c语言编写井字棋游戏代码

㈠ 用c语言编辑井字棋有一步不知道什么意思，请求解答！！

while(1)或者while(true)之类的写法都表示无限循环，要跳出循环就需要在循环中用break或者return

㈡ 一个关于用C语言写井字棋的游戏的问题 怎么让他重新开始的问题 代码如下如果他平局了 怎么重新开始呢

不知道你在问什么？？

重新开始就写循环啊,用循环把那些东西包起来

检测胜利用函数就写个函数呗

intv(inti){
if((board[0][0]==board[1][1]&&board[1][1]==board[2][2])||(board[0][2]==board[1][1]&&board[1][1]&&board[1][1]==board[2][0]))
{
printf("Congratulations,player%disthewinner
",i);

return0;

}
if(board[ROW][0]==board[ROW][1]&&board[ROW][1]==board[ROW][2])
{
printf("Congratulations,player%disthewinner
",i);

return0;

}
if(board[0][COL]==board[1][COL]&&board[1][COL]==board[2][COL])
{
printf("Congratulations,player%disthewinner
",i);

return0;

}
return1;
}

㈢ 用C语言，编程，求助大神

时间仓促，做得比较简陋。

#include<stdio.h>
#include<windows.h>
#include<conio.h>
voidmenu();
voidgame();
intmain()
{
intn=0;
menu();
while(scanf("%d",&n))
{
system("cls");
menu();
switch(n)
{
case1:game();break;
case2:return0;
default:printf("请重新输入：");
}

}
}
voidmenu()
{
printf("**********
");
printf("*井字棋小游戏*
");
printf("*按1开始游戏*
");
printf("*按2退出游戏*
");
printf("**********
");
}
voidgame()
{
printf("请在小键盘输入1-9");
intstep_number=1;
intboard[3][3]={0};
while(1)
{
charnum;
num=getch();
system("cls");
if(num=='1')board[2][0]+=step_number++;
if(num=='2')board[2][1]+=step_number++;
if(num=='3')board[2][2]+=step_number++;
if(num=='4')board[1][0]+=step_number++;
if(num=='5')board[1][1]+=step_number++;
if(num=='6')board[1][2]+=step_number++;
if(num=='7')board[0][0]+=step_number++;
if(num=='8')board[0][1]+=step_number++;
if(num=='9')board[0][2]+=step_number++;
for(inti=0;i<3;i++)//打印棋盘
{
printf("
");
for(intj=0;j<3;j++)
{
if(board[i][j]==0)printf("_");
elseif(board[i][j]%2==1)printf("X");
elseif(board[i][j]%2==0)printf("O");
}
}

for(inti=0;i<3;i++)
{
if(board[i][0]!=0&&board[i][1]!=0&&board[i][2]!=0)
{
if(board[i][0]%2==1)
{
if(board[i][0]%2==board[i][1]%2&&board[i][0]%2==board[i][2]%2)
printf("
X方胜！
");
}
if(board[i][0]%2==0)
{
if(board[i][0]%2==board[i][1]%2&&board[i][0]%2==board[i][2]%2)
printf("
O方胜！
");
}
}


}
for(intj=0;j<3;j++)
{
if(board[0][j]!=0&&board[1][j]!=0&&board[2][j]!=0)
{
if(board[0][j]%2==1)
{
if(board[0][j]%2==board[1][j]%2&&board[0][j]%2==board[2][j]%2)
printf("
X方胜！
");
}
if(board[0][j]%2==0)
{
if(board[0][j]%2==board[1][j]%2&&board[0][j]%2==board[2][j]%2)
printf("
O方胜！
");
}
}

}
if(board[0][0]!=0&&board[1][1]!=0&&board[2][2]!=0)
{
if(board[0][0]%2==1)
{
if(board[0][0]%2==board[1][1]%2&&board[0][0]%2==board[2][2]%2)
printf("
X方胜！
");
}
if(board[0][0]%2==0)
{
if(board[0][0]%2==board[1][1]%2&&board[0][0]%2==board[2][2]%2)
printf("
O方胜！
");
}
}
if(board[0][2]!=0&&board[1][1]!=0&&board[2][0]!=0)
{
if(board[0][2]%2==1)
{
if(board[0][2]%2==board[1][1]%2&&board[0][2]%2==board[2][0]%2)
printf("
X方胜！
");
}
if(board[0][2]%2==0)
{
if(board[0][2]%2==board[1][1]%2&&board[0][2]%2==board[2][0]%2)
printf("
O方胜！
");
}
}

if(step_number>10)
{
printf("平局
");
printf("按任意键回主菜单");
if(getchar())break;
}
}

}

㈣ C++ 井字棋 (注释) 一定要给注释!!!

