C语言版井字棋.docx

十五、井字棋【问题描述】设计一个井字棋的游戏程序【基本要求】游戏规则是：从一个空的棋盘开始，人为x一方，计算机为o一方，人机双方轮流放置棋子，人下的时候，字符x将被放入棋盘中指定的位置，轮到计算机下时，程序将扫描棋盘，并将字符o放入扫描到的第一个空格，某一方有3枚棋子占据了一横行、一竖行或一对角线则获胜，若整个棋盘被占满还没有一方获胜则为和局。截图：代码：//@author GXU-pengcheng#includestdio.h#includewindows.h#includeconio.hvoid Help();//帮助int Play(char player,int choice);//对弈void Init();//初始化棋盘void Display();//打印棋盘void Black(int choice);//黑棋void White(int choice);//白棋void Block(int choice);//添加选择框void Clear(int choice);//清空之前的选择框int Judge(int choice,char symbol);//判断输赢 返回值0为无结果,1为获胜,2为平局int Funcx(int choice);//将标号转换为行号int Funcy(int choice);//将标号转换为列号void End(char symbol);//结束int Found();//返回第一个没有棋子的位置的编号char a[31][64];//用数组存放棋盘和棋子，a[行][列]int b=0;//棋子个数int main(){char player;int choice; system(title 井字棋);//设置标题system(mode con cols=64 lines=35);//设置窗口大小system(color 70);//设置颜色while(1){printf(\n\n\n\n\n\n\t\t\t井\t字\t棋);printf(\n\n\n\n\n\n\n\n\t\t\t1. 玩 家 先 手\n\n\t\t\t2. 电 脑 先 手\n\n\t\t\t3. 帮 助\n\n\t\t\t4. 退 出\n\n\n\n\n\t\t请输入: );player=getch();if(player==1){Init();Block(5);choice=5;Play(player,choice);}else if(player==2){Init();Play(player,choice);}else if(player==3){Help();getch();system(cls);continue;}else if(player==4){return 0;}else{printf(\n\n\t\t 输入错误请重新输入！);Sleep(1000);system(cls);continue;}}return 0;}void Help(){system(cls);printf(\n\n\n\n\n\n\n\n\n\t\t\t 帮助\n\n\n);printf(\t W上移, S下移, A左移, D右移 \n\n);printf(\t\t 游戏中按4退出);printf(\n\n\n\n\t\t 按任意键退出);}int Play(char player,int choice){//对弈char get;char symbol;int c=0;//Judge得出的结果while(1){system(cls);//每次循环清屏一次Display();if(player==1){//玩家下棋while(1){//确定要下的位置if((get=getch())!= ){if(get==4){system(cls);return 0;}else if( (get==w||get==W) choice-3=1){Clear(choice);choice =choice-3;}else if((get==s||get==S)choice+3=9){Clear(choice);choice = choice+3;}else if((get==a||get==A)(choice+2)/3==(choice+1)/3){Clear(choice);choice -= 1;}else if((get==d||get==D)(choice-1)/3==choice/3){Clear(choice);choice += 1;}else{continue;}Block(choice);system(cls);Display();}else{