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lianna [129]
2 years ago
14

Assume that getPlayer2Move works as specified, regardless of what you wrote in part (a) . You must use getPlayer1Move and getPla

yer2Move appropriately to receive full credit. Complete method playGame below. /** Plays a simulated game between two players, as described in part (b). */ public void playGame()
Computers and Technology
1 answer:
kakasveta [241]2 years ago
3 0

Answer:

(1)

public int getPlayer2Move(int round)

{

  int result = 0;

 

  //If round is divided by 3

  if(round%3 == 0) {

      result= 3;

  }

  //if round is not divided by 3 and is divided by 2

  else if(round%3 != 0 && round%2 == 0) {

      result = 2;

  }

  //if round is not divided by 3 or 2

  else {

      result = 1;

  }

 

  return result;

}

(2)

public void playGame()

{

 

  //Initializing player 1 coins

  int player1Coins = startingCoins;

 

  //Initializing player 2 coins

  int player2Coins = startingCoins;

 

 

  for ( int round = 1 ; round <= maxRounds ; round++) {

     

      //if the player 1 or player 2 coins are less than 3

      if(player1Coins < 3 || player2Coins < 3) {

          break;

      }

     

      //The number of coins player 1 spends

      int player1Spends = getPlayer1Move();

     

      //The number of coins player 2 spends

      int player2Spends = getPlayer2Move(round);

     

      //Remaining coins of player 1

      player1Coins -= player1Spends;

     

      //Remaining coins of player 2

      player2Coins -= player2Spends;

     

      //If player 2 spends the same number of coins as player 2 spends

      if ( player1Spends == player2Spends) {

          player2Coins += 1;

          continue;

      }

     

      //positive difference between the number of coins spent by the two players

      int difference = Math.abs(player1Spends - player2Spends) ;

     

      //if difference is 1

      if( difference == 1) {

          player2Coins += 1;

          continue;

      }

     

      //If difference is 2

      if(difference == 2) {

          player1Coins += 2;

          continue;

      }

     

     

  }

 

  // At the end of the game

  //If player 1 coins is equal to player two coins

  if(player1Coins == player2Coins) {

      System.out.println("tie game");

  }

  //If player 1 coins are greater than player 2 coins

  else if(player1Coins > player2Coins) {

      System.out.println("player 1 wins");

  }

  //If player 2 coins is grater than player 2 coins

  else if(player1Coins < player2Coins) {

      System.out.println("player 2 wins");

  }

}

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Explanation:

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Assume the input data is structured as follows: first there is a non-negative integer specifying the number of employee timeshee
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Answer:

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int a,b;     //set two integer type variable

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Here, we define four integer type variable "a", "b", "wage" "hour" and assign value in "wage" to 0 and "hours" to 0.

Then, we get input from the user in the variable "employees".

Then, we set the for loop and pass the condition inside it we get input from the user in the variable "wage".

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In mathematics, the notation n! represents the factorial of the nonnegative integer n. The factorial of n is the product of all
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The program to this question can be given as:

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Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

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No. of sectors per track = 600

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To find:

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Total Average time to satisfy a request.

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Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

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b) The rotations per minute are 7200

1 min = 60 sec

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If the greedy solution is non-optimal, then there exists another solution { t'_i }, with the same constraints, s.t. t'_n = m' < t_n = m.

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t_k != t'_k => Either

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   So the greedy solution must be optimal.

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