#include <stdio.h>
#include <stdlib.h>

#define SIZE 3
typedef enum {CBLANK, CBLACK, CWHITE} CHESS;
typedef enum {GM_WIN, GM_LOST, GM_UNKNOW, GM_ERROR} GAMEFLAG;

void init_board(CHESS board[SIZE][SIZE]) //初始化
{
int i, j;
for (i = 0; i < SIZE; i++)
{
for (j = 0; j < SIZE; j++)
{
board[i][j] = CBLANK;
}
}
}

void print_chess(CHESS board[SIZE][SIZE]) //打印棋盘
{
int i, j;

putchar(' ');
for (i=0; i < SIZE; i++)
{
printf("%2d", i+1);
}
putchar('\n');

for (i=0; i < SIZE; i++)
{
printf("%-2d", i+1);
for (j=0; j < SIZE; j++)
{
switch (board[i][j])
{
case CWHITE:
putchar('O');
break;
case CBLACK:
putchar('*');
break;
case CBLANK:
putchar('_');
break;
default:
putchar('?');
break;
}
putchar(' ');
}
putchar('\n');
}
}

void swc(CHESS chess, int *black, int *white, int *bmax, int *wmax) //判断
{
switch (chess)
{
case CBLACK:
*white = 0;
(*black)++;
break;
case CWHITE:
*black = 0;
(*white)++;
break;
case CBLANK:
*black = 0;
*white = 0;
break;
default:
break;
}

if (*black > *bmax) *bmax = *black;
if (*white > *wmax) *wmax = *white;
}

GAMEFLAG res(CHESS board[SIZE][SIZE]) //判断输赢
{
int i, j;
int win[4] = {0, 0, 0, 0};
int rblack, rwhite, cblack, cwhite,
loblack = 0, lowhite = 0,
roblack = 0, rowhite = 0,
bmax = 0, wmax = 0;

for (i=0; i < SIZE; i++)
{
rblack = 0;
rwhite = 0;
cblack = 0;
cwhite = 0;

swc(board[i][i], &loblack, &lowhite, &bmax, &wmax);
swc(board[i][SIZE-i-1], &roblack, &rowhite, &bmax, &wmax);

for (j=0; j < SIZE; j++)
{
swc(board[i][j], &rblack, &rwhite, &bmax, &wmax);
swc(board[j][i], &cblack, &cwhite, &bmax, &wmax);

}

}

if (bmax >= 3)
{
if (wmax >= 3)
{
return GM_ERROR;
}
else
{
return GM_WIN;
}
}
else
{
if (wmax >= 3)
{
return GM_LOST;
}
else
{
return GM_UNKNOW;
}
}

}

int move(CHESS board[SIZE][SIZE], CHESS chs, int x, int y)
{
int bs = 1;
if (board[x][y])
bs = 0;
else if (y >= SIZE || y < 0 || x >= SIZE || x < 0)
bs = 0;
else
board[x][y] = chs;

return bs;
}

int main()
{
CHESS b[SIZE][SIZE];
char *msg[] = {"BLACK WIN!\n", "WHITE LOST!", "NOT YET", "ERROR!!"};
char *plr[] = {"NON", "BLACK", "WHITE"};
CHESS p = CBLACK;
GAMEFLAG flg;

init_board(b);
while ((flg = res(b)) == GM_UNKNOW)
{
int x, y, bmv = 1;
system("cls");
print_chess(b);
while (bmv)
{
printf("%s回合,输入坐标:", plr[p]);
scanf("%d%d", &x, &y);
bmv = !move(b,p,x-1,y-1);
}
p = (CHESS)(CWHITE + CBLACK - p);
}

printf("%s", msg[flg]);
system("pause");

return 0;
}

有加分有注释!!

