C语言版井字棋

1、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#include#include#includevoid help();/帮助int play(char player,int choice);/对弈void init();/初始化棋盘void di

2、splay();/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c

3、har a3164;/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(nnnnnnttt井t字t棋);printf(nnnnnnnnttt1. 玩 家 先 手nnttt2. 电 脑 先 手nnttt3. 帮 助nnttt4. 退 出nnnnntt请输入: );player=getch();if(player=1)init(

4、);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;elseprintf(nntt 输入错误请重新输入！);sleep(1000);system(cls);continue;return 0;void help()system(cls);printf(nnnnnnnnnttt 帮助nnn);printf(t w上移, s下移

5、, a左移, d右移 nn);printf(tt 游戏中按4退出);printf(nnnntt 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);

6、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;elsecontinue;block(choice);system(cls);display();elseif(a(choice-1)/3)*10+2(c

7、hoice-1)%3)*21+4!= )printf(该位置已有其他棋子！);continue;elseclear(choice);break;/while结束b+;black(choice);player=0;symbol=#;/if结束else/电脑下棋choice=found();block(choice);b+;white(choice);player=1;symbol=*;if(c=judge(choice,symbol)!=0)/判断输赢if(c=2)symbol=;c=0;end(symbol);break;return 0;void init()/初始化棋盘int x=0,y=

8、0;int i=1;for(x=0;x31;x+)for(y=0;y64;y+)if(y=funcy(1)|y=funcy(2)|y=funcy(3)|y=funcy(2)+funcy(3)if(x=funcx(1)|x=funcx(4)|x=funcx(7)|x=funcx(4)+funcx(7)axy=+;elseaxy=|;else if(x=funcx(1)|x=funcx(4)|x=funcx(7)|x=funcx(4)+funcx(7)axy=-;elseaxy= ;/for(x=1;x=7;x=x+3)/在每个格子的左上角添加序号，范围19/for(y=1;y=3;y+)/a(f

9、uncx(x)+1)(funcy(y)+1)=i;/i+;/void display()/打印棋盘int x=0,y=0;for(x=0;x31;x+)for(y=0;y64;y+)printf(%c,axy);void block(int choice)/添加选择框int x,y;for(x=funcx(choice)+1;xfuncx(choice)+10;x+)for(y=funcy(choice)+2;yfuncy(choice)+20;y+)if(x=funcx(choice)+1|x=funcx(choice)+9)if(y=funcy(choice)+2|y=funcy(choi

10、ce)+19)axy=+;else if(yfuncy(choice)+14)axy=-;else if(xfuncx(choice)+6)if(y=funcy(choice)+2|y=funcy(choice)+19)axy=|;void clear(int choice)/清空选择框int x,y,i;for(i=1;i=9;i+)if(afuncx(choice)+1funcy(choice)+2=+&i!=choice)for(x=funcx(choice)+1;xfuncx(choice)+10;x+)for(y=funcy(choice)+2;yfuncy(choice)+20;y

11、+)if(x=funcx(choice)+1|x=funcx(choice)+9)if(y=funcy(choice)+2|y=funcy(choice)+19)axy= ;else if(yfuncy(choice)+14)axy= ;else if(xfuncx(choice)+6)if(y=funcy(choice)+2|y=funcy(choice)+19)axy= ;void black(choice)/添加黑棋int x,y;for(x=funcx(choice)+2;xfuncx(choice)+9;x+)for(y=funcy(choice)+4;yfuncy(choice)+

12、18;y+)if(x=funcx(choice)+2|x=funcx(choice)+8)axy=#;else if(y=funcy(choice)+4|y=funcy(choice)+17)axy=#;void white(choice)/添加白棋int x;int d=4;for(x=funcx(choice)+2;x=funcx(choice)+8;x+)axfuncy(choice)+d=*;axfuncy(choice)+20-d=*;d=d+2;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if(afuncx(ch

13、oice)+2funcy(choice)+4=afuncx(choice+3)+2funcy(choice)+4&afuncx(choice+3)+2funcy(choice)+4=afuncx(choice+6)+2funcy(choice)+4)|(afuncx(choice)+2funcy(choice)+4=afuncx(choice)+2funcy(choice+1)+4&afuncx(choice)+2funcy(choice+1)+4=afuncx(choice)+2funcy(choice+2)+4)|afuncx(5)+2funcy(5)+4=symbol&(afuncx(1

14、)+2funcy(1)+4=afuncx(5)+2funcy(5)+4&afuncx(5)+2funcy(5)+4=afuncx(9)+2funcy(9)+4)|(afuncx(3)+2funcy(3)+4=afuncx(5)+2funcy(5)+4&afuncx(5)+2funcy(5)+4=afuncx(7)+2funcy(7)+4)return 1;if(b=9)return 2;return 0;int funcx(int choice)/将标号转换为行号return (choice-1)/3)%3)*10;int funcy(int choice)/将标号转换为列号return (choice-1)%3)*21;void end(char symbol)/结束system(cls);display();i