c语言井字游戏

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<stdio.h>#include<windows.h>#include<conio.h>void 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{if(a[((choice-1)/3)*10+2][((choice-1)%3)*21+4]!=' '){ printf("该位置已有其他棋子！");continue;}else{Clear(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=0;int i='1';for(x=0;x<31;x++){for(y=0;y<64;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))a[x][y]='+';elsea[x][y]='|';}elseif(x==Funcx(1)||x==Funcx(4)||x==Funcx(7)||x==Funcx(4)+Funcx(7))a[x][y]='-';elsea[x][y]=' ';}//for(x=1;x<=7;x=x+3)//在每个格子的左上角添加序号，范围1~9 //{// for(y=1;y<=3;y++)// {// a[(Funcx(x)+1)][(Funcy(y)+1)]=i;// i++;// }//}}void Display(){//打印棋盘int x=0,y=0;for(x=0;x<31;x++){for(y=0;y<64;y++){printf("%c",a[x][y]);}}}void Block(int choice){//添加选择框int x,y;for(x=Funcx(choice)+1;x<Funcx(choice)+10;x++){for(y=Funcy(choice)+2;y<Funcy(choice)+20;y++)if(x==Funcx(choice)+1||x==Funcx(choice)+9){if(y==Funcy(choice)+2||y==Funcy(choice)+19)a[x][y]='+';else if(y<Funcy(choice)+7||y>Funcy(choice)+14)a[x][y]='-';}else if(x<Funcx(choice)+4||x>Funcx(choice)+6)if(y==Funcy(choice)+2||y==Funcy(choice)+19)a[x][y]='|';}}void Clear(int choice){//清空选择框int x,y,i;for(i=1;i<=9;i++){if(a[Funcx(choice)+1][Funcy(choice)+2]=='+'&&i!=choice){ for(x=Funcx(choice)+1;x<Funcx(choice)+10;x++){for(y=Funcy(choice)+2;y<Funcy(choice)+20;y++)if(x==Funcx(choice)+1||x==Funcx(choice)+9){if(y==Funcy(choice)+2||y==Funcy(choice)+19)a[x][y]=' ';else if(y<Funcy(choice)+7||y>Funcy(choice)+14)a[x][y]=' ';}else if(x<Funcx(choice)+4||x>Funcx(choice)+6)if(y==Funcy(choice)+2||y==Funcy(choice)+19)a[x][y]=' ';}}}}void Black(choice){//添加黑棋int x,y;for(x=Funcx(choice)+2;x<Funcx(choice)+9;x++){for(y=Funcy(choice)+4;y<Funcy(choice)+18;y++)if(x==Funcx(choice)+2||x==Funcx(choice)+8)a[x][y]='#';else if(y==Funcy(choice)+4||y==Funcy(choice)+17)a[x][y]='#';}}void White(choice){//添加白棋int x;int d=4;for(x=Funcx(choice)+2;x<=Funcx(choice)+8;x++){a[x][Funcy(choice)+d]='*';a[x][Funcy(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(((a[Funcx(choice)+2][Funcy(choice)+4]==a[Funcx(choice+3)+2][Funcy(choice)+4]&&a[Funcx(choice+3)+2][Funcy(choice)+4]==a[Funcx(choice+6)+2][Funcy(choice) +4])||(a[Funcx(choice)+2][Funcy(choice)+4]==a[Funcx(choice)+2][Funcy(choice+1)+4]&&a[Funcx(choice)+2][Funcy(choice+1)+4]==a[Funcx(choice)+2][Funcy(choice+2) +4]))||a[Funcx(5)+2][Funcy(5)+4]==symbol&&((a[Funcx(1)+2][Funcy(1)+4]==a[Funcx(5)+2][Funcy(5)+4]&&a[Funcx(5)+2][Funcy(5)+4]==a[Funcx(9)+2][Funcy(9)+4])||(a[Funcx(3)+2][Funcy(3)+4]==a[Funcx(5)+2][Funcy(5)+4]&&a[Funcx(5)+2][Funcy(5)+4]==a[Funcx(7)+2][Funcy(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();if(symbol=='*')printf("\t\t\t 电脑胜利！\n\n");else if(symbol=='#')printf("\t\t\t 玩家胜利！\n\n");elseprintf("\t\t\t 平局！\n\n");b=0;printf("\t\t\t按任意键返回菜单");getch();system("cls");}int Found(){//返回第一个没有棋子的位置的编号int i;for(i=1;i<=9;i++){if(a[Funcx(i)+2][Funcy(i)+4]==' ')return i;}}如有侵权请联系告知删除，感谢你们的配合！。

C语⾔实现井字棋游戏（⼈机对弈）井字棋游戏：即三⼦棋，英⽂名叫Tic-Tac-Tic，是⼀种在3*3格⼦上进⾏的连珠游戏，和五⼦棋⽐较类似，由于棋盘⼀般不画边线框，格线排成井字故得名。

题⽬分析：要完成该游戏的编写，我们需要先分析出完成整个游戏过程都需要⼲什么？1.⾸先，需要定义出⼀个3*3的棋盘，根据相关知识，我们可以以⼆维数组的⽅式将棋盘表⽰出来；2.棋盘定义出来后，需要将棋盘初始化，将3*3⼆维数组的每⼀个位置初始化为‘ ’（空格）；3.有了棋盘，我们就可以开始进⾏下棋了，⾸先要确定是玩家先下还是电脑先下。