㈤ C语言井字游戏

http://www.pudn.com/downloads112/sourcecode/game/49636984jzGame.rar
还有这个
井字游戏:为双人对弈游戏,双方轮流放子,单任一行,列,斜线为三个相同的 子时即为胜利!
http://download.pudn.com/downloads63/sourcecode/game/57578890xtkdige.rar
还有
http://download.pudn.com/downloads76/sourcecode/game/55593407402922602.rar
还有这个
一个用c语言写的“井字游戏”，经过turbo c编译通过。
#include<stdio.h>
#include<graphics.h>
#include<conio.h>
#include<bios.h>
#include<alloc.h>
#include<stdlib.h>
#define x1 150
#define x2 250
#define x3 350
#define x4 450
#define y1 100
#define y2 200
#define y3 300
#define y4 400
/* 1表示O, 2表示X */

void *buf_yuan,*buf_cha,*buf;
FILE *fp;
int a[4][4];
int flag=0;
/*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void main(void)
{
int gd=DETECT,gm;
int END=0;
int i,j,h,kai=0,restart,key;
int heng=2,lie=2;
int temp=1;
void SystemInitial(void);/*初始化游戏界面*/
void SiJiao(int heng,int lie);/*显示将要走步的位格的边框*/
void hua(int heng,int lie,int type);/*走步，即画圆或画叉*/
void display(int type);/*显示谁赢了*/
void hz(int x,int y,int a,int COL,int b,char *s);/*显汉字，只可以显示中文*/
void computer(void);/*单人游戏时，电脑走步*/
void hanzi(int x,int y,char *p,int colour);/*也是显汉字，可以中文与英语混合显示*/
initgraph(&amt;gd,&amt;gm,"");
if ((fp=fopen("hzk16","rb"))==NULL)
{
printf("Can't open hzk16,Please add it");
getch();
closegraph();
exit(0);
}
cleardevice();
hz(220,100,40,2,YELLOW,"井字游戏");
setcolor(BLUE);
hanzi(400,460,"作者：04计科3班 邓永华 08号",BLUE);
setcolor(WHITE);
setfillstyle(SOLID_FILL,WHITE);
flag=0;
if(flag==0)/*单人游戏*/
{
circle(195,242,4);
floodfill(195,242,WHITE);
}
else/*双人游戏*/
{
circle(195,262,4);
floodfill(195,262,WHITE);
}
outtextxy(200,220,"Please select:");
setcolor(RED);
outtextxy(205,240,"Play with computer");
outtextxy(205,260,"Two people play");

hanzi(30,350,"游戏说明：双方轮流走步，先连成三个符号一直线（横、竖、斜）就Win",GREEN);

while(temp)
{
switch(bioskey(0))
{
case 0x1c0d:/*回车*/
{
temp=0;
break;
}
case 0x11b:/*Esc*/
exit(0);
case 0x4800:/*上*/
if(flag==0)
break;
else
{
flag=0;
setcolor(BLACK);
setfillstyle(SOLID_FILL,BLACK);
circle(195,262,4);
floodfill(195,262,BLACK);

setcolor(WHITE);
setfillstyle(SOLID_FILL,WHITE);
circle(195,242,4);
floodfill(195,242,WHITE);
}
break;
case 0x5000:/*下*/
if(flag==1)
break;
else
{
flag=1;
/*清除*/
setcolor(BLACK);
setfillstyle(SOLID_FILL,BLACK);
circle(195,242,4);
floodfill(195,242,BLACK);

setcolor(WHITE);
setfillstyle(SOLID_FILL,WHITE);
circle(195,262,4);
floodfill(195,262,WHITE);
break;
}
}
}
cleardevice();
/*begin to play*/
SystemInitial();
while(END!=1)
{
restart=0;
switch(bioskey(0))/*按键*/
{
case 0x11b:/*Esc退出*/
END=1;
break;
case 0x3920:/*space*/
if(kai==1)
break;
if(a[heng][lie]) break;
kai=1;
hua(heng,lie,1);
a[heng][lie]=1;

