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);