在这⾥我们规定玩家先下且玩家的下棋⽅式为‘x’，电脑下棋⽅式为‘o’；4.每⼀次下完棋后需要进⾏检测，判断该下棋位置是否合法、判断是否胜利等等。

根据上述分析，可以⼤致定义出以下函数窗⼝：void InitGame();//初始化游戏（棋盘）void PrintChess();//输出棋盘void PlayerMove();//玩家下棋void ComputerMove();//电脑下棋char CheckGameOver();//判断游戏是否结束（玩家胜/电脑胜/和棋）初始化棋盘：将3*3的⼆维数组棋盘的每个位置初始化为‘ ’void InitGame(){for (int i = 0; i < ROW; i++){for (int j = 0; j < COL; j++)chess_board[i][j] = ' ';}}输出棋盘：输出棋盘时，棋盘的风格可以根据⾃⼰的喜好来设计void PrintfChess()//输出棋盘，棋盘的设计可以根据⾃⼰的喜好设计{for (int i = 0; i < ROW; i++){printf("| %c | %c | %c |\n", chess_board[i][0], chess_board[i][1], chess_board[i][2]);if (i < ROW - 1)printf("|---|---|---|\n");}}玩家下棋：玩家输⼊下棋位置后，需要判断该位置是否合法、输⼊位置是否已被占⽤void PlayerMove()//玩家下棋{printf("玩家落⼦.\n");int row, col;while (1){printf("请输⼊⼀组坐标(下棋位置):>");scanf("%d %d", &row, &col);//检查坐标的有效性if (row < 0 || row > ROW || col < 0 || col > COL){printf("输⼊⾮法，请重新输⼊...");continue;}if (chess_board[row][col] != ' '){printf("输⼊的位置已被占⽤，请重新输⼊...");continue;}chess_board[row][col] = 'x';//x代表玩家下的棋break;}}电脑下棋：电脑下棋时，下棋的位置利⽤srand函数随机产⽣void ComputerMove()//电脑下棋{srand(time(0));while (1){int row = rand() % ROW;int col = rand() % COL;if (chess_board[row][col] != ' '){continue;}chess_board[row][col] = 'o';//o代表电脑下的棋break;}}检查棋盘：在检测棋盘时，分别判断⾏、列、对⾓线，在这⾥我规定：'x'代表玩家赢 'o'代表电脑赢 'h'代表和棋 'c'代表继续char CheckGameOver()//检测游戏是否结束{//检查⾏for (int i = 0; i < ROW; i++){if (chess_board[i][0] != ' '&& chess_board[i][0] == chess_board[i][1]&& chess_board[i][0] == chess_board[i][2])return chess_board[i][0];}//检查列for (int j = 0; j < COL; j++){if (chess_board[0][j] != ' '&& chess_board[0][j] == chess_board[1][j]&& chess_board[0][j] == chess_board[2][j])return chess_board[0][j];}//检查对⾓线if (chess_board[0][0] != ' '&& chess_board[0][0] == chess_board[1][1]&& chess_board[0][0] == chess_board[2][2])return chess_board[0][0];if (chess_board[0][2] != ' '&& chess_board[0][2] == chess_board[1][1]&& chess_board[0][2] == chess_board[2][0])return chess_board[0][2];//判断是否和棋if (ChessFull())return 'h';return 'c';}⾄此，主要的功能函数均已编写完毕，整个程序的流程如下所⽰：1.初始化棋盘；2.输出棋盘；3.玩家下棋；4.检测棋盘；5.电脑下棋；6.检测棋盘#define _CRT_SECURE_NO_WARNINGS#include <stdio.h>#include <time.h>#include <stdbool.h>#include <stdlib.h>#define START 1#define QUIT 0#define ROW 3#define COL 3static char chess_board[ROW][COL];//定义棋盘void StartGame();void InitGame();void PrintfChess();void PlayerMove();void ComputerMove();char CheckGameOver();bool ChessFull();int main(int argc, char* argv[]){int select = 1;while (select){printf("*********************\n");printf("* [1] Start Game *\n");printf("* [2] Over Game *\n");printf("*********************\n");printf("请选择:>");scanf_s("%d", &select);if (select == QUIT)break;if (select != START){printf("输⼊有错，请重新输⼊.....\n"); continue;}StartGame();}printf("GoodBye.....");return 0;}void StartGame(){char winner;//1 初始化游戏（棋盘）InitGame();//2 进⼊游戏while (1){//3 输出棋盘PrintfChess();//4玩家下棋PlayerMove();//5检查结果winner = CheckGameOver();if (winner != 'c')break;//6电脑下棋ComputerMove();//7检查结果CheckGameOver();winner = CheckGameOver();if (winner != 'c')break;}if (winner == 'x')printf("玩家赢.\n");printf("电脑赢.\n");if (winner == 'h')printf("和棋.\n");}void InitGame(){for (int i = 0; i < ROW; i++){for (int j = 0; j < COL; j++)chess_board[i][j] = ' ';}}void PrintfChess()//输出棋盘，棋盘的设计可以根据⾃⼰的喜好设计{for (int i = 0; i < ROW; i++){printf("| %c | %c | %c |\n", chess_board[i][0], chess_board[i][1], chess_board[i][2]); if (i < ROW - 1)printf("|---|---|---|\n");}}void PlayerMove()//玩家下棋{printf("玩家落⼦.\n");int row, col;while (1){printf("请输⼊⼀组坐标(下棋位置):>");scanf("%d %d", &row, &col);//检查坐标的有效性if (row < 0 || row > ROW || col < 0 || col > COL){printf("输⼊⾮法，请重新输⼊...");continue;}if (chess_board[row][col] != ' '){printf("输⼊的位置已被占⽤，请重新输⼊...");continue;}chess_board[row][col] = 'x';//x代表玩家下的棋break;}}void ComputerMove()//电脑下棋{srand(time(0));while (1){int row = rand() % ROW;int col = rand() % COL;if (chess_board[row][col] != ' '){continue;}chess_board[row][col] = 'o';//o代表电脑下的棋break;}}/** 'x'代表玩家赢* 'o'代表电脑赢* 'h'代表和棋* 'c'代表继续*/char CheckGameOver()//检测游戏是否结束{//检查⾏for (int i = 0; i < ROW; i++){if (chess_board[i][0] != ' '&& chess_board[i][0] == chess_board[i][1]&& chess_board[i][0] == chess_board[i][2])return chess_board[i][0];}//检查列for (int j = 0; j < COL; j++){if (chess_board[0][j] != ' '&& chess_board[0][j] == chess_board[1][j]&& chess_board[0][j] == chess_board[2][j])return chess_board[0][j];}//检查对⾓线if (chess_board[0][0] != ' '&& chess_board[0][0] == chess_board[1][1]&& chess_board[0][0] == chess_board[2][2])return chess_board[0][0];if (chess_board[0][2] != ' '&& chess_board[0][2] == chess_board[1][1]&& chess_board[0][2] == chess_board[2][0])return chess_board[0][2];//判断是否和棋if (ChessFull())return 'h';return 'c';}bool ChessFull(){for (int i = 0; i < ROW; i++){for (int j = 0; j < COL; j++){if (chess_board[i][j] == ' ')return false;}}return true;}运⾏测试图：程序的运⾏界⾯我们还可以利⽤system("cls")对界⾯进⾏优化，这样可以使界⾯更加美观。

