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aliina [53]
3 years ago
15

Find the outputI need it immediately​

Computers and Technology
1 answer:
kupik [55]3 years ago
3 0

This is a while loop that prints the value of I each time squared, after which the I value is incremented by one. Example in the attached screenshot.

Hope this helps!

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3 years ago
In most programming languages, the compiler carries a preprocessing step to determine if certain statements will compile. For in
MakcuM [25]

Answer:

See explaination for program code

Explanation:

package com.company;

import java.io.IOException;

import java.nio.file.Files;

import java.nio.file.Paths;

import java.util.Stack;

/*

* A program which reads input from given text filename and prints

* whether it passed the pre-processing stage or not

* Sample input

* int A = (a + b);static void display(int x)

* {

* //A sample input file

* }

* */

public class MyPreprocessor {

/*

* Reads input from filename and returns a string

* containing whole text

* */

public String readInput(String filename) {

String content = "";

try {

content = new String(Files.readAllBytes(Paths.get(filename)));

} catch (IOException e) {

e.printStackTrace();

}

return content;

}

/**

* Filters input since we are only concerned with

* a subset of input = '{}[]()/*'

*/

public String filterText(String text) {

String output = "";

for(int i = 0;i<text.length();++i) {

switch (text.charAt(i)) {

case '[':

case ']':

case '{':

case '}':

case '(':

case ')':

case '*':

case '/':

output += text.charAt(i);

}

}

return output;

}

/*

* Uses stack to determine if input is valid or not

* */

public boolean parse(String input) {

Stack<Character> St = new Stack<Character>();

// '$' is special symbol to represent bottom of stack

St.push('$');

for(int i = 0;i<input.length();++i) {

Character symbol = input.charAt(i);

if(St.peek().equals(getOpenSymbol(symbol)))

St.pop();

else St.push(symbol);

}

return St.peek() == '$';

}

private static Character getOpenSymbol(Character symbol) {

switch (symbol) {

case '}': return '{';

case ']': return '[';

case ')': return '(';

case '*': return '*';

case '/': return '/';

default: return ' ';

}

}

public static void main(String[] args) {

MyPreprocessor preprocessor = new MyPreprocessor();

String inputText = preprocessor.readInput("src/input.txt");

String filteredInput = preprocessor.filterText(inputText);

if(preprocessor.parse(filteredInput))

System.out.println("preprocessing passed.");

else System.out.println("preprocessing failed!");

}

}

6 0
3 years ago
Write a generator function named count_seq that doesn't take any parameters and generates a sequence that starts like this: 2, 1
-BARSIC- [3]

Answer:

#required generator function

def count_seq():

   #starting with number 2. using string instead of integers for easy manipulation

   n='2'

   #looping indefinitely

   while True:

       #yielding the integer value of current n

       yield int(n)

       #initializing an empty string

       next_value=''

       #looping until n is an empty string

       while len(n)>0:

           #extracting first digit (as char)

           first=n[0]

           #consecutive count of this digit

           count=0

           #looping as long as n is non empty and first digit of n is same as first

           while len(n)>0 and n[0]==first:

               #incrementing count

               count+=1

               #removing first digit from n

               n=n[1:]

           #now appending count and first digit to next_value

           next_value+='{}{}'.format(count,first)

       #replacing n with next_value

       n=next_value

#testing, remove if you don't need this

if __name__ == '__main__':

   #creating a generator from count_seq()

   gen=count_seq()

   #looping for 10 times, printing next value

   for i in range(10):

       print(next(gen))

Explanation:

6 0
3 years ago
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