if((a[1][1]==1&amt;&amt;a[1][2]==1&amt;&amt;a[1][3]==1)/*判断是否赢了*/
||(a[2][1]==1&amt;&amt;a[2][2]==1&amt;&amt;a[2][3]==1)
||(a[3][1]==1&amt;&amt;a[3][2]==1&amt;&amt;a[3][3]==1)
||(a[1][1]==1&amt;&amt;a[2][1]==1&amt;&amt;a[3][1]==1)
||(a[1][2]==1&amt;&amt;a[2][2]==1&amt;&amt;a[3][2]==1)
||(a[1][3]==1&amt;&amt;a[2][3]==1&amt;&amt;a[3][3]==1)
||(a[1][1]==1&amt;&amt;a[2][2]==1&amt;&amt;a[3][3]==1)
||(a[1][3]==1&amt;&amt;a[2][2]==1&amt;&amt;a[3][1]==1))
{
display(1);
END=1;
}
if(!END)
{
h=0;
for(i=1;i<4;i++)
for(j=1;j<4;j++)
if(a[i][j])
h++;
if(h==9)
{
display(3);
END=1;
}
}
if(END==1)
{
outtextxy(260,450,"play again? Y/N ");
while(1)
{
key=bioskey(0);
if(key==0x1579||key==0x1559)/*y的大小写*/
{
END=0;
restart=1;
break;
}
else if((key==0x316e)||(key==0x314e))/*n的大小写*/
break;
else continue;
}
}
break;
case 0x5230:/*0*/
if(kai==2) break;
if(a[heng][lie]) break;
kai=2;
hua(heng,lie,2);
a[heng][lie]=20;

if((a[1][1]==20&amt;&amt;a[1][2]==20&amt;&amt;a[1][3]==20)/*判断是否赢了*/
||(a[2][1]==20&amt;&amt;a[2][2]==20&amt;&amt;a[2][3]==20)
||(a[3][1]==20&amt;&amt;a[3][2]==20&amt;&amt;a[3][3]==20)
||(a[1][1]==20&amt;&amt;a[2][1]==20&amt;&amt;a[3][1]==20)
||(a[1][2]==20&amt;&amt;a[2][2]==20&amt;&amt;a[3][2]==20)
||(a[1][3]==20&amt;&amt;a[2][3]==20&amt;&amt;a[3][3]==20)
||(a[1][1]==20&amt;&amt;a[2][2]==20&amt;&amt;a[3][3]==20)
||(a[1][3]==20&amt;&amt;a[2][2]==20&amt;&amt;a[3][1]==20))

{
display(2);
END=1;
}
if(!END)
{ h=0;
for(i=1;i<4;i++)
for(j=1;j<4;j++)
if(a[i][j])
h++;
if(h==9)
{
display(3);
END=1;
}
}
if(END==1)
{
outtextxy(260,450,"play again? Y/N ");
while(1)
{
key=bioskey(0);
if(key==0x1579||key==0x1559)
{
END=0;
restart=1;
break;
}
if(key==0x316e||key==0x314e)
break;
else continue;
}
}
break;
case 0x4800:
case 0x1177:
heng--;
if(heng<1)
heng=1;
SiJiao(heng,lie);
break;
case 0x5000:
case 0x1f73:
heng++;
if(heng>3)
heng=3;
SiJiao(heng,lie);
break;
case 0x4b00:
case 0x1e61:
lie--;
if(lie<1)
lie=1;
SiJiao(heng,lie);
break;
case 0x4d00:
case 0x2064:
lie++;
if(lie>3)
lie=3;
SiJiao(heng,lie);
break;
}
if(flag==0&amt;&amt;kai==1&amt;&amt;END!=1&amt;&amt;restart!=1)
{
computer();
kai=2;
if((a[1][1]==20&amt;&amt;a[1][2]==20&amt;&amt;a[1][3]==20)/*判断是否赢了*/
||(a[2][1]==20&amt;&amt;a[2][2]==20&amt;&amt;a[2][3]==20)
||(a[3][1]==20&amt;&amt;a[3][2]==20&amt;&amt;a[3][3]==20)
||(a[1][1]==20&amt;&amt;a[2][1]==20&amt;&amt;a[3][1]==20)
||(a[1][2]==20&amt;&amt;a[2][2]==20&amt;&amt;a[3][2]==20)
||(a[1][3]==20&amt;&amt;a[2][3]==20&amt;&amt;a[3][3]==20)
||(a[1][1]==20&amt;&amt;a[2][2]==20&amt;&amt;a[3][3]==20)
||(a[1][3]==20&amt;&amt;a[2][2]==20&amt;&amt;a[3][1]==20))

