Which advertising medium has the widest reach on a global front?

Define the instance method inc_num_kids() for PersonInfo. inc_num_kids increments the member data num_kids. Sample output for th

kow [346]

Answer:

This is the instance method inc_num_kids()

def inc_num_kids(self):

self.num_kids += 1

Explanation:

The first line is the definition of function inc_num_kids() which has a parameter self. self is basically an instance of class PersonInfo

self.num_kids += 1 this statement incremented the data member num_kids by 1. This means that the instance method inc_num_kids, using instance self, increments the value of num_kids data member by 1.

The complete program is as following:

class PersonInfo: #class name PersonInfo

def __init__(self): #constructor of the class

self.num_kids = 0 # data member num_kids

def inc_num_kids(self): # method inc_num_kids

self.num_kids += 1 #increments num_kids by 1

person1 = PersonInfo() #creates object person of class PersonInfo

print('Kids:', person1.num_kids) # prints num_kids value before increment

person1.inc_num_kids() #calls inc_num_kids method to increment num_kids

print('New baby, kids now:', person1.num_kids) #prints the value of num_kids after increment by inc_num_kids method

The person1 object first access the initial value of num_kids initialized by the constructor. So the first value of num_kids is 0. self.num_kids = 0 This statement assigns the value 0 to num_kids data member. Next the method inc_num_kids() using the object person1. Now when this method is called the value of data member num_kids is incremented to 1.

So previously the value was 0 which now increments by 1 and becomes 1. Last print statement displays this new value which is incremented by the inc_num_kids() method.

Hence the output is:

Kids: 0

New baby, kids now: 1

The program along with its output is attached.

7 0

2 years ago

Write two methods: encrypt and decrypt. encrypt should #take as input a string, and return an encrypted version #of it according

Harman [31]

Answer:

The code is given below with appropriate comments

Explanation:

CIPHER = (("D", "A", "V", "I", "O"),

("Y", "N", "E", "R", "B"),

("C", "F", "G", "H", "K"),

("L", "M", "P", "Q", "S"),

("T", "U", "W", "X", "Z"))

# Add your code here!

def encrypt(plaintext):

theList = []

for char in plaintext:

if char.isalpha():

char = char.upper()

if char == "J":

char = "I"

theList.append(char)

if len(theList) % 2 == 1:

theList.append("X")

for i in range(0, len(theList), 2):

if theList[i] == theList[i + 1]:

theList[i + 1] = "X"

findex = [-1, -1]

sindex = [-1, -1]

for j in range(len(CIPHER)):

for k in range(len(CIPHER)):

if theList[i] == CIPHER[j][k]:

findex = [j, k]

if theList[i + 1] == CIPHER[j][k]:

sindex = [j, k]

# same row

if (findex[0] == sindex[0]):

findex[1] += 1

sindex[1] += 1

if findex[1] == 5:

findex[1] = 0

if sindex[1] == 5:

sindex[1] = 0

theList[i] = CIPHER[findex[0]][findex[1]]

theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

# same column

elif (findex[1] == sindex[1]):

findex[0] += 1

sindex[0] += 1

if findex[0] == 5:

findex[0] = 0

if sindex[0] == 5:

sindex[0] = 0

theList[i] = CIPHER[findex[0]][findex[1]]

theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

else:

theList[i] = CIPHER[findex[0]][sindex[1]]

theList[i + 1] = CIPHER[sindex[0]][findex[1]]

return "".join(theList)

def decrypt(ciphertext):

theString = ""

findex = [-1, -1]

sindex = [-1, -1]

for i in range(0, len(ciphertext), 2):

for j in range(len(CIPHER)):

for k in range(len(CIPHER)):

if ciphertext[i] == CIPHER[j][k]:

findex = [j, k]

if ciphertext[i + 1] == CIPHER[j][k]:

sindex = [j, k]

if (findex[0] == sindex[0]):

findex[1] -= 1

sindex[1] -= 1

if findex[1] == -1:

findex[1] = 4

if sindex[1] == -1:

sindex[1] = 4

theString += CIPHER[findex[0]][findex[1]]

theString += CIPHER[sindex[0]][sindex[1]]

# same column

elif (findex[1] == sindex[1]):

findex[0] -= 1

sindex[0] -= 1

if findex[0] == -1:

findex[0] = 4

if sindex[0] == -1:

sindex[0] = 4

theString += CIPHER[findex[0]][findex[1]]

theString += CIPHER[sindex[0]][sindex[1]]

else:

theString += CIPHER[findex[0]][sindex[1]]

theString += CIPHER[sindex[0]][findex[1]]

return theString

# Below are some lines of code that will test your function.

# You can change the value of the variable(s) to test your

# function with different inputs.

#

# If your function works correctly, this will originally

# print: QLGRQTVZIBTYQZ, then PSHELXOWORLDSX

print(encrypt("PS. Hello, worlds"))

print(decrypt("QLGRQTVZIBTYQZ"))

5 0

2 years ago

Assume that you have a list of n home maintenance/repair tasks (numbered from 1 to n ) that must be done in numeric order on you

tangare [24]

Answer:

Explanation:

(a) The recurrence relation for the given problem is :

T(n) = T(n-1) + T(n-4) + 1

(b) The O(n) time recursive algorithm with memoization for the above recurrence is given below :

Create a 1-d array 'memo' of size, n (1-based indexing) and initialize its elements with -1.

func : a recursive function that accepts the cost array and startingJobNo and returns the minimum cost for doing the jobs from startingJobNo to n.

