我正在尝试制作一个寻宝板程序，在该程序中，用户将探索一个 10 x 10 的板，其中将包含宝藏、海盗和离岛逃生。目标是带着尽可能多的宝藏逃离岛屿。宝藏(T) - 随着发现更多宝藏而增加（即，第一个宝藏是 +100 金达布隆，第二个是 +200 达布隆，依此类推） 海盗 (P) - 您遇到的海盗越多，他们可以拿走的金币就越多。 （即第一组是-100金达布隆，第二组是-200金达布隆，依此类推）但是，如果玩家没有宝藏，就没有什么可失去的。 两者 (B)- 如果有宝藏和海盗，玩家将获得一半的可用宝藏。 逃生 (X) - 当远处看到一艘船时，用户可以离开岛屿。如何在二维数组中显示金达布隆的值？
public static void main(String[] args) throws FileNotFoundException {
Scanner s = new Scanner(System.in);
// create 10x10 arrays
final int ROWS = 10, COLS = 10;
char[][] displayMap = new char[ROWS][COLS]; //stores the map
char[][] board = new char[ROWS][COLS];// this will display the board
//variales for how many pirates
char pirate = 'P';
//sets the location of the pirates on the board
int upperBound = 5, lowerBound = 0;
int pirateRow = (int) ((Math.random() * (upperBound - lowerBound + 1))
+ lowerArea); // the pirates location
int pirateColumn = (int) ((Math.random() * (upperBound - lowerBound + 1))
+ lowerBound); // pirates location
int treasures = 5;
//call loadMap to read in the map and return it
displayMap = loadMap();
//this will take the board and resets the board
clearBoard(board);
//passes board and pirate informations to displayBoard then display it
displayBoard(board, pirate, pirateRow, pirateColumn, treasures);
int selection = 0;
int playAgain = 0;
while (selection != 4) {
// displays the menu
System.out.println("WELCOME TO TREASURE HUNT");
System.out.println("1. Play Game ");
System.out.println("2. How to Play");
System.out.println("3. Legend");
System.out.println("4. Exit");
// take the input from menu
System.out.println("Enter your selection.");
selection = s.nextInt();
switch (selection) {
case 1:
PlayGame();
break;
case 2:
HowToPlay();
break;
case 100:
Legend();
break;
case 4:
System.out.println("See you again next time!");
default:
System.out.println("ERORR");
break;
}
// when user input = 1
if (selection == 1) {
// test for valid input
int rowNum = 0, columnNum = 0;
boolean isValid = false;
while ( !isValid) {
//prompt user
}
System.out.println("Pick a position to explore or enter 100 to"
+ "see the legend. ");
char row;
// check the row input
Scanner sc = new Scanner(System.in); // decalres scanner
row = sc.next().charAt(0); // this will store the character input
if (row >= 0&& row <= 99) { // this has to be a range of 0-99
rowNum = testRowInput(row); // this will test and return what is in the meethod
if (rowNum >= 0 && rowNum <= 9) { // tests if rowNum is within the range
// prompt user for column number / position
System.out.println("Pick a position to explore or enter 100 to"
+ "see the legend. ");
columnNum = getIntRange(1, 10);
columnNum --; // this will convert column range to 0-9
/**
* check if the spot has not yet been revealed, otherwise
* it has been explored already
*/
if ( testRevealed (board, rowNum, columnNum)) {
sc.nextLine(); // empty row input from buffer
System.out.println("You have already explored there.");
} else {
isValid = true;
}
} else {
sc.nextLine(); // empty row input
System.out.println("Nothing to see here.");
}
}
} // finished checking user input
// caught by pirate / losses gold doubloons
if ( rowNum == pirateRow ) && ( columnNum == pirateColumn)){
board[pirateRow][[pirateColumn]] = pirate;
}
}
}
/**
*
*/
public static void PlayGame() {
Scanner sc = new Scanner(System.in);
int pirates = 0;
int treasures = 0;
System.out.println("You chose option 1. Enter the number of pirates");
pirates = sc.nextInt();
System.out.println("Enter the number of treasures.");
treasures = sc.nextInt();
System.out.println(displayBoard);
}
/**
*
*/
public static void HowToPlay() {
System.out.println("You chose option 2");
System.out.println("For this game, the user will be going on a treasure"
+ " hunt. They will explore a 10 x 10 board that will contain "
+ "treasure, pirates and an escape off the island. The goal is "
+ "to escape the island with the most treasure possible. \n"
+ "Treasure(T) - increases as more treasure is found (ie, the first treasure is "
+ "+100 gold doubloons, the second is +200 gold doubloons and so on)\n"
+ "Pirates (P) - the more pirates you encounter the more gold doubloons they can take."