{
display(2);
END=1;
}
if(!END)
{ h=0;
for(i=1;i<4;i++)
for(j=1;j<4;j++)
if(a[i][j])
h++;
if(h==9)
{
display(3);
END=1;
}
}
if(END==1)
{
outtextxy(260,450,"play again? Y/N ");
while(1)
{
key=bioskey(0);
if(key==0x1579||key==0x1559)
{
END=0;
restart=1;
break;
}
if(key==0x316e||key==0x314e)
break;
else continue;
}
}
}
if(restart==1)
{
cleardevice();
SystemInitial();
kai=0;
heng=2;
lie=2;
}
}
free(buf_yuan);
free(buf_cha);
free(buf);
fclose(fp);
closegraph();
}
/*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void SystemInitial(void)
{
void hz(int x,int y,int a,int COL,int b,char *s);
void SiJiao(int heng,int lie);
int size,i,j;

for(i=1;i<4;i++)
for(j=1;j<4;j++)
a[i][j]=0;

SiJiao(2,2);
hz(240,30,40,2,YELLOW,"井字游戏");
setcolor(GREEN);
outtextxy(10,200,"1P");
setcolor(WHITE);
outtextxy(10,220,"up: w");
outtextxy(10,240,"down: s");
outtextxy(10,260,"left: a");
outtextxy(10,280,"right: d");
outtextxy(10,300,"fill: space");
outtextxy(10,320,"exit: Esc");
if(flag==1)
{
setcolor(GREEN);
outtextxy(520,200,"2P");
setcolor(WHITE);
outtextxy(520,220,"up:");
outtextxy(520,240,"down:");
outtextxy(520,260,"left: ");
outtextxy(520,280,"right: ");
outtextxy(520,300,"fill: 0");
outtextxy(520,320,"exit: Esc");
hz(585,220,25,1,WHITE,"↑");
hz(585,240,25,1,WHITE,"↓");
hz(585,260,25,1,WHITE,"←");
hz(585,280,25,1,WHITE,"→");
}
line(x1,y1,x1,y4);
line(x1,y1,x4,y1);
line(x4,y1,x4,y4);
line(x1,y4,x4,y4);
line(x2,y1,x2,y4); /*shu*/
line(x3,y1,x3,y4);
line(x1,y2,x4,y2); /*heng*/
line(x1,y3,x4,y3);

circle((x2+x3)/2,(y2+y3)/2,(y3-y2)/2-10);/*hua yuan*/
size=imagesize((x2+x3)/2-(y3-y2)/2+9,(y2+y3)/2-(y3-y2)/2+9,(x2+x3)/2+(y3-y2)/2-9,(y2+y3)/2+(y3-y2)/2-9);
buf_yuan=malloc(size);
if(!buf_yuan) exit(1);
getimage((x2+x3)/2-(y3-y2)/2+9,(y2+y3)/2-(y3-y2)/2+9,(x2+x3)/2+(y3-y2)/2-9,(y2+y3)/2+(y3-y2)/2-9,buf_yuan);
setcolor(BLACK);
circle((x2+x3)/2,(y2+y3)/2,(y3-y2)/2-10);

setcolor(WHITE); /*hua cha*/
line(x2+10,y2+10,x3-10,y3-10);
line(x2+10,y3-10,x3-10,y2+10);
buf_cha=malloc(size);
getimage((x2+x3)/2-(y3-y2)/2+9,(y2+y3)/2-(y3-y2)/2+9,(x2+x3)/2+(y3-y2)/2-9,(y2+y3)/2+(y3-y2)/2-9,buf_cha);
setcolor(BLACK);
line(x2+10,y2+10,x3-10,y3-10);
line(x2+10,y3-10,x3-10,y2+10);
}
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void SiJiao(int heng,int lie)
{
int a1,b1,a2,b2;
void huahei(void);/*把原来用白色显示的将要走步的边框用黑色覆盖*/
huahei();

setcolor(WHITE);
switch(heng)
{
case 1:
b1=y1;
b2=y2;
break;
case 2:
b1=y2;
b2=y3;
break;
case 3:
b1=y3;
b2=y4;
break;
}
switch(lie)
{
case 1:
a1=x1;
a2=x2;
break;
case 2:
a1=x2;
a2=x3;
break;
case 3:
a1=x3;
a2=x4;
break;
}
line(a1+3,b1+3,a1+30,b1+3);
line(a1+3,b1+3,a1+3,b1+30);
line(a1+3,b2-3,a1+3,b2-30);
line(a1+3,b2-3,a1+30,b2-3);
line(a2-30,b1+3,a2-3,b1+3);
line(a2-3,b1+3,a2-3,b1+30);
line(a2-30,b2-3,a2-3,b2-3);
line(a2-3,b2-30,a2-3,b2-3);