Algorithm :

func(costArr[], startingJobNo){

if(startingJobNo>n)

then return 0

END if

if(memo[startingJobNo] != -1)

then return memo[startingJobNo];

END if

int ans1 = func(costArr, startingJobNo+1) + costArr[startingJobNo]

int ans2 = func(costArr, startingJobNo+4) + h

memo[startingJobNo] = min(ans1,ans2);

return memo[startingJobNo];

}

(c)

First, Create a 1-d array 'dp' of size, N+1.

dp[0] = 0

bottomUp(int c[]){

for i=1 till i = n

DO

dp[i] = min(dp[i-1] + c[i], dp[max(0,i-4)] + h);

END FOR

return dp[n];

}

(d)

Modifying the algorithm given in part (b) as follows to know which job to do yourself and in which jobs we need to hire a handyman.

First, Create a 1-d array 'memo' of size, n (1-based indexing) and initialize its elements with -1.

Next, Create another 1-d array 'worker' of size,n (1-based indexing) and initialize its elements with character 'y' representing yourself.

Algorithm :

func(costArr[], startingJobNo){

if(startingJobNo>n)

then return 0

END if

if(memo[startingJobNo] != -1)

then return memo[startingJobNo];

END if

int ans1 = func(costArr, startingJobNo+1) + costArr[startingJobNo]

int ans2 = func(costArr, startingJobNo+4) + h

if(ans2 < ans1)

THEN

for (i = startingJobNo; i<startingJobNo+4 and i<=n; i++)

DO

// mark worker[i] with 'h' representing that we need to hire a mechanic for that job

worker[i] = 'h';

END for

END if

memo[startingJobNo] = min(ans1,ans2);

return memo[startingJobNo];

}

//the worker array will contain 'y' or 'h' representing whether the ith job is to be done 'yourself' or by 'hired man' respectively.

4 0

2 years ago

How can migrating to an enterprise platform solution help a business improve employee productivity?

Lorico [155]

Answer:

By allowing all employees to access and share the same stored dats and by increasing the number of wmployees in business activities

Explanation:

#<em>carry on learning</em>

5 0

2 years ago

This program will calculate the rise in ocean levels over 5, 10, and 50 years, Part of the program has been written for you. The

GREYUIT [131]

Answer:

Here is the C++ program:

#include <iostream> //to use input output functions

using namespace std; //to identify objects cin cout

int main(){ //start of main method

double risingLevel; //declares a double type variable to hold rising level

cin>>risingLevel; //reads the value of risingLevel from user

cout<<"Level: "<<risingLevel<<endl; //displays the rising level

cout << "Years: 5, Rise: " << risingLevel * 5<<endl; //displays the rise in ocean level over 5 years

cout << "Years: 10, Rise: " << risingLevel * 10<<endl; //displays the rise in ocean level over 10 years

cout << "Years: 50, Rise: " << risingLevel * 50<<endl; //displays the rise in ocean level over 50 years

}

Explanation:

The program works as follows:

Lets say the user enters rising level 2.1 then,

risingLevel = 2.1

Now the first print (cout) statement :

cout<<"Level: "<<risingLevel<<endl; prints the value of risingLevel on output screen. So this statement displays:

Level: 2.1

Next, the program control moves to the statement:

cout << "Years: 5, Rise: " << risingLevel * 5<<endl;

which computes the rise in ocean levels over 5 years by formula:

risingLevel * 5 = 2.1 * 5 = 10.5

It then displays the computed value on output screen. So this statement displays:

Years: 5, Rise: 10.5 Next, the program control moves to the statement:

cout << "Years: 10, Rise: " << risingLevel * 10<<endl;

which computes the rise in ocean levels over 10 years by formula:

risingLevel * 10 = 2.1 * 10 = 21

It then displays the computed value on output screen. So this statement displays:

Years: 10, Rise: 21

Next, the program control moves to the statement:

cout << "Years: 50, Rise: " << risingLevel * 50<<endl;

which computes the rise in ocean levels over 50 years by formula:

risingLevel * 50 = 2.1 * 50 = 105

It then displays the computed value on output screen. So this statement displays:

Years: 50, Rise: 105 Hence the output of the entire program is:

Level: 2.1 Years: 5, Rise: 10.5 Years: 10, Rise: 21 Years: 50, Rise: 105

The screenshot of the program and its output is attached.

4 0

2 years ago