MODE —— 两个人在计算机上玩圈叉游戏|井字游戏（知识点：二维数组）

问题描述：

        让两个人在计算机上玩井字游戏（也称圈叉游戏）。

        井字游戏就是一个3X3的方格，两个人轮流在方格中输入标记X 或者 O 。谁先使自己的3个标记链接成水平，垂直或对角线。谁就是赢家。 

运行结果：

代码部分（代码说明）：

#include <stdio.h>
int main()
{	
	int player = 0; 		//Current player number -1 or 2?
	int winner = 0;			//The winning player number
	unsigned int i = 0;
	int choice = 0;			//Chosen square
	unsigned int row = 0;		//Row index for a square
	unsigned int column = 0;	//Column index for a square
	unsigned int line  = 0;

	char board[3][3] = {		//The board
		{'1','2','3'},		//Initial values are characters '1' to '9'
		{'4','5','6'},		//used to select a vacant square
		{'7','8','9'}		//for a player`s turn				
			};
	//The main game loop, The game continues for up to 9 turns
	//as loong as there is no winner
	for(i = 0;i < 9 && winner ==0;++i)
	{
		//Display the board
		printf("\n");
		printf(" %c | %c | %c \n",board[0][0],board[0][1],board[0][2]);
		printf("---+---+---\n");
		printf(" %c | %c | %c \n",board[1][0],board[1][1],board[1][2]);
		printf("---+---+---\n");
		printf(" %c | %c | %c \n",board[2][0],board[2][1],board[2][2]);
		
 //有一种让两个玩家轮流输入标记的办法，将两个玩家识别为1和2，编号为1的玩家先玩。然后根据轮流的次数决定输入标记的玩家的号码。
 //轮到奇数号码时候  就由玩家1输入标记。轮到偶数号时，就由玩家2输入标记。
		player = i%2 + 1;	//Select player  1 or 2
		
		//Get valid player square selection 
		do{
			printf("Player %d,please enter a valid square number"
				"for where you want to place your %c: ",player,(player == 1) ? 'X' : 'O');
			

//轮到一个玩家输入标记时，需要一种方法标记选择出来的方格。可以用1~9的数字标记这9个方格。玩家只需要输入要选择的方格数字。
                      scanf("%d",&choice);

row = --choice/3; //Get row index of squarecolumn = choice % 3; //Get column index of square}while(choice < 0 || choice > 8 || board[row][column] > '9'); //有三种可能导致选择无效：*输入的方格数小于0； *输入的方格数大于8； *选择已包含X或者O的方格（因为X O 的字符码都大于9）//Insert player symbloboard[row][column] = (player == 1) ? 'X' : 'O';//Check for a winning line - diagonals first if((board[0][0] == board[1][1] && board[0][0] == board[2][2]) || (board[0][2] == board[1][1] && board[0][2]== board[2][0]))winner = player;else{//Check row and columns for a winning line for(line = 0;line <= 2;++line ){if((board[line][0] == board[line][1] && board[line][0] == board[line][2]) || (board[0][line] == board[1][line] && board[0][line] == board[2][line]))winner = player;}}/*Code to check for a winner*/}/*Code to output the game*///Game is over so display the final boardprintf("\n");printf(" %c | %c | %c \n",board[0][0],board[0][1],board[0][2]);printf("---+---+---\n");printf(" %c | %c | %c \n",board[1][0],board[1][1],board[1][2]);printf("---+---+---\n");printf(" %c | %c | %c \n",board[2][0],board[2][1],board[2][2]);//Display result message if(winner)printf("\nCongratulation,player %d,YOU ARE THE WONNER!\n",winner);elseprintf("\nHow boring , it is a draw\n");return 0;}