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VladimirAG [237]
3 years ago
13

Write a Python program stored in a file q1.py to play Rock-Paper-Scissors. In this game, two players count aloud to three, swing

ing their hand in a fist each time. When both players say three, the players throw one of three gestures: Rock beats scissors Scissors beats paper Paper beats rock Your task is to have a user play Rock-Paper-Scissors against a computer opponent that randomly picks a throw. You will ask the user how many points are required to win the game. The Rock-Paper-Scissors game is composed of rounds, where the winner of a round scores a single point. The user and computer play the game until the desired number of points to win the game is reached. Note: Within a round, if there is a tie (i.e., the user picks the same throw as the computer), prompt the user to throw again and generate a new throw for the computer. The computer and user continue throwing until there is a winner for the round.
Computers and Technology
1 answer:
never [62]3 years ago
8 0

Answer:

The program is as follows:

import random

print("Rock\nPaper\nScissors")

points = int(input("Points to win the game: "))

player_point = 0; computer_point = 0

while player_point != points and computer_point != points:

   computer = random.choice(['Rock', 'Paper', 'Scissors'])

   player = input('Choose: ')

   if player == computer:

       print('A tie - Both players chose '+player)

   elif (player.lower() == "Rock".lower() and computer.lower() == "Scissors".lower()) or (player.lower() == "Paper".lower() and computer.lower() == "Rock".lower()) or (player == "Scissors" and computer.lower() == "Paper".lower()):

       print('Player won! '+player +' beats '+computer)

       player_point+=1

   else:

       print('Computer won! '+computer+' beats '+player)

       computer_point+=1

print("Player:",player_point)

print("Computer:",computer_point)

Explanation:

This imports the random module

import random

This prints the three possible selections

print("Rock\nPaper\nScissors")

This gets input for the number of points to win

points = int(input("Points to win the game: "))

This initializes the player and the computer point to 0

player_point = 0; computer_point = 0

The following loop is repeated until the player or the computer gets to the winning point

while player_point != points and computer_point != points:

The computer makes selection

   computer = random.choice(['Rock', 'Paper', 'Scissors'])

The player enters his selection

   player = input('Choose: ')

If both selections are the same, then there is a tie

<em>    if player == computer: </em>

<em>        print('A tie - Both players chose '+player) </em>

If otherwise, further comparison is made

<em>    elif (player.lower() == "Rock".lower() and computer.lower() == "Scissors".lower()) or (player.lower() == "Paper".lower() and computer.lower() == "Rock".lower()) or (player == "Scissors" and computer.lower() == "Paper".lower()): </em>

If the player wins, then the player's point is incremented by 1

<em>        print('Player won! '+player +' beats '+computer) </em>

<em>        player_point+=1 </em>

If the computer wins, then the computer's point is incremented by 1

<em>    else: </em>

<em>        print('Computer won! '+computer+' beats '+player) </em>

<em>        computer_point+=1 </em>

At the end of the game, the player's and the computer's points are printed

<em>print("Player:",player_point) </em>

<em>print("Computer:",computer_point)</em>

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Answer:

solution

Explanation:

The correct option is - solution

Reason -

To solve a problem,

Firstly we give input , then system will process that input which then gives output.

Solution is not a part of the process.

So, Solution is not a significant part of a simple problem.

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3 years ago
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siniylev [52]

Answer:

B. 1 6 3

Explanation:

Given function definition for calc:

void calc (int a, int& b)

{

int c;

c = a + 2;

a = a * 3;

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}

Function invocation:

x = 1;

y = 2;

z = 3;

calc(x, y);

cout << x << " " << y << " " << z << endl;

  • Since x is passed by value, its value remains 1.
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Tracing the function execution:

c=3

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b=c+a = 6;

But b actually corresponds to y. So y=6 after function call.

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3 years ago
2.8 Code Practice: Question 1
-BARSIC- [3]

Answer:

The program to this question can be given as:

Program:

#include <stdio.h> //define header file

int main() //define main method.

{

int a1,b1,c1; //define variable

printf("Enter three numbers:\n"); //message

scanf("%d",&a1); //input numbers.

scanf("%d",&b1);//input numbers.

scanf("%d",&c1);//input numbers.

if (a1>b1) //check condition

{

//check condition

if(a1>c1)//inner if block

{

printf("The largest number: %d", a1); //message

}

else  //inner else block

{

printf("The largest number: %d", c1);//message

}

}

else //else block

{

//check condition

if(b1>c1) //inner if block.

{

printf("The largest number: %d", b1);//message

}

else //else block

{

printf("The largest number: %d",c1);//message

}

}

return 0;  //return 0.

}

Output:

Enter three numbers:

22

44

11

The largest number: 44

Explanation:

The description of the above program can be given as:

  • In the C language program firstly we define the header file. Then we define the main function. In the main function, we define three variables that are "a1, b1 and c1". In this variable, we take input from the user.
  • Then we define the conditional statements. In the if block we check if the value of a1 is greater then the value of b1 if this condition is true so we use another condition that if the value of a1 is greater then the value of c1. so, it will print the largest value. If the above condition is not true then it will go to else section.
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4 0
3 years ago
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3 years ago
What is the space complexity of the algorithm?ArithmeticSeries(list, listSize) { i = 0 arithmeticSum = 0 while (i &lt; listSize)
blondinia [14]

Answer:

O(n) which is a linear space complexity

Explanation:

Space complexity is the amount of memory space needed for a program code to be executed and return results. Space complexity depends on the input space and the auxiliary space used by the algorithm.

The list or array is an integer array of 'n' items, with the memory size 4*n, which is the memory size of an integer multiplied by the number of items in the list. The listSize,  i, and arithmeticSum are all integers, the memory space is 4(3) = 12. The return statement passes the content of the arithmetic variable to another variable of space 4.

The total space complexity of the algorithm is "4n + 16" which is a linear space complexity.

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2 years ago
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