Java 硬币游戏。如何让玩家在满足特定标准的情况下失去下一回合
我目前正在制作一个程序,有4名玩家轮流投掷3枚硬币。第一个获得16分的玩家获胜。玩家每次掷硬币都会获得点数。他得到的点数等于他掷的人头数。如果他没有掷头,那么他就失去了下一回合。如果他翻了3个头,那么他就多赚了一个回合,并再次掷硬币。如果他掷的球少于3个,那么轮到下一个球员了。一名球员必须获得16分才能获胜。如果一名球员得了14分,抛了2个头,他就赢了,但是如果他抛了n个头,并且超过16分,那么他就失去了一半的分数,也失去了轮次。他必须有16分才能赢 如何让玩家在下一回合被跳过?每个玩家目前都按顺序进行。汤姆,汉克,汉娜,蒂娜。如果汉克掷0个头球或超过16分,他将输掉下一轮,下一轮的顺序是汤姆,汉娜,蒂娜,汤姆,汉克,汉娜,蒂娜。我已经发布了我的代码,我需要找出如何编辑nextPlayer()方法以满足我的需要。我相信我在computerState()方法中所做的逻辑和代码是正确的。任何帮助都将不胜感激 游戏类Java 硬币游戏。如何让玩家在满足特定标准的情况下失去下一回合,java,Java,我目前正在制作一个程序,有4名玩家轮流投掷3枚硬币。第一个获得16分的玩家获胜。玩家每次掷硬币都会获得点数。他得到的点数等于他掷的人头数。如果他没有掷头,那么他就失去了下一回合。如果他翻了3个头,那么他就多赚了一个回合,并再次掷硬币。如果他掷的球少于3个,那么轮到下一个球员了。一名球员必须获得16分才能获胜。如果一名球员得了14分,抛了2个头,他就赢了,但是如果他抛了n个头,并且超过16分,那么他就失去了一半的分数,也失去了轮次。他必须有16分才能赢 如何让玩家在下一回合被跳过?每个玩家目前都按
import java.util.Random;
public class Game
{
private Random randomizer;
private final int n_players;
private final int m_coins;
private final int p_points;
private int player_index;
private boolean game_over;
public Game()
{
n_players = 4;
m_coins = 3;
p_points = 16;
game_over = false;
randomizer = new Random();
player_index = randomizer.nextInt(n_players);
}
public Game(int new_m_coins, int new_n_players, int new_p_points)
{
n_players = new_n_players;
m_coins = new_m_coins;
p_points = new_p_points;
game_over = false;
randomizer = new Random();
player_index = randomizer.nextInt(n_players);
}
public int getPlayerIndex()
{
return player_index;
}
public void setPlayerIndex()
{
player_index = randomizer.nextInt(n_players);
}
public boolean gameOver()
{
return game_over;
}
public int nextPlayer(Player[] players)
{
//player_index = (player_index + 1) % n_players;
if(players[player_index].getState() == State.EXTRA_TURN)
{
players[player_index].setState(State.NORMAL);
}
else if(players[player_index].getState() == State.LOSE_TURN)
{
player_index = (player_index + 1) % n_players;
}
else
{
player_index = (player_index + 1) % n_players;
}
/*while(players[player_index].getState() != State.NORMAL)
{
players[player_index].setState(State.NORMAL);
player_index = (player_index + 1) % n_players;
}*/
return player_index;
}
public void computeState(Player player, int m_heads, int oldPoints, int newPoints)
{
int player_points = player.getPoints();
if(player_points == p_points)
game_over = true;
else if(player_points > p_points)
{
player.setPoints(player_points / 2);
player.setState(State.LOSE_TURN);
}
else if(m_heads == 0)
{
player.setState(State.LOSE_TURN);
}
else if(m_heads == 3)
{
player.setState(State.EXTRA_TURN);
}
else if(m_heads == 3 && player_points > p_points)
{
player.setState(State.NORMAL);
}
else
player.setState(State.NORMAL);
}
}
TestCoinGame
public class testcoingame
{
public static void main(String[] args)
{
try
{
int m_coins = 3;
int n_players = 4;
int p_points = 16;
String [] names = {"Hank", "Tina", "Hannah", "Tom"};
Player [] players = new Player[n_players];
for(int index = 0; index < players.length; index++)
players[index] = new Player(names[index]);
Coins coins = new Coins();
Game game = new Game();
int player_index;
do
{
player_index = game.nextPlayer(players);
System.out.printf("It is %s's turn\n", players[player_index].getName());
System.out.printf("%s has %d points\n", players[player_index].getName(),
players[player_index].getPoints());
coins.tossCoins();
int n_heads = coins.getNHeads();
System.out.printf("%s tossed %d heads\n",
players[player_index].getName(), n_heads);
int old_points = players[player_index].getPoints();
int new_points = old_points + n_heads;
players[player_index].setPoints(new_points);
game.computeState(players[player_index], n_heads, old_points, new_points);
System.out.printf("%s has %d points\n", players[player_index].getName(),players[player_index].getPoints());
}
while(!game.gameOver());
System.out.printf("%s wins!\n", players[player_index].getName());
}
catch(Exception ex)
{
}
}
}
公共类testcoingame
{
公共静态void main(字符串[]args)
{
尝试
{
国际货币单位=3;
国际n_玩家=4;
int p_点=16;
String[]name={“Hank”、“Tina”、“Hannah”、“Tom”};
玩家[]玩家=新玩家[n_玩家];
for(int index=0;index
在玩家类中可以有一个布尔实例变量,用于跟踪玩家是否跳过下一回合。当您要检查该值是真是假,并且它的计算结果为真(即是,她跳过了她的回合),然后跳过您通常会为回合执行的操作,但在您继续下一个玩家
或回合
或您拥有的任何操作之前,将布尔值设置为假
编辑(基于用户响应):
伪代码:
我假设你通过数组、列表或类似的东西来跟踪轮到谁。我将把它称为playerList。例如,如果我们有三个玩家,那么我们将继续循环玩家列表[0]
,玩家列表[1]
,玩家列表[2]
while (gameNotOver())
{
if (!playerList[currentPlayer].skipNextTurn())
{
//do what you would normally do
}
else
{
playerList[currentPlayer].setSkipNextTurn(false);
}
}
您可以从注释代码开始;使用注释“隐藏”代码或与使用过的代码完全相同的注释并不重要。为了解释您收到的反对票:您的问题需要5分钟以上的投入。这超过了SO的典型阈值,所以我们否决了投票,继续前进。看起来家庭作业也很自然,这可能会让人们不太可能提供帮助,除非你说得对,并提出一个小问题。