}
/*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void huahei(void)
{
int i,j,a1,b1,a2,b2;

setcolor(BLACK);
for(i=1;i<4;i++)
for(j=1;j<4;j++)
{
switch(i)
{
case 1:
b1=y1;
b2=y2;
break;
case 2:
b1=y2;
b2=y3;
break;
case 3:
b1=y3;
b2=y4;
break;
}
switch(j)
{
case 1:
a1=x1;
a2=x2;
break;
case 2:
a1=x2;
a2=x3;
break;
case 3:
a1=x3;
a2=x4;
break;
}

line(a1+3,b1+3,a1+30,b1+3);
line(a1+3,b1+3,a1+3,b1+30);
line(a1+3,b2-3,a1+3,b2-30);
line(a1+3,b2-3,a1+30,b2-3);
line(a2-30,b1+3,a2-3,b1+3);
line(a2-3,b1+3,a2-3,b1+30);
line(a2-30,b2-3,a2-3,b2-3);
line(a2-3,b2-30,a2-3,b2-3);
}

}
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void hua(int heng,int lie,int type)/*type的值为1表示圆,2表示叉*/
{
int x,y;
setcolor(WHITE);
switch(lie)
{
case 1:
x=(x1+x2)/2-(x2-x1)/2+9;
break;
case 2:
x=(x2+x3)/2-(x3-x2)/2+9;
break;
case 3:
x=(x3+x4)/2-(x4-x3)/2+9;
break;
}

switch(heng)
{
case 1:
y=(y1+y2)/2-(y2-y1)/2+9;
break;
case 2:
y=(y2+y3)/2-(y3-y2)/2+9;
break;
case 3:
y=(y3+y4)/2-(y4-y3)/2+9;
break;
}
switch(type)
{
case 1:
buf=buf_yuan;
a[heng][lie]=1;
break;
case 2:
buf=buf_cha;
a[heng][lie]=20;
break;
}
putimage(x,y,buf,COPY_PUT);
}
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void display(int type)
{
if(type==1)
outtextxy(270,430,"O Win");
if(type==2)
outtextxy(270,430,"X Win");
if(type==3)
outtextxy(270,430,"draw");

}
/*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void hz(int x,int y,int a,int COL,int b,char *s)/*x,y为显示的坐标，a为字与字之间的间隔，b为字的大小，s为指向为显示的汉字的指针*/
{
int ROW;
char buffer[32];
register m,n,i1,j1,k;
unsigned char qh,wh;
unsigned long offset;
ROW=COL;
while(*s)
{
qh=*(s)-0xa0;/*汉字区位码*/
wh=*(s+1)-0xa0;
offset=(94*(qh-1)+(wh-1))*32L;/*计算该汉字在字库中偏移量*/
fseek(fp,offset,SEEK_SET);
fread(buffer,32,1,fp);/*取出汉字32字节的点阵字模存入buffer中（一个汉字） */
for (i1=0;i1<16;i1++)/*将32位字节的点阵按位在屏幕上打印出来(1:打印,0:不打印),显示汉字 */
for(n=0;n<ROW;n++)
for(j1=0;j1<2;j1++)
for(k=0;k<8;k++)
for(m=0;m<COL;m++)
if (((buffer[i1*2+j1]>>(7-k))&amt;0x1)!=NULL)
putpixel(x+8*j1*COL+k*COL+m,y+i1*ROW+n,b);
s+=2;/*因为一个汉字内码占用两个字节,所以s必须加2*/
x+=a;
}
}
/**************************************************************************/
void computer(void)
{
int i,j;
for(i=1;i<4;i++)/*横向判断*/
if(a[i][1]+a[i][2]+a[i][3]==40)
for(j=1;j<4;j++)
if(a[i][j]==0)
{
hua(i,j,2);
return;
}
for(i=1;i<4;i++)/*横向判断*/
if(a[i][1]+a[i][2]+a[i][3]==2)
for(j=1;j<4;j++)
if(a[i][j]==0)
{
hua(i,j,2);
return;
}

