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nordsb [41]
4 years ago
6

In this chapter, you saw an example of how to write an algorithm that determines whether a number is even or odd. Write a progra

m that generates 100 random numbers and keeps a count of how many of those random numbers are even, and how many of them are odd.]

Computers and Technology
2 answers:
vagabundo [1.1K]4 years ago
8 0

Answer:

// Program is written in Java Programming Language

// Comments are used for explanatory purpose

// Program starts here

public class RandomOddEve {

/** Main Method */

public static void main(String[] args) {

int[] nums = new int[100]; // Declare an array of 100 integers

// Store the counts of 100 random numbers

for (int i = 1; i <= 100; i++) {

nums[(int)(Math.random() * 10)]++;

}

int odd = 0, even = 0; // declare even and odd variables to 0, respectively

// Both variables will serve a counters

// Check for odd and even numbers

for(int I = 0; I<100; I++)

{

if (nums[I]%2 == 0) {// Even number.

even++;

}

else // Odd number.

{

odd++;

}

}

//.Print Results

System.out.print("Odd number = "+odd);

System.out.print("Even number = "+even);

}

yarga [219]4 years ago
3 0

Answer:

Complete python code along with comments to explain the whole code is given below.

Python Code with Explanation:

# import random module to use randint function

import random

# counter variables to store the number of even and odd numbers

even = 0

odd = 0

# list to store randomly generated numbers

random_num=[]

# A for loop is used to generate 100 random numbers

for i in range(100):

# generate a random number from 0 to 100 and append it to the list

   random_num.append(random.randint(0,100))

# check if ith number in the list is divisible by 2 then it is even

   if random_num[i] % 2==0:

# add one to the even counter

        even+=1

# if it is not even number then it must be an odd number

   else:

# add one to the odd counter

        odd+=1

# finally print the count number of even and odd numbers

print("Total Even Random Numbers are: ",even)

print("Total Odd Random Numbers are: ",odd)

Output:

Total Even Random Numbers are: 60

Total Odd Random Numbers are: 40

Total Even Random Numbers are: 54

Total Odd Random Numbers are: 46

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4 years ago
Read 2 more answers
-What does VI indicate when talking about LabView?
BaLLatris [955]

Answer:

The VI in LabView indicates c-Virtual Instrument

Explanation:

The VI in LabView is a program-subroutine. The VI stands for Virtual Instrument. The VI is composed of a Block diagram, Connector panel and a Front Panel.

4 0
3 years ago
2. Research, find and use a better (different from the simple Shift Cipher) encryption technique to encrypt a string of your cho
Kruka [31]

Answer:

There are two better techniques for encryption as compared to shift cipher. One is block cipher. Such techniques use one key for both encryption and decryption. For example Advanced Encryption Standard (AES). This is based on symmetric key encryption.

The other way is to use asymmetric key encryption technique which based on two keys i.e. public and private key for encryption and decryption. For example, Rivest-Shamir-Adleman (RSA).

Explanation:

What is AES?

Advanced Encryption Standard (AES) is a symmetric key block cipher  which uses 128, 192 or 256-bit keys  length to encrypt and decrypt a block of plain text or message. It is stronger and faster than Shift Cipher encryption technique. It can encrypt data blocks of 128 bit using the above mentioned bit key lengths. The greater then key length the greater the security. the number of rounds in AES is variable and depends on the length of the key. The length of these keys determines the number of rounds for example 128 bit keys has 10 rounds, 12 rounds for 192 bit keys and 14 rounds for 256 bit keys. Each rounds involves some processing. This process is explained below:

How AES works?

The data is divided into each of size 128 bits makes a matrix of 4x4 columns of 16 bytes. (bytes because AES takes 128 bit plain text block as 16 bytes)

Substitution: First the data is substituted using a fixed substitution table which is predetermined. This makes a matrix of rows and columns.

Shifting: Each rows of the matrix is shifted to the left and the dropped entries in row are inserted on the right. Shift is carried out as follows −  

Mixing columns: Use a mathematical method/equation to transform the columns where the input of the method is each column of data which is then replaced by this function into a different new matrix of data(bytes). This is basically the first round to transform plain text to cipher text. Now consider these bytes as 128 bits, the first round key is added to these resultant column bits by XOR. A separate portion of the encryption key length is used to perform the transformation on each column. The number of rounds depend on the the key length.

After the first round key comes the second round key and so on and when the last round key is added then this whole process goes back to Substitution phase, then shifting rows phase, then mixing columns and another round key is added after this. For example, if 128 bit key is used this means there will be nine such rounds.

Advantages

AES supports large key sizes which makes it more secure and stronger than shift cipher encryption techniques which can be cracked because of their small key space. If someone wants to decrypt a message and knows it's encrypted by a Shift Cipher technique then the key space of all possible keys is only the size of the alphabet. This is a very small scale.  An attack such brute force attack attempting all 26 keys or even using exhaustive technique will easily crack this. So AES having large key size is more robust against cracking. For example in AES in order to crack an encryption for 128 bit around 2128 attempts are needed. Also for a 256 bit key, 2256 different combinations attempts are needed to ensure the right one is included which makes it harder to crack and makes is more secure.

I have used an internet solver to encrypt a string and i am attaching the result of the encryption. The plain text is: Two One Nine Two

The key is Thats my Kung Fu

and the cipher text in hexadecimal after using AES is:

29 c3 50 5f 57 14 20 f6 40 22 99 b3 1a 02 d7 3a

8 0
4 years ago
For an activity with more than one immediate predecessor activity, which of the following is used to compute its earliest finish
laila [671]

Answer:

The correct option is A

Explanation:

In project management, earliest finish time for activity A refers to the earliest start time for succeeding activities such as B and C to start.

Assume that activities A and B comes before C, the earliest finish time for C can be arrived at by computing the earliest start-finish (critical path) of the activity with the largest EF.

That is, if two activities (A and B) come before activity C, one can estimate how long it's going to take to complete activity C if ones knows how long activity B will take (being the activity with the largest earliest finish time).

Cheers!

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3 years ago
Black and white squares codehs, i need the whole code (40 points for correct answer)
ludmilkaskok [199]

Answer:

speed(0)

penup()

setposition(-100,0)

count=0

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penup()

pendown()

for i in range(6):

pendown()

make_squares(i)

penup()

forward(35)

Explanation:

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3 years ago
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