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Pavel [41]
3 years ago
12

WAP to find area of circle​

Computers and Technology
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

program by LET statement

Explanation:

CLS

REM to find the area of circle

LET r=10

LET a=22/7*2*r^2

PRINT "area=;"a

END  

                                                                                         press f5

output:

a=628.57

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Int add( int, int );
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All of the following are potential benefits of blogging except
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C is the right answer
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2. Using the Enhanced for Statement, write an application that uses an enhanced for statement to sum the double values passed by
solniwko [45]

Answer:

Here is the program:

public class DoubleValue

{ public static void main(String[] args) { //start of the main function

 double sum = 0.0;  // initialize double type sum by 0.0

//for statement to sum the double values passed by the command-line //arguments

       for (String str : args)        {  

         sum += Double.parseDouble(str);  

//returns double representation of the passed str argument and takes the //sum of these double values         }

       System.out.printf("The sum of the double values passed in from the command line is %.1f\n", sum);  }  }      

//prints the above message with output 0.0                                  

Explanation:

The above program has a static method parseDouble of class Double to convert a string str to a double value and the for statement is used to sum the double values passed by command line arguments. This sum is stored in sum variable and is displayed in the output. The program along with its output is attached.

If you want to get the input from the user you can use the following code. Just use the Scanner class to take input from the user.

import java.util.Scanner;

public class DoubleValue

{ public static void main(String[] args) {

//two string type variables

 String str1;

               String str2;

//scans and reads input from user

Scanner input = new Scanner(System.in);

System.out.println("Enter value of str1: ");

str1 = input.next();

System.out.println("Enter value of str2: ");

str2 = input.next();

double sum = 0.0;

//sum double values and parseDouble function is used to convert string to //double value

sum += Double.parseDouble(str1) + Double.parseDouble(str2);

System.out.printf("The sum of the double values passed in from the command line is %.1f\n", sum); } }  //prints result of the sum

5 0
3 years ago
g You are looking to rob a jewelry store. You have been staking it out for a couple of weeks now and have learned the weights an
ololo11 [35]

Answer:

A python code (Python recursion) was used for this given question

Explanation:

Solution

For this solution to the question, I am attaching code for these 2 files:

item.py

code.py

Source code for item.py:

class Item(object):

def __init__(self, name: str, weight: int, value: int) -> None:

  self.name = name

  self.weight = weight

  self.value = value

def __lt__(self, other: "Item"):

  if self.value == other.value:

if self.weight == other.weight:

  return self.name < other.name

else:

  return self.weight < other.weight

  else:

   return self.value < other.value

def __eq__(self, other: "Item") -> bool:

  if is instance(other, Item):

return (self.name == other.name and

self.value == other.value and

self.weight == other.weight)

  else:

return False

def __ne__(self, other: "Item") -> bool:

  return not (self == other)

def __str__(self) -> str:

  return f'A {self.name} worth {self.value} that weighs {self.weight}'

Source code for code.py:

#!/usr/bin/env python3

from typing import List

from typing import List, Generator

from item import Item

'''

Inductive definition of the function

fun3(0) is 5

fun3(1) is 7

fun3(2) is 11

func3(n) is fun3(n-1) + fun3(n-2) + fun3(n-3)

Solution 1: Straightforward but exponential

'''

def fun3_1(n: int) -> int:

result = None

if n == 0:

result = 5 # Base case

elif n == 1:

result = 7 # Base case

elif n == 2:

result = 11 # Base case

else:

result = fun3_1(n-1) + fun3_1(n-2) + fun3_1(n-3) # Recursive case

return result

''

Solution 2: New helper recursive function makes it linear

'''

def fun3(n: int) -> int:

''' Recursive core.

fun3(n) = _fun3(n-i, fun3(2+i), fun3(1+i), fun3(i))

'''

def fun3_helper_r(n: int, f_2: int, f_1: int, f_0: int):

result = None

if n == 0:

result = f_0 # Base case

elif n == 1:

result = f_1 # Base case

elif n == 2:

result = f_2 # Base case

else:

result = fun3_helper_r(n-1, f_2+f_1+f_0, f_2, f_1) # Recursive step

return result

return fun3_helper_r(n, 11, 7, 5)

''' binary_strings accepts a string of 0's, 1's, and X's and returns a generator that goes through all possible strings where the X's

could be either 0's or 1's. For example, with the string '0XX1',

the possible strings are '0001', '0011', '0101', and '0111'

'''

def binary_strings(string: str) -> Generator[str, None, None]:

def _binary_strings(string: str, binary_chars: List[str], idx: int):

if idx == len(string):

yield ''.join(binary_chars)

binary_chars = [' ']*len(string)

else:

char = string[idx]

if char != 'X':

binary_chars[idx]= char

yield from _binary_strings(string, binary_chars, idx+1)

else:

binary_chars[idx] = '0'

yield from _binary_strings(string, binary_chars, idx+1)

binary_chars[idx] = '1'

yield from _binary_strings(string, binary_chars, idx+1)

binary_chars = [' ']*len(string)

idx = 0

yield from _binary_strings(string, binary_chars, 0)

''' Recursive KnapSack: You are looking to rob a jewelry store. You have been staking it out for a couple of weeks now and have learned

the weights and values of every item in the store. You are looking to

get the biggest score you possibly can but you are only one person and

your backpack can only fit so much. Write a function that accepts a

list of items as well as the maximum capacity that your backpack can

hold and returns a list containing the most valuable items you can

take that still fit in your backpack. '''

def get_best_backpack(items: List[Item], max_capacity: int) -> List[Item]:

def get_best_r(took: List[Item], rest: List[Item], capacity: int) -> List[Item]:

if not rest or not capacity: # Base case

return took

else:

item = rest[0]

list1 = []

list1_val = 0

if item.weight <= capacity:

list1 = get_best_r(took+[item], rest[1:], capacity-item.weight)

list1_val = sum(x.value for x in list1)

list2 = get_best_r(took, rest[1:], capacity)

list2_val = sum(x.value for x in list2)

return list1 if list1_val > list2_val else list2

return get_best_r([], items, max_capacity)

Note: Kindly find an attached copy of the code outputs for python programming language below

5 0
3 years ago
Editing is the process of cutting out the bad parts. True or false?
Maurinko [17]

Answer:

true

Explanation:

your welcome

7 0
3 years ago
Read 2 more answers
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