Xx学院
计算机信息学院《C语言课程设计(短一)》
指
导
书
年
附件二封面格式
学院
C语言课程设计总结报告
题目井字棋
指导教师
院系计算机与信息工程学院
专业计算机科学与技术
班级
学号
姓名
成绩
xx年xx月xx日
附件三目录格式
目录
第一章系统功能说明 (1)
第二章使用说明 (2)
2．1 安装手册 (3)
2．2 使用手册 (4)
第三章程序结构 (5)
3．1 程序结构说明 (6)
3．2 重要数据说明 (7)
3．3 函数清单 (8)
第四章系统设计难点及其解决方法 (9)
第五章不足之处 (10)
第一章系统功能说明
井字游戏的游戏界面是一个3*3的格子，系统提示玩家输入棋子的坐标来表示要下的棋子的位置，如果玩家输入的坐标所在位置已经有棋子或者该坐标已经超出了游戏坐标的范围，就提示玩家重新输入坐标，。

当有三个棋子在一条直线上的时候就结束游戏，并输出哪个玩家赢了。

c语言井字游戏第一篇：c语言井字游戏#include void main(){ int hang,lie,count=0;int player,ying=0;char weizhi[3][3]={ {'-','-','-'}, {'-','-','-'}, {'-','-','-'} };for(int i=0;i<=9&&ying==0;i++){printf(“nn”);printf(“§0 1 2n”);printf(“0 %c %c %cn”,weizhi[0][0],weizhi[0][1],weizhi[0][2]);printf(“1 %c %c %cn”,weizhi[1][0],weizhi[1][1],weizhi[1][2]);printf(“2 %c %c %cn”,weizhi[2][0],weizhi[2][1],weizhi[2][2]);player=i%2+1;do{printf(“n 玩家%d:请输入您的棋子位置(%c)”,player,(player==1)? '1':'2');scanf(“%d%d”,&hang,&lie);if(weizhi[hang][lie]=='1'||weizhi[hang][lie]=='2')printf(“n对不起!这个位置已经有棋子了,请您再输入棋子的位置:n”);}:while(hang<0||hang>2||lie<0||lie>2||weizhi[hang][lie]=='1'|| weizhi[hang][lie]=='2');weizhi[hang][lie]=(player==1)? '1':'2';count++;if(count==9)break;if((weizhi[0][0]==weizhi[1][1]&&weizhi[0][0]==weizhi[2][2] &&weizhi[2][2]!='-')||(weizhi[0][2]==weizhi[1][1]&&weizhi[0][2]==weizhi[2][0]& &weizhi[2][0]!='-')||(weizhi[0][0]==weizhi[1][0]&&weizhi[2][0]==weizhi[1][0]&&weizhi[1][0]!='-')||(weizhi[0][1]==weizhi[1][1]&&weizhi[2][1]==weizhi[1][1]& &weizhi[1][1]!='-')||(weizhi[0][2]==weizhi[1][2]&&weizhi[2][2]==weizhi[0][2]& &weizhi[0][2]!='-')||(weizhi[0][0]==weizhi[0][1]&&weizhi[0][0]==weizhi[0][2]& &weizhi[0][2]!='-')||(weizhi[1][0]==weizhi[1][1]&&weizhi[1][2]==weizhi[1][1]& &weizhi[1][1]!='-')||(weizhi[2][0]==weizhi[2][1]&&weizhi[2][2]==weizhi[2][1]& &weizhi[2][1]!='-')){ying=player;break;}elseif(count==9)break;}printf(“nn”);printf(“%c %c %cn”,weizhi[0][0],weizhi[0][1],weizhi[0][2]);printf(“%c %c %cn”,weizhi[1][0],weizhi[1][1],weizhi[1][2]);printf(“%c %c %cn”,weizhi[2][0],weizhi[2][1],weizhi[2][2]);if(ying==0)printf(“n实力相当!此局平局!n”);elseprintf(“n恭喜玩家%d!了!nn”,ying);}您赢补充：第二篇：猜字游戏横向：1、世乒赛中与王励勤一起拿下混双金牌的女选手2、即APEC3、我国一种民间工艺4、欧阳修的《生查子》中“月上柳梢头”的下句5、中德合资的汽车有限公司，车标是“Volks Wagenwerk”的两个首字母6、《倚天屠龙记》中男主人公的义父7、著名国画大师徐悲鸿的一幅名作8、由用餐者自取菜肴、主食的用餐方式9、自愿服兵役的士兵，我国专指服满一定年限的兵役后自愿继续服役的士兵10、家庭内部代代相传的有价值之物11、塞黑国家足球队球员，现效力于国际米兰12、在病人自愿的前提下，由医务人员实施的加速其死亡的一种行为13、无机化合物，成分是碳酸氢钠(NaHCO3)，可以灭火或制焙粉纵向：一、《西游记》主角之一二、形容文学作品叙述描写真实生动三、美国一所著名的私立大学四、魏源提出“师夷长技以制夷”的一本书五、美国乔治亚州的首府，1996年奥运会的举办城市六、一部网络小说，在Google和百度的2004年十大中文搜索关键词中分别位列第三和第十七、为买卖双方撮合或代他人进行买卖而取得佣金的人八、曾因兴奋剂而冠军旁落的短跑名将九、由阿里巴巴公司投资创办的个人交易网上平台十、一个棒球术语十一、“雷锋精神”的一个方面十二、达·芬奇的一幅名画横列1.由商业银行发行，专供消费者购买商品和支付费用的凭证。