+ " (ie. the first crew is -100 gold doubloons, the second crew "
+ "is -200 gold doubloons and so on) BUT, if the player has no"
+ " treasure, there is nothing to be lost.\n"
+ "Both (B)- If there is treasure AND pirates, the player gets half the treasure "
+ "available.\n"
+ "Escape (X) - When a ship is seen in the distance, the user can leave the island.");
}
/**
*
*/
public static void Legend() {
System.out.println("****************************************");
System.out.println("%20s " + "LEGEND");
System.out.println("****************************************");
System.out.println("0 - indicates that spot has not been explored yet\n"
+ "T - indicates treasure, 100 gold doubloon\n"
+ "P - indicates pirates, lose 100 gold doubloon\n"
+ "B - indicates both treasure and pirates, 50 gold doubloon\n"
+ "X - escape the island");
}
These codes are not yet finished and I am still in process in doing it. There are
more methods to work on but what I am trying to work on right now, is how I can
display how much gold doubloons the user has every time a treasure is found and
every time they encounter a pirate or both.
The user's position entry must be between 0 and 99, inclusive. Should look like
this.
Positions [00 - 09]: 0 0 0 0 0 0 0 0 0 0
Positions [10 - 19]: 0 0 0 0 0 0 0 0 0 0
Positions [20 - 29]: 0 0 0 0 0 0 0 0 0 0
Positions [30 - 39]: 0 0 0 0 0 0 0 0 0 0
Positions [40 - 49]: 0 0 0 0 0 0 0 0 0 0
Positions [50 - 59]: 0 0 0 0 0 0 0 0 0 0
Positions [60 - 69]: 0 0 0 0 0 0 0 0 0 0
Positions [70 - 79]: 0 0 0 0 0 0 0 0 0 0
Positions [80 - 89]: 0 0 0 0 0 0 0 0 0 0
Positions [90 - 99]: 0 0 0 0 0 0 0 0 0 0
答案 0 :(得分：0)
在本例中，让 x 代表所选单元格的 x 坐标，y 代表所选单元格的 y 坐标。您的代码似乎相当不完整，因此让 doubloons 表示玩家在游戏期间持有的达布隆总数。
// We need two fields to track the number of treasures and pirates.
// These fields are effectively multipliers for the rewards and penalties
// for each treasure or pirate encountered as the game progresses
int doubloons = 0;
int treasureCount = 0;
int pirateCount = 0;
if(board[x][y] == 'T')
{
treasureCount++;
doubloons = doubloons + (treasureCount * 100);
}
else if(board[x][y] == 'P')
{
pirateCount++;
doubloons = doubloons - (pirateCount * 100);
}
else if(board[x][y] == 'B')
{
treasureCount++;
pirateCount++;
doubloons = doubloons + (treasureCount * 50);
}
else if(board[x][y] == 'X')
{
// whatever your end game scenario is supposed to be
}
else
{
// whatever is supposed to happen when there's no treasure, no
// pirates and no escape
}
请将任何后续问题限制在如何跟踪宝藏和海盗的具体问题上，因为这似乎是手头的问题。 Stack Overflow 试图坚持“一个问题，一个接受的答案”格式，因此如果您有更多不相关（甚至半相关）的问题，请将它们作为单独的问题发布，以免混淆演示文稿。