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3 years ago

Will the Python code below print something? And will it terminate?

Computers and Technology

1 answer:

77julia77 [94]3 years ago

3 0

Answer:

Yes and Yes

Explanation:

When programs are written without indentation; it becomes a problem to read;

I'll rewrite your program as follows:

def print_n(s, n):

if n > 0:

print(s)

print_n(s, n-1)

return n

n = 3

while print_n("hi", n):

print_n("there!", n)

n = 0

Using the above as a point of reference,

Yes, the program will run and Yes, it'll terminate;

Because n is initialized to 3 at line 6 (where the main starts);

The following is passed to the function; s = "hi" and n=3

This prints "hi", three times

s = "there" and n=3 is then passed to the function print_n, afterwards;

The function then prints "there" three times

Then, the program terminates

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You are hired by a game design company and one of their most popular games is The Journey. The game has a ton of quests, and for

Kamila [148]

Answer:

Explanation:

Let's describe the connected part for the very first time.

Component with good connections:-

A semi in any graph 'G' is defined by a way to align element if this can be crossed from the beginning of any link in that pixel or if this can be stated that there is a path across each organized node pair within this subgram.

Consecrated pair means (ni, nj) and (nj, ni) that 2 different pairs would be regarded.

In point a:

We're going to be using contradictions for this segment. They start with the assumption that two lucky journeys are not even in the same strongly interconnected component.

For instance, qi and qj are providing special quests that were not in the very strongly linked element and which implies that, if we start to cross from qi or qi, then qj cannot be approached if we begin to move through qj and as we established if qi or qj is fortunate contests, we may reach any other hunts. Which implies qj from qi or conversely should be reachable. Or we might claim that qi is part of the strongly linked portion of qj or vice versa with this situation.

Consequently, we would not be capable of forming part of a different and strongly linked element for two successful scientists; they must have the same strongly related to the element.

In point b:

Its definition of its strong, line segment indicates that all other searches within such a strongly coordinate system can be made possible from any quest. Because if all quests are accessible from any quest then a lucky search is named, and all other quests can be accessed from any quest in a very coordinated system. So, all contests are fortunate contests in a strongly connected element.

Algorithm:

Build 'n' size range named 'visited' wherein 'n' is the number of graphic nodes that keep records of nodes already frequented. Running DFS out of every unknown vertex or mark all edges as seen. The lucky search will be the last unexplored peak.

Psuedo-Code:

Requires this functionality to also be called Solution, it requires 2 reasons as an input, V is a set of objects in graph 'G' and 'G' is a chart itself.

Solution(V, G):

visited[V] = {false}//assign value

for v: 0 to 'V'-1: //using for loop

if not visited[v]: //use if block

DFS(v, visited, V, G) //use DFS method

ans = v //holding value

return ans //return value

3 0

3 years ago

Read 2 more answers

On laptops, wireless cards tend to be attached to which panel?

ELEN [110]

Answer:

LCD panel

Explanation:

Regularly these wireless cards tend to be attached to the LCD panel because the wireless cards have a Wi-Fi antenna to get a better signal.

Regularly the wireless card has two cables, those cables go between the hinges until arrives at the display panel (LCD panel).

There are wireless cards that are connected by the USB port, this is so useful because there are some cards welded in the motherboard.

8 0

4 years ago

When the POST is complete,t he operating systems is loaded into?

Lina20 [59]

After the POST is successfully completed the CPU searches Storage devices for special files that indicate the beginning of the OS and then loads it into memory. 

The sequence in which the storage devices are searched is part of the configuration set-up stored in BIOS.

Once the Operating System (OS) is located, the CPU copies what’s called the 'boot record' from the OS into DRAM.

Control is then passed to the 'boot record' in DRAM, which then continues loading the rest of the OS.

This process continues until the OS load is completed.

When the OS load is completed the Desktop appears in the display and waits for you, the user, to tell it what you want it to do.

4 0

3 years ago

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brilliants [131]

Tell me the question

8 0

4 years ago

Write a program (Console or GUI) that determines the change to be dispensed from a vending machine. An item in the machine can c

Stells [14]

Answer:

Here is the JAVA program:

import java.util.Scanner; //to accept input from user

public class Main { //class name

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

Scanner input =new Scanner(System.in); // creates Scanner class object

int amount, quarters, dimes, nickels, pennies,change; //declare variables

System.out.println("Enter the price of the(from 25 cents to a dollar, in 5-cent increments): "); //prompts user to enter the price

amount=input.nextInt(); //reads amount from user

change= 100-amount; //computes change

System.out.println("You bought an item for " + amount+" cents and gave me a dollar, so your change is :");

quarters=change/25; //computes quarters

change=change%25; //computes quarter remaining

if(quarters == 1) // if quarter is equal to 1

System.out.println(quarters+ " quarter"); //prints quarter value in singular

else if (quarters>1) // if value of quarters is greater than 1

System.out.println(quarters+" quarters"); //prints plural quarters value

dimes=change/10; //computes dimes

if(dimes == 1) // if value of dime is equal to 1

System.out.println(dimes+ " dime"); //prints single value of dimes

else if (dimes>1) //if value of dimes is greater than 1

System.out.println(dimes+" dimes"); //prints plural value of dimes

change=change%10; //computes dimes remaining

nickels=change/5; //computes nickels

if(nickels == 1) //if value of nickels is equal to 1

System.out.println(nickels+ " nickel"); //prints single value of nickels

else if (nickels>1) //if value of nickels is greater than 1

System.out.println(nickels+" nickels"); //prints plural value of nickels

change=change%5; //computes nickels remaining

pennies=change; //computes pennies

if(pennies == 1) //if value of pennies is equal to 1

System.out.println(pennies+ " penny"); //prints single value of pennies

else if (pennies>1) //if value of pennies is greater than 1

System.out.println(pennies+" pennies"); } } //prints plural value of pennies

Explanation:

I will explain the program with an example:

Suppose amount= 75

Then change = 100 - amount= 100 - 75 = 25

change = 25

quarters = change/25 = 25/25 = 1

quarters = 1

change = change % 25 = 25%25 = 0

dimes = 0/10 = 0

Now all other values are 0.

Now the following line is printed on screen:

You bought an item for 75 cents and gave me a dollar. Your change is:

Now program moves to if part

if quarters == 1

This is true because value of quarter is 1

sot "quarter" is displayed with the value of quarters

Now the following line is printed on screen:

1 quarter

So the complete output of the program is:

You bought an item for 75 cents and gave me a dollar. Your change is:

1 quarter

3 0

4 years ago