井字棋游戏代码（Tic tac toe game code）#包含iostream > <#包括<<程序。

”#包括<<时间。

”使用名称空间；字符“=”；字符；国际jiru1 [ 9 ] = { 0 }；//人人游戏时用来记录是否在此处下过棋int = i 1；字符游戏[ 3 ] [ 3 ]；//用来下棋int，y；a [ 8 ]；//人机游戏时记录每行，列，斜行有多少子b [ 8 ]；//人机游戏时记录机器每行，列，斜行有多少子int c [ 8 ]；//人机游戏时记录人每行，列，斜行有多少子int吉如[ 3 ] [ 3 ] = { 0 }；//人机游戏时用来记录是否在此处下过棋int齐[ 3 ] [ 3 ] = { 0 }；//人人游戏时用来记录机器是否在此处下过棋int任[ 3 ] [ 3 ] = { 0 }；//人人游戏时用来记录人是否在此处下过棋无效的转化（int t）/把输入的1 ~ 9数字转化成坐标{整数h；h = % 3；G =（TH）/ 3；= g；Y = H-1；}无效suan() /人机游戏时计算每行，列，斜行有多少子{一个[ 1 ] =吉如[ 0 ] [ 0 ] +吉如[ 0 ] [ 1 ] +吉如[ 0 ] [ 2 ]；一个[ 2 ] =吉如[ 1 ] [ 0 ] +吉如[ 1 ] [ 1 ] +吉如[ 1 ] [ 2 ]；一个[ 3 ] =吉如[ 2 ] [ 0 ] +吉如[ 2 ] [ 1 ] +吉如[ 2 ] [ 2 ]；一个[ 4 ] =吉如[ 0 ] [ 0 ] +吉如[ 1 ] [ 0 ] +吉如[ 2 ] [ 0 ]；一个[ 5 ] =吉如[ 0 ] [ 1 ] +吉如[ 1 ] [ 1 ] +吉如[ 2 ] [ 1 ]；一个[ 6 ] =吉如[ 0 ] [ 2 ] +吉如[ 1 ] [ 2 ] +吉如[ 2 ] [ 2 ]；一个[ 7 ] =吉如[ 0 ] [ 0 ] +吉如[ 1 ] [ 1 ] +吉如[ 2 ] [ 2 ]；一个[ 8 ] =吉如[ 0 ] [ 2 ] +吉如[ 1 ] [ 1 ] +吉如[ 2 ] [ 0 ]；b [ 1 ] =齐[ 0 ] 0 [齐] [ 0 ] 1 [齐] [ 0 ] [ 2 ]；b [ 2 ] =齐[ 1 ] 0 [齐] [ 1 ] 1 [齐] [ 1 ] [ 2 ]；b [ 3 ] =齐[ 2 ] 0 [齐] [ 2 ] 1 [齐] [ 2 ] [ 2 ]；b [ 4 ] =齐[ 0 ] 0 [齐] [ 1 ] 0 [齐] [ 2 ] [ 0 ]；b [ 5 ] =齐[ 0 ] 1 [齐] [ 1 ] 1 [齐] [ 2 ] [ 1 ]；b [ 6 ] =齐[ 0 ] 2 [齐] [ 1 ] 2 [齐] [ 2 ] [ 2 ]；b [ 7 ] =齐[ 0 ] 0 [齐] [ 1 ] 1 [齐] [ 2 ] [ 2 ]；b [ 8 ] =齐[ 0 ] 2 [齐] [ 1 ] 1 [齐] [ 2 ] [ 0 ]；C [ 1 ] =任[ 0 ] [ 0 ] +任[ 0 ] [ 1 ]任任[ 0 ] [ 2 ]；C [ 2 ] =任[ 1 ] [ 0 ] +任[ 1 ] [ 1 ]任任[ 1 ] [ 2 ]；C [ 3 ] =任[ 2 ] [ 0 ] +任[ 2 ] [ 1 ]任任[ 2 ] [ 2 ]；C [ 4 ] =任[ 0 ] [ 0 ] +任[ 1 ] [ 0 ]任任[ 2 ] [ 0 ]；C [ 5 ] =任[ 0 ] [ 1 ] +任[ 1 ] [ 1 ]任任[ 2 ] [ 1 ]；C [ 6 ] =任[ 0 ] [ 2 ] +任[ 1 ] [ 2 ]任任[ 2 ] [ 2 ]；C [ 7 ] =任[ 0 ] [ 0 ] +任[ 1 ] [ 1 ]任任[ 2 ] [ 2 ]；C [ 8 ] =任[ 0 ] [ 2 ] +任[ 1 ] [ 1 ]任任[ 2 ] [ 0 ]；}无效chushihua() /把数据初始化{我= 1；k；对于（j＝0；j＜3；j + +）{对于（k＝0；k＜3；k +）{游戏[ K ]；}}对于（j＝0；j＜3；j + +）{对于（k＝0；k＜3；k +）{吉如[ J ] [ k ] = 0；气[钾]＝0；任志强[ J＝0；}}}无效shuru() /输入数据{k，j；对于（k＝0；k＜3；k +）{对于（j＝0；j＜3；j + +）{cout <<游戏[ k ] [ J ] <<”；}cout << endl；}}void（int，int的任侠）/人落子的情况{如果（i = 2 = 1）{游戏[不] [的] = 'x'；吉如[不] [的] = 1；仁[ T ] = 1；系统（“CLS”）；shuru()；}其他的{游戏[不] [的] = O；吉如[不] [的] = 1；仁[ T ] = 1；系统（“CLS”）；shuru()；}}无效huiqi1（int，int s）/人机游戏时人悔棋的情况{游戏；吉如[不] [的] = 0；仁[ T ] = 0；系统（“CLS”）；shuru()；}无效suiji() /产生随机数{int的选择；srand（（unsigned int）时间（空））；选择= rand() % 10；整数h；如果（选择= 10）{选择＝9；}否则如果（选择= 0）{选择＝1；}选择% 3；G =（choice-h）/ 3；游戏[克] [ 1 ] = 'x'；吉如[克] [ 1 ] = 1；齐[克] [ 1 ] = 1；}无效jixia() /人机游戏时机器下子{如果（i = 1）{suiji()；系统（“CLS”）；shuru()；}否则如果（i = 3）{int p；对于（p＝1，p＜10，p＋＋）{整数h；h = p % 3；G =（P-H）/ 3；如果（吉如[克] [ 1 ] = = 0）{游戏[克] [ 1 ] = 'x'；吉如[克] [ 1 ] = 1；齐[克] [ 1 ] = 1；打破；}}系统（“CLS”）；shuru()；}其他的{suan()；int，v＝0；对于（u＝1；u＜4；u +）{如果（（一个[ ] = = 2）&（（B [u] = = 2）| |（C [u] = = 2）））{变量z；对于（z＝0；z＜3；z + +）如果（吉如[ 1 ] [ Z ] = = 0）{游戏[ 1 ] [ Z ] = 'x'；吉如[ 1 ] [ Z ] = 1；齐[ 1 ] [ Z ] = 1；}}系统（“CLS”）；shuru()；++；打破；}}如果（= = 0）{对于（u＝4；u＜7；u +）{if (([u] = 2 & = & ((b) [u] = = 2) | | (c [u] = = 2))) {int z;for (z = 0,; z < 3; z + +){if (jiru [z], [u] = = 4 - 0){game [z], [u] = - 4 'x';jiru [z], [u] = 1 - 4;qi [z] [u] = 1 - 4;}}system ("cls");shuru ();v + +;break;}}if (v = = 0){if (([7] = = 2) & & ((b = = 2 (7) (c) | | [7] = = 2))) {if (jiru [0], [0] = = 0){game [0], [0] = 'x';jiru [0], [0] = 1;qi [0], [0] = 1;}else if (jiru [1] [1] = = 0){game [1], [1] = 'x';jiru [1] [1] = 1;qi [1] [1] = 1;}else if (jiru [2], [2] = = 0) {game [2], [2] = 'x';jiru [2], [2] = 1;qi [2], [2] = 1;}system ("cls");shuru ();v + +;}if (v = = 0){if (([8] = = 2) & & ((b [8] = = 2) | | (c [8] = = 2))) {if (jiru [0], [2] = = 0){game [0], [2] = 'x';jiru [0], [2] = 1;qi [0], [2] = 1;}else if (jiru [1] [1] = = 0){game [1], [1] = 'x';jiru [1] [1] = 1;qi [1] [1] = 1;}else if (jiru [2], [0] = = 0) {game [2], [0] = 'x';jiru [2], [0] = 1;qi [2], [0] = 1;}system ("cls");shuru ();v + +;}if (v = = 0){for (u = 1; u < 4; u + +) {if (([u] = = 1 & & (b) [u] = = 1)) {int z;for (z = 0,; z < 3; z + +){if (u jiru [- 1] [z] = = 0){game [n - 1] [z] = 'x';jiru [n - 1] [z] = 1;qi [n - 1] [z] = 1;system ("cls");shuru ();v + +;break;}}}}if (i = = 0){for (u = 4; u < 7; u + +){if (a [n] = = 1) & & (b [n] = = 1) {int z;for (z = 0 and z < 3; z + +){if (jiru [z] [u4] = = 0){game [z] [u4] = 'x';jiru [z] [u4] = 1;qi [z] [u4] = 1;system ("cls");shuru ();v + +;break;}}}}if (i = = 0){if (a [7] = = 1) & & (b [7] = = 1) & & (c) [7] = = 0){if ((jiru [0] [0] = = 1) & & (jiru [1] [1] = = 0) {game [1] [1] = 'x'; jiru [1] [1] = 1; qi [1] [1] = 1;}else if ((jiru [1] [1] = = 1) & & (jiru [2] [3] = = 0) {game [1] [2] = 'x'; jiru [1] [2] = 1; qi [1] [2] = 1;}else if ((jiru [2] [3] = = 1) & & (jiru [1] [1] = = 0) {game [1] [1] = 'x'; jiru [1] [1] = 1; qi [1] [1] = 1;}system ("cls");shuru ();v + +;}if (i = = 0){if (a [8] = = 1) & & (b [8] = = 1) & & (c) [8] = = 0){if ((jiru [0], [2] = = 1) & & (jiru [1] [1] = = 0) {game [1] [1] = 'x'; jiru [1] [1] = 1; qi [1] [1] = 1;}else if ((jiru [1] [1] = = 1) & & (jiru [0], [2] = = 0) {game [0], [2] = 'x'; jiru [0], [2] = 1; qi [0], [2] = 1;}else if ((jiru [2] [0] = = 1) & & (jiru [1] [1] = = 0) {game [1] [1] = 'x'; jiru [1] [1] = 1; qi [1] [1] = 1;}system ("cls");shuru ();v + +;}if (i = = 0){int p;for (p = 1; p < 9; p + +){int h, g;h = p% 3;g = (p h) / 3;if ((jiru [g] [h - 1] = = 0) & & (pure [g] [h - 1] = = 0) {game [g] [h - 1] = 'x';jiru [g] [h - 1] = 1;qi [g] [h - 1] = 1;system ("cls"); shuru ();v + +;break;}}system ("cls"); shuru ();}}}}}}}}void tishi () / / 提示{if (i = = 1){cout < < "建议输入1,3,7,9中的一个" < < final;}else if (a = = 2){if ((qi [0] [0] = = 1) | | (qi [0], [2] = = 1) | | (qi [2] [0] = = 1) | | (qi [2], [2] = = 1) & & (i = = 2)){cout < < "建议输入5" < < final;}else if (qi [1] [1] = = 1)cout < < "建议输入7" < < final; }else if (qi [0] [1] = = 1){cout < < "建议输入5" < < final; }else if (qi [2] [1] = = 1){cout < < "建议输入5" < < final; }else if (qi [1] [0] = = 1){cout < < "建议输入5" < < final; }else if (qi [1] [2] = = 1){cout < < "建议输入5" < < final;}}else if ((qi [0] [0] = = 1) & & (jiru [1] [0] = = 0) & & (qi [2] [0] = = 1)){cout < < "建议输入4" < < final;}else if ((qi [0] [0] = = 1) & & (jiru [0] [1] = = 0) & & (qi [0], [2] = = 1)){cout < < "建议输入2" < < final;}else if ((pure [1] [1] = = 1) & & (jiru [0], [2] = = 0) & & (jiru [2] [0] = = 0) & & (qi [2] [0] = = 0) & & (qi [1] [0] = = 0) & & (qi [2] [1] = = 0) & & (qi [2], [2] = = 0)cout < < "建议输入3" < < final;}else if ((pure [1] [1] = = 1) & & (jiru [0] [0] = = 0) & & (jiru [2], [2] = = 0){cout < < "请在正向的斜排未落子处落子" < < final;}else{suan ();int u, v = 0;for (u = 1; d < 4; d + +){if (a [d] = = 2 & & ((b [d] = = 2) | | (c [u] = = 2))){cout < < "请在第" < < d < < "行未落子处落子" < < final;v + +;break;}}if (v = = 0){for (u = 4; d < 7; d + +){if (a [d] = = 2 & & ((b [d] = = 2) | | (c [u] = = 2))){cout < < "请在第" < < (d 3) < < "列未落子处落子" < < final; v + +;break;}}if (v = = 0){if (a [8] = = 2 & & ((b [8] = = 2) | | (c [8] = = 2))) {cout < < "请在反向的斜排未落子处落子" < < final;v + +;}if (v = = 0){if (a [7] = = 2 & & ((b [7] = = 2) | | (c [7] = = 2))) {cout < < "请在正向的斜排未落子处落子" < < final;v + +;}if (v = = 0)for (u = 1; d < 4; d + +){if (a [d] = = 1 & & c [u] = = 1 & & b [d] = = 0){cout < < "请在第" < < d < < "行未落子处落子" < < final; v + +;break;}}if (v = = 0){for (u = 4; d < 7; d + +){if (a [d] = = 1 & & c [u] = = 1 & & b [d] = = 0){cout < < "请在第" < < (d 3) < < "列未落子处落子" < < final; v + +;break;}}if (v = = 0){if (a [8] = = 1 & & (b [8] = = 0 & & c [8] = = 1)){cout < < "请在反向的斜排未落子处落子" < < final;v + +;}if (v = = 0){if (a [7] = = 1 & & (b [7] = = 0 & & c [7] = = 1))cout < < "请在正向的斜排未落子处落子" < < final; v + +;}if (v = = 0){cout < < "请在未落子处落子" < < final;}}}}}}}}}}void jixianzou () / / 人机对战的情况{if((game[0][0]=='x'&&game[0][1]=='x'&&game[0][2]=='x')||(ga