for(i=1;i<4;i++)/*竖向判断*/
if(a[1][i]+a[2][i]+a[3][i]==40)
{
for(j=1;j<4;j++)
if(a[j][i]==0)
{
hua(j,i,2);
return;
}
}
for(i=1;i<4;i++)/*竖向判断*/
if(a[1][i]+a[2][i]+a[3][i]==2)
for(j=1;j<4;j++)
if(a[j][i]==0)
{
hua(j,i,2);
return;
}

if(a[1][1]+a[2][2]+a[3][3]==40)/*左上角到右下角判断*/
{
for(i=1;i<4;i++)
if(a[i][i]==0)
{
hua(i,i,2);
return;
}
}
else if(a[1][1]+a[2][2]+a[3][3]==2)
for(i=1;i<4;i++)
if(a[i][i]==0)
{
hua(i,i,2);
return;
}

if(a[3][1]+a[2][2]+a[1][3]==40)/*右上角到左下角判断*/
{
for(i=1;i<4;i++)
if(a[i][4-i]==0)
{
hua(i,4-i,2);
return;
}
}
else if(a[3][1]+a[2][2]+a[1][3]==2)
for(i=1;i<4;i++)
if(a[i][4-i]==0)
{
hua(i,4-i,2);
return;
}

for(i=1;i<4;i++)/*随便找到空位填上去*/
for(j=1;j<4;j++)
if(a[i][j]==0)
{
hua(i,j,2);
return;
}

}
void hanzi(int x,int y,char *p,int colour)
{
FILE *fp;
char buffer[32];
register i,j,k;
unsigned char qh,wh;
unsigned long location;
if((fp=fopen("hzk16","rb"))==NULL)
{
printf("Can't open hzk16!");
getch();
exit(0);
}
while(*p)
{
if(((unsigned char)*p>=0xa1&amt;&amt;(unsigned char)*p<=0xfe)&amt;&amt;((unsigned char)*(p+1)>=0xal&amt;&amt;(unsigned char)*(p+1)<=0xfe))
{
qh=*p-0xa0;
wh=*(p+1)-0xa0;
location=(94*(qh-1)+(wh-1))*32L;
fseek(fp,location,SEEK_SET);
fread(buffer,32,1,fp);
for(i=0;i<16;i++)
for(j=0;j<2;j++)
for(k=0;k<8;k++)
if(((buffer[i*2+j]>>(7-k))&amt;0x1)!=NULL)
putpixel(x+8*j+k,y+i,colour);
p+=2;
x+=18;
if(x>600)
{
x=15;y+=18;
}
}
else
{
char q[2];
moveto(x,y);
*q=*p;
*(q+1)='\0';
outtextxy(x,y+4,q);
x+=8+1;p++;
}
}
fclose(fp);
}
都是井字游戏，C语言的，你参考看看。

㈥ C语言 怎么编程井字棋

一两句话说不明白，你先做个简单的流程规划，把这个问题细化，然后再考虑每一步都需要怎么做，比如需要建立几个类，需要定义什么变量，如何存贮等问题。这个必须一步步来，哪有你这样解决问题的

㈦ C语言程序设计 井字棋 求一段电脑下棋部分 下面给出的是人下的

没图我怎么回答？

㈧ C语言编写井字棋游戏 代码已有半成品

你初始化一个字符数组，里面都给他一样的初始值 E
X 下了就改成X ，O 下了就改成 O，下完以后判断横竖斜三条线有没有一样的，

一共就九个格子，下一步就少一步 ，下之前判断一下，如果等于E ,就是空的，可以放子。
下完一步总步数减一，你这样不就能确定还有几步可以走了。
九步都下完如果没有横竖斜都一样的不就是平局吗，所有数组元素的值都不等于E了不就结束了，或者九步完了也结束了

