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2 years ago

In Java please:

Computers and Technology

1 answer:

Nana76 [90]2 years ago

6 0

Answer:

See explaination

Explanation:

public class PalindromeChange {

public static void main(String[] args) {

System.out.println("mom to non palindrom word is: "+changTONonPalindrom("mom"));

System.out.println("mom to non palindrom word is: "+changTONonPalindrom("aaabbaaa"));

}

private static String changTONonPalindrom(String str)

{

int mid=str.length()/2;

boolean found=false;

char character=' ';

int i;

for(i=mid-1;i>=0;i--)

{

character = str.charAt(i);

if(character!='a')

{

found=true;

break;

}

if(!found)

{

for(i=mid+1;i<str.length();i++)

{

character = str.charAt(i);

if(character!='a')

{

found=true;

break;

}

// This gives the character 'a'

int ascii = (int) character;

ascii-=1;

str = str.substring(0, i) + (char)ascii+ str.substring(i + 1);

return str;

}

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Write a for loop that iterates from 1 to numbersamples to double any element's value in datasamples that is less than minvalue.

CaHeK987 [17]

Answer:

function dataSamples=AdjustMinValue(numberSamples, userSamples, minValue)

dataSamples=userSamples;

%for loop

for i=1:numberSamples

%checking if dataSamples value at index,i

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if dataSamples(i)<minValue

%set double of dataSamples value

dataSamples(i)= 2*dataSamples(i);

end

Explanation:

The given code is in MATLAB.

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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

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findex[1] = 4

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sindex[1] = 4

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

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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]]

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# 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"))

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2 years ago

What is the value of the variable named myNum after the statements that follow are executed? var myNum = 14; var yourNum = 4; my

Lostsunrise [7]

Answer:

Explanation:

Initially, myNum is equal to 14 and yourNum is equal to 4

Then, myNum is incremented by 1 and becomes 15

Also, yourNum is decremented by 1 and becomes 3

Finally, myNum is set to myNum x yourNum, 15 x 3 = 45

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2 years ago

5.15 LAB: Output values below an amount Write a program that first gets a list of integers from input. The input begins with an

IgorC [24]

Answer:

# the terminal display waiting for the user to enter the input

# the received input is assigned to user_input

user_input = input()

# the user_input is splitted based on space and is assigned to integerlist

integerlist = user_input.split(" ")

# the last element in the integerlist is assigned as threshold

threshold = int(integerlist[len(integerlist) - 1])

# a for loop that loop from index 1 to second to the last element in the list

# the loop compare each element with the threshold

# if element is less than threshold, it is displayed

# the loop start from index 1 because index 0 represent number of element

for i in range(1, (len(integerlist) - 1)):

if int(integerlist[i]) < threshold:

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Explanation:

The program is written in Python and well commented.

An sample of program output when it is executed is attached.

5 0

2 years ago

Read 2 more answers

What is the distance rn between the point of application of n⃗ and the axis of rotation? what is the distance rw between the poi

Mandarinka [93]

Complete Question.

A. Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical?

B. What is the distance rn between the point of application of n⃗ and the axis of rotation?

C. What is the distance rw between the point of application of w⃗ and the axis? enter your answers in meters separated by a comma?

Answer:

A. â = - 17rad/s²

B. rn = 0, meters

C. rw = 0.074, meters

Explanation:

Where;

l = length of the pencil

θ = angle between the vertical line and the pencil

â = angular acceleration.

The torque due to weight about the pivotal point is negative since it's clockwise;

t = - mg(l/2)sinØ ..... equation 1

torque with respect to angular acceleration;

t = Iâ ..... equation 2

The moment of Inertia of the pencil about one end;

I = (ml²)/3

Substituting I into equation 2;

t = [(ml²)/3]×â

Equating the equation, we have;

[(ml²)/3]×â = - mg(l/2)sinØ

â = (-3gsinØ)/2l

â = (-3*9.8*sin10°)/(2*0.15)

Angular acceleration, â = -17rad/s²

B. The normal force is acting at the normal force only, so the distance r n between the point of application of n ⃗ and the axis of rotation is zero because the axis of pencil is passing the same point of application w.

C. rn = (lcosØ)/2

rn = (0.15 * cos10°)/2

rn = (0.15 * 0.9848)/2

rn = (0.1477)/2

rn = 0.074m.

Since the gravity acts at exactly half of the length of pencil, distance r w between the point of application of w⃗ and the axis equals 0.074m.

8 0

2 years ago