me[1][0]=='x'&&game[1][1]=='x'&&game[1][2]=='x')||(game[2][ 0]=='x'&&game[2][1]=='x'&&game[2][2]=='x')||(game[0][0]=='x '&&game[1][0]=='x'&&game[2][0]=='x')||(game[0][1]=='x'&&gam e[1][1]=='x'&&game[2][1]=='x')||(game[0][2]=='x'&&game[1][2 ]=='x'&&game[2][2]=='x')){cout <<“玩家失败”<< endl；}else if（（游戏[ 0 ] [ 0 ] = = 'x'和游戏[ 1 ] [ 1 ] = = 'x'和游戏[ 2 ] [ 2 ] = = 'x'）| |（游戏[ 0 ] [ 2 ] = = 'x'和游戏[ 1 ] [ 1 ] = = 'x'和游戏[ 2 ] [ 0 ] = = 'x'））{cout <<“玩家失败”<< endl；}elseif((game[0][0]=='o'&&game[0][1]=='o'&&game[0][2]=='o')||(ga me[1][0]=='o'&&game[1][1]=='o'&&game[1][2]=='o')||(game[2][ 0]=='o'&&game[2][1]=='o'&&game[2][2]=='o')||(game[0][0]=='o '&&game[1][0]=='o'&&game[2][0]=='o')||(game[0][1]=='o'&&gam e[1][1]=='o'&&game[2][1]=='o')||(game[0][2]=='o'&&game[1][2 ]=='o'&&game[2][2]=='o')){cout <<“玩家获胜”<< endl；}else if（（游戏[ 0 ] [ 0 ] = = O &游戏[ 1 ] [ 1 ] = = O &游戏[ 2 ] [ 2 ] = = O）| |（游戏[ 0 ] [ 2 ] = = O &游戏[ 1 ] [ 1 ] = = O &游戏[ 2 ] [ 0 ] = = O））{cout <<“玩家获胜”<< en dl；}否则如果（i = 10）{cout <<“平局”<< endl；}其他的{如果（i = 2 = 1）{jixia()；++；jixianzou()；}其他的{cout <<“是否需要提示（0,1）“<< endl；国际问；q值；如果（q＝0）{tishi()；}int t；CIN；转化（T）；如果（（t＜0）| | t > 10 | |吉如[X] [Y] = = 1）{cout <<“输入错误，请重新输入”<< endl；jixianzou()；}其他的{任侠（x，y）；cout <<“是否悔棋（Y或N），是否重新开始（C、N），是否结束（T 或N）“<< endl；CIN；如果（F = = Y）{huiqi1（x，y）；jixianzou()；}如果（F = =“c”）{系统（“CLS”）；chushihua()；shuru()；jixianzou()；}else if (f = = 't') {}elserenxia (x, y);in + +;jixianzou ();}}}}}void ren1 (int choice) / / 人人对战时人下子{if (in% 2 = = 1){int h, g;h = choice% 3;g = (choice h) / 3;game [g] [h - 1] = 'x'; jiru1 [choice - 1] = 1; system ("cls");shuru ();}else{int h, g;h = choice% 3;g = (choice h) / 3; game [g] [h - 1] = 'o'; jiru1 [choice - 1] = 1; system ("cls");shuru ();}void huiqi (int choice) / / 人人对战时人悔棋{int h, g;h = choice% 3;g = (choice h) / 3;game [g] [h - 1] = '/';jiru1 [choice - 1] = 0;system ("cls");shuru ();}void xiaqi () / / 人人对战时的下棋情况{if((game[0][0]=='x'&&game[0][1]=='x'&&game[0][2]=='x')||(ga me[1][0]=='x'&&game[1][1]=='x'&&game[1][2]=='x')||(game[2][ 0]=='x'&&game[2][1]=='x'&&game[2][2]=='x')||(game[0][0]=='x '&&game[1][0]=='x'&&game[2][0]=='x')||(game[0][1]=='x'&&gam e[1][1]=='x'&&game[2][1]=='x')||(game[0][2]=='x'&&game[1][2]=='x'&&game[2][2]=='x')){cout < < "玩家1获胜" < < final;}else if ((game [0] [0] = = 'x' & & game [1] [1] = = 'x' & & game [2], [2] = = 'x') | | (game [0], [2] = = 'x' & & game [1] [1] = = 'x' & & game [2] [0] = = 'x')){cout < < "玩家1获胜" < < final;}elseif((game[0][0]=='o'&&game[0][1]=='o'&&game[0][2]=='o')||(ga me[1][0]=='o'&&game[1][1]=='o'&&game[1][2]=='o')||(game[2][ 0]=='o'&&game[2][1]=='o'&&game[2][2]=='o')||(game[0][0]=='o '&&game[1][0]=='o'&&game[2][0]=='o')||(game[0][1]=='o'&&gam e[1][1]=='o'&&game[2][1]=='o')||(game[0][2]=='o'&&game[1][2 ]=='o'&&game[2][2]=='o')){cout < < "玩家2获胜" < < final;}else if ((game [0] [0] = = 'o' & & game [1] [1] = = 'o' & & game [2], [2] = = o) | | (game [0], [2] = = 'o' & & game [1] [1] = = 'o' & & game [2] [0] = = o)){cout < < "玩家2获胜" < < final;}else if (a = = 10){cout < < "平局" < < final;}else{int m;子宫内膜异位囊肿；如果（m＞9 | | M＜1 | | jiru1 [ ] = = 1 M-1）{cout <<“输入错误，请重新输入”<< endl；xiaqi()；}其他的{任（M）；cout <<“（是否悔棋Y或N）或（是否重新开始C或N）或（是否结束游戏T或N）“<< endl；CIN；如果（R = = Y）{惠奇（M）；xiaqi()；}如果（R = =“c”）{我= 1；k；对于（j＝0；j＜3；j + +）{对于（k＝0；k＜3；k +）{游戏[ K ]；}}对于（j＝0；j＜9；j + +）{jiru1 [ J ] = 0；}系统（“CLS”）；shuru()；xiaqi()；}否则如果（r = =）{}其他的{++；系统（“CLS”）；shuru()；xiaqi()；}}}}国际main() /主函数cout <<“欢迎进入井字棋游戏！”<< endl；cout <<“本游戏用数字1 ~ 9表示下棋位置”<< endl；（1）{如果（（R = =不）| |（F = =不））{打破；}cout <<“是否下棋Y或N”<< endl；字符Y；CIN；如果（y = = n）{打破；}其他的{我= 1；k；对于（j＝0；j＜3；j + +）{对于（k＝0；k＜3；k +）{游戏[ K ]；}}对于（j＝0；j＜9；j + +）{jiru1 [ J ] = 0；}cout <<“人机下棋或人人下棋（1或2）”<< en dl；int u；宫颈囊肿；如果（u＝2）{shuru()；xiaqi()；}如果（u＝1）{chushihua()；shuru()；jixianzou()；}}}返回0；}。