另外，少用GOTO ,变量定义最好有意义，写点注释

㈨ c语言井字棋双人对战

3x3的棋盘输入0~8的数字来确定落子位置，简单的程序啊 scanf("%d",&choice); 以下一句3的整倍数确定玩家落棋的行数 row = --choice / 3; 确定行数还不行，必须确定列数，所以除3的余数就是列数 column = choice % 3;

㈩ 用C写一个井字棋程序，但第（1,1）格总是显示5，代码如下，望解答

#include<stdio.h>
#include<ctype.h>
#include<conio.h>
int n=9, z=0, qp[10]={0};
int chkwin(int t[], int w)
{if (t[1]==w && t[1]==t[2] && t[2]==t[3]) return(w);
if (t[4]==w && t[4]==t[5] && t[5]==t[6]) return(w);
if (t[7]==w && t[7]==t[8] && t[8]==t[9]) return(w);
if (t[1]==w && t[1]==t[4] && t[4]==t[7]) return(w);
if (t[2]==w && t[2]==t[5] && t[5]==t[8]) return(w);
if (t[3]==w && t[3]==t[6] && t[6]==t[9]) return(w);
if (t[1]==w && t[1]==t[5] && t[5]==t[9]) return(w);
if (t[3]==w && t[3]==t[5] && t[5]==t[7]) return(w);
return(0);
}

long search(int n, int k, int t[])
{int i, h, f, g;
long j;
if (n==0) return(chkwin(t,k));
for (f=0, j=0,i=1; i<10; i++)
if (t[i]==0)
{t[i]=k; h=chkwin(t,k);
if (h==k) f++;
else j+=search(n-1,k*-1,t);
t[i]=0;
}
if (f>z) for (j=k, g=n; g>0; j*=g--);
return(j);
}
/* 电脑选位置走棋，择位原则1.当前位置必须为空 */
/* 2.若走当前位置电脑能赢，则走这一步，否则转（3） */
/* 3.若当前位置被人走电脑会输，则走这一步，否则转（4） */
/* 4.试走当前位置并调用 search（）函数，求走当前位置电脑的有利程度 */
/* 5.找对电脑有利程度最大的位置走 */
void cgo()
{int i, ti=0;
long j=-8000000, t=0;
for (ti=1; ti>-2; ti-=2) /*ti=1，先看自己能否赢；ti=-1,看对受能否赢*/
for (i=1; i<10; i++)
if (qp[i]==0)
{qp[i]=ti;
if (chkwin(qp, ti)!=0) {n--; qp[i]=1; return;}
qp[i]=0;
}
for (i=1; i<10;i++)
if (qp[i]==0)
{qp[i]=1; t=search(n-1,-1,qp);
if (t>j) {j=t; ti=i;}
qp[i]=0;
}
n--; qp[ti]=1;
}
/* 函数mgo(),人输入走棋位置 */
void mgo()
{int c=0;
printf ("\nPlease enter the Num to go: ");
for (c=getche(); ; printf("\n"), c=getche() )
if (isdigit(c) && c!='0' && qp[c-48]==0)
{n--; qp[c-48]=-1; return;
} }
/* 屏幕输出函数display,在屏幕上输出当前的棋盘 */
void display(int x)
{int i;
char t[10]={0};
for (i=1; i<10; i++)
{if (qp[i]>0) t[i]=88;
if (qp[i]<0) t[i]=79;
}
printf ("\n%c|%c|%c\n-----\n%c|%c", t[1], t[2], t[3], t[4], t[5]);
printf ("|%c\n-----\n%c|%c|%c\n", t[6], t[7], t[8], t[9]);
if (x==0) printf("\ndraw! \n");
if (x==1) printf("\ncomputer win!\n");
if (x==2) printf("\ncontinue \n");
}
main()
{char c;
printf ("\nGo first? [Y/N]:"); /*选择谁先走*/
for (c=getche(); c!='Y'&&c!='y'&&c!='N'&&c!='n'; c=getche());
if (c=='N'||c=='n') {cgo(); z=1; display(2);}
while (1)
{mgo(); if (!n) {display(0); break;} /*人走，若不是最后一步，继续；否则跳出*/
cgo(); if (chkwin(qp,1)) {display(1); break;} /*电脑走，若没赢，继续；否则跳出*/
if (n) display(2); /*还没走到最后一步，继续；否则跳出*/
else {display(0); break;}
}
getchar();
return 0;
}