where you can store a specific contact with other associated information such as a phone number, email

NikAS [45]

Answer:

C++.

Explanation:

<em>Code snippet.</em>

#include <map>

#include <iterator>

cin<<N;

cout<<endl;

/////////////////////////////////////////////////

map<string, string> contacts;

string name, number;

for (int i = 0; i < N; i++) {

cin<<name;

cin<<number;

cout<<endl;

contacts.insert(pair<string, string> (name, number));

}

/////////////////////////////////////////////////////////////////////

map<string, string>::iterator it = contacts.begin();

while (it != contacts.end()) {

name= it->first;

number = it->second;

cout<<word<<" : "<< count<<endl;

it++;

}

/////////////////////////////////////////////////////////////////////////////////////////////////////////

I have used a C++ data structure or collection called Maps for the solution to the question.

Maps is part of STL in C++. It stores key value pairs as an element. And is perfect for the task at hand.

8 0

2 years ago

Which of the following is not one of the four criteria for evaluating websites?

Flura [38]

Answer:validity

Explanation:

Because it dont sound right

5 0

2 years ago

Suppose that the data mining task is to cluster points (with (x, y) representing location) into three clusters, where the points

solong [7]

Answer:

Explanation:

K- is the working procedure:

It takes n no. of predefined cluster as input and data points.

It also randomly initiate n centers of the clusters.

In this case the initial centers are given.

Steps you can follow

Step 1. Find distance of each data points from each centers.

Step 2. Assign each data point to the cluster with whose center is nearest to this data point.

Step 3. After assigning all data points calculate center of the cluster by taking mean of data points in cluster.

repeat above steps until the center in previous iteration and next iteration become same.

A1(4,8), A2(2, 4), A3(1, 7), B1(5, 4), B2(5,7), B3(6, 6), C1(3, 7), C2(7,8)

Centers are X1=A1, X2=B1, X3=C1

A1 will be assigned to cluster1, B1 will be assigned to cluster2 ,C1 will be assigned to cluster3.

Go through the attachment for the solution.

5 0

2 years ago

5. Assume a computer has a physical memory organized into 64-bit words. Give the word address and offset within the word for eac

spin [16.1K]

Answer:

see explaination and attachment

Explanation:

5.

To convert any byte address, By, to word address, Wo, first divide By by No, the no. of bytes/word, and ignores the remainder. To calculate a byte offset, O, in word, calculate the remainder of By divided by No.

i) 0 : word address = 0/8 = 0 and offset, O = 0 mod 8 = 0

ii) 9 : word address = 9/8 = 1 and offset, O = 9 mod 8 = 1

iii) 27 : word address = 27/8 = 3 and offset, O = 27 mod 8 = 3

iv) 31 : word address = 31/8 = 3 and offset, O = 31 mod 8 = 7

v) 120 : word address = 120/8 = 15 and offset, O = 120 mod 8 = 0

vi) 256 :word address = 256/8 = 32 and offset, O = 256 mod 8 = 0

6. see attachment for the python programming screen shot and output

4 0

2 years ago