c 井字棋课课程设计一、教学目标本节课旨在通过学习井字棋游戏的设计与实现，让学生掌握以下知识目标：理解井字棋的规则；掌握Python编程基础，如条件语句、循环语句等。

在技能目标方面，学生能够运用Python编程实现井字棋游戏，提高编程能力。

在情感态度价值观目标方面，学生通过小组合作完成游戏设计，培养团队协作精神和问题解决能力。

二、教学内容本节课的教学内容主要包括井字棋游戏的规则介绍、Python编程基础以及井字棋游戏的实现。

首先，介绍井字棋的规则，让学生了解游戏的基本玩法。

其次，教授Python编程基础，包括条件语句、循环语句等，为学生实现井字棋游戏打下基础。

最后，指导学生运用所学的Python编程知识，分组设计并实现井字棋游戏。

三、教学方法为了提高学生的学习兴趣和主动性，本节课采用讲授法、讨论法、案例分析法和实验法等多种教学方法。

在讲授井字棋规则时，采用生动的语言和实例进行讲解，让学生更容易理解和接受。

在教授Python编程基础时，通过案例分析和实验操作，让学生在实践中掌握编程知识。

在设计井字棋游戏环节，鼓励学生分组讨论、合作解决问题，培养团队协作精神。

四、教学资源为了支持教学内容和教学方法的实施，丰富学生的学习体验，本节课准备以下教学资源：教材《Python编程：从入门到实践》、参考书《Python核心编程》、多媒体资料（井字棋游戏教学视频、编程实例演示等）、实验设备（计算机、投影仪等）。

这些教学资源将有助于学生更好地掌握井字棋游戏的设计与实现，提高编程能力。

五、教学评估本节课的教学评估将采用多元化评价方式，全面客观地评价学生的学习成果。

评估主要包括以下几个方面：1.平时表现：观察学生在课堂上的参与程度、提问回答、小组讨论等，以了解学生的学习态度和实际操作能力。

2.作业：布置相关的编程练习，要求学生在规定时间内完成，以检验学生对井字棋游戏设计和Python编程知识的掌握程度。

3.考试：设置期末考试，包括选择题、填空题和编程题，全面测试学生对本节课知识点的理解和应用能力。

#include <iostream.h>#include <stdlib.h>void PrintBoard();void PrintInfo();void PrintStats(int, int, int);void IfError(int&, int&);void ChoiceOfChar(char&);void PromptTurnO(int&, int&);void PromptTurnX(int&, int&);char board[3][3];int main(){int ROW;int COLUMN;int FirstPlayer;int SecondPlayer;int Draws;FirstPlayer = 0;SecondPlayer = 0;Draws = 0;char choice;choice = 'Y';PrintInfo();while(choice == 'Y'){for (ROW = 0; ROW < 3; ROW++)for (COLUMN = 0; COLUMN < 3; COLUMN++) board[ROW][COLUMN]= ' ';int x;int y;int SpotsOnBoard;SpotsOnBoard = 0;while ( SpotsOnBoard <= 9){if (SpotsOnBoard == 0)PrintBoard();PromptTurnO(x, y);IfError(x, y);board[x - 1][y - 1]= 'O';SpotsOnBoard++;PrintBoard();if (board[0][0]== 'O' && board[1][1]== 'O' && board[2][2]== 'O') {cout << " O player wins the game" << endl;FirstPlayer++;break;}else if (board[2][0]== 'O' && board[1][1]== 'O' && board[0][2]== 'O') {cout << "O player wins the game" <<endl;FirstPlayer++;break;}else if (board[0][0]== 'O' && board[1][0]== 'O' && board[2][0] == 'O') {cout << "O player wins the game" << endl;FirstPlayer++;break;}else if (board[0][1]== 'O' && board[1][1]== 'O' && board[2][1]== 'O') {cout << "O player wins the game" << endl;FirstPlayer++;break;}else if (board[0][2]== 'O' && board[1][2]== 'O' && board[2][2]== 'O') {cout << "O player wins the game" << endl;FirstPlayer++;break;}else if (board[0][0]== 'O' && board[0][1]== 'O' && board[0][2]== 'O') {cout << "O player wins the game" << endl;FirstPlayer++;break;}else if (board[1][0]== 'O' && board[1][1]== 'O' && board[1][2]== 'O') {FirstPlayer++;break;}else if (board[2][0]== 'O' && board[2][1]== 'O' && board[2][2]== 'O') {cout << "O player wins the game" <<endl;FirstPlayer++;break;}else if (SpotsOnBoard == 9){cout << "Draw!" <<endl;Draws++;break;}PromptTurnX(x, y);IfError(x, y);board[x - 1][y - 1]= 'X';SpotsOnBoard++;PrintBoard();if (board[0][0]== 'X' && board[1][1]== 'X' && board[2][2]== 'X') {cout << "X player is victorious" << endl;SecondPlayer++;}else if (board[2][0]== 'X' && board[1][1]== 'X' && board[0][2]== 'X') {cout << "X player is victorious" << endl;SecondPlayer++;}else if (board[0][0]== 'X' && board[1][0]== 'X' && board[2][0]== 'X') {cout << "X player is victorious" << endl;SecondPlayer++;}else if (board[0][1]== 'X' && board[1][1]== 'X' && board[2][1]== 'X') {cout << "X player is victorious" << endl;SecondPlayer++;}else if (board[0][2]== 'X' && board[1][2]== 'X' && board[2][2]== 'X') {SecondPlayer++;}else if (board[0][0]== 'X' && board[0][1]== 'X' && board[0][2]== 'X'){cout << "X player is victorious" << endl;SecondPlayer++;}else if (board[1][0]== 'X' && board[1][1]== 'X' && board[1][2]== 'X'){cout << "X player is victorious" << endl;SecondPlayer++;}else if (board[2][0]== 'X' && board[2][1]== 'X' && board[2][2]== 'X'){cout << "X player is victorious" << endl;SecondPlayer++;}}ChoiceOfChar(choice);}PrintStats(FirstPlayer, SecondPlayer, Draws);system("PAUSE");return 0;}void PrintBoard(){cout << endl;cout << " 1 2 3 " << endl;cout << "1 " << board[0][0] << " | " << board[0][1]<<" | "<< board[0][2]<< endl; cout << " ---|---|---" << endl;;cout << "2 " << board[1][0]<< " | " << board[1][1]<< " | " << board[1][2]<< endl; cout << " ---|---|---" << endl;cout << "3 " << board[2][0]<< " | " << board[2][1]<< " | " << board[2][2]<< endl; cout << endl;cout << endl;}void PrintInfo(){cout << "First player is represented by O, second by X." << endl;cout << "Enter the coordinates for Rows and Columns" << endl;cout << "Strike enter when Done." << endl;cout << endl;}void PrintStats(int FirstPlayer, int SecondPlayer, int Draws){cout << "The O player have won " << FirstPlayer << " times" << endl; cout << "The X player have won " << SecondPlayer << " times" << endl; cout << "There was " << Draws << " draws" << endl;cout << "Thanks for using my program" << endl;}void IfError(int& x, int& y){while (x > 3 || x < 1 || y > 3 || y < 1 ||('O'== board[x - 1][y - 1]||'X'== board[x - 1][y - 1])){cout << "This coordinate is not allowed, try again" << endl;cout << "row: ";cin >> x;cout << "column: ";cin >> y;}}void ChoiceOfChar(char& choice){cout << endl;cout << " Press CAPITAL Y if you want to play again." << endl;cout << " Otherwise, press any other letter key to exit the loop." << endl; cin >> choice;}void PromptTurnO(int& x, int& y){cout << "Turn of the first player (O), enter the coordinates" << endl; cout << "Row: ";cin >> x;cout << "Column: ";cin >> y;}void PromptTurnX(int& x, int& y){cout << "Turn of the second player (X), enter the coordinates"<< endl; cout << "Row: ";cin >> x;cout << "Column: ";cin >> y;}。

c语言井字棋课程设计一、课程目标知识目标：1. 理解C语言基本语法结构，掌握变量定义、数据类型、运算符和表达式等基础知识。

2. 学习使用C语言实现数组操作，掌握二维数组的使用方法。

3. 掌握C语言控制结构，如顺序结构、选择结构（if-else）、循环结构（for、while）等，并能运用到实际编程中。

技能目标：1. 培养学生运用C语言解决问题的能力，学会分析问题、设计算法、编写代码、调试程序。

2. 培养学生团队合作意识，学会在团队中分工协作，共同完成项目。

3. 提高学生逻辑思维能力，学会通过编程解决实际问题。

情感态度价值观目标：1. 培养学生积极的学习态度，激发学生对编程的兴趣和热情。

2. 培养学生面对问题敢于挑战、勇于克服困难的精神，增强自信心。

3. 培养学生遵守编程规范，养成良好的编程习惯。

本课程针对初中年级学生，结合C语言学科特点，注重培养学生的编程兴趣和实际动手能力。

课程内容以井字棋游戏为载体，引导学生学习C语言基础知识，掌握编程技巧，同时强调团队合作和逻辑思维能力的培养。

通过本课程的学习，使学生能够在实践中巩固所学知识，提高编程技能，培养积极向上的情感态度。

二、教学内容1. C语言基础知识回顾：变量定义、数据类型、运算符和表达式。

2. 数组概念及使用：一维数组、二维数组，重点掌握二维数组在井字棋游戏中的应用。

3. 控制结构：顺序结构、选择结构（if-else）、循环结构（for、while），结合井字棋游戏逻辑进行讲解。

4. 函数定义与调用：介绍函数的概念，编写并调用函数完成井字棋游戏功能模块。

5. 井字棋游戏设计：a. 游戏界面设计：使用二维数组表示棋盘，设计美观易用的游戏界面。

b. 玩家输入与输出：实现玩家输入落子位置，并在棋盘上显示落子。

c. 判断胜负：编写函数判断当前棋局是否分出胜负，并输出结果。

d. 人工智能对手：引入简单的算法，实现电脑自动落子，与玩家对弈。

6. 项目实践：分组进行项目实践，每组完成一个完整的井字棋游戏。

十五、井字棋ﻫ【问题描述】ﻫ设计一个井字棋得游戏程序【基本要求】ﻫ游戏规则就是:从一个空得棋盘开始,人为ｘ一方,计算机为ｏ一方,人机双方轮流放置棋子,人下得时候,字符x将被放入棋盘中指定得位置,轮到计算机下时,程序将扫描棋盘,并将字符o放入扫描到得第一个空格,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与局、截图:代码:／／autｈor GXＵ-pengｃｈeng＃inｃluｄe〈ｓｔｄio、h>#incｌude〈ｗindoｗｓ。

h>#inｃlude<ｃｏｎio、ｈ＞void Heｌp();//帮助inｔPlay(cｈａr player,int ｃｈoiｃe);/／对弈voｉｄＩｎit();//初始化棋盘void Display();//打印棋盘vｏｉd Bｌaｃk(ｉnt cｈｏice);//黑棋voiｄWｈite(inｔｃｈｏiｃe);//白棋ｖoidＢlocｋ(inｔchoiｃｅ);//添加选择框voｉｄClear(int choiｃｅ);//清空之前得选择框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(ｉnｔｃｈｏice);//将标号转换为行号iｎt Fｕncy(ｉnt choｉce);/／将标号转换为列号vｏidＥｎd(ｃhar symbol);//结束int Found();／／返回第一个没有棋子得位置得编号cｈar a［31］［64］;//用数组存放棋盘与棋子,a[行]［列]int ｂ=0;//棋子个数iｎt mａin(){chａr playｅr;ﻩiｎt 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。