1.三字棋存在只能演算法，雖然程式較笨，希望大佬指導。
2.利用for迴圈巢狀if判斷，每一行有三種堵棋的情況。
3.玩家或者電腦每下一步棋都要進行判斷輸贏和列印棋盤
game.h 標頭檔案
#ifndef __game_h__
#define __game_h__
#include#include#include#define row 3
#define col 3
void printmenu();
void printboard(char board[row][col],int row,int col);
void initboard(char board[row][col], int row, int col);
void playermove(char board[row][col], int row, int col);
void computermove(char board[row][col], int row, int col);
char checkwin(char board[row][col], int row, int col);
#endif __game_h__
test.c
#include"game.h"
void game()
; char win = ' ';
initboard(board,row,col);
printboard(board,row,col);
do while (1);
if (win == 'x')
else if (win == '0')
else if (win == 'q')
printboard(board, row, col);
} int main()
} while (choose);
system("pause");
}
game.c#include"game.h"void printmenu()
void printboard(char board[row][col], int row, int col)
} printf("---------------\n");
}void initboard(char board[row][col], int row, int col) }}
static int isfull(char board[row][col], int row, int col)
} }return 1;
}void playermove(char board[row][col], int row, int col)
else
}else
}}void computermove(char board[row][col], int row, int col)
else if (board[i][1] == board[i][2] && board[i][1] == 'x')
else if (board[i][0] == board[i][2] && board[i][0] == 'x')
} //行堵棋
if ((board[0][i] == board[1][i] || board[0][i] == board[2][i] || board[1][i] == board[2][i]) && (board[0][i] == 'x' || board[1][i] == 'x' || board[2][i] == 'x'))
else if (board[1][i] == board[2][i] && board[1][i] == 'x')
else if (board[0][i] == board[2][i] && board[0][i] == 'x')
}//列堵棋
if ((board[0][0] == board[1][1] || board[0][0] == board[2][2] || board[1][1] == board[2][2]) && (board[0][0] == 'x' || board[1][1] == 'x' || board[2][2] == 'x'))
else if (board[1][1] == board[2][2] && board[1][1] == 'x')
else if (board[0][0] == board[2][2] && board[0][0] == 'x')
}//主對角線堵棋
if ((board[0][2] == board[1][1] || board[1][1] == board[2][0] || board[0][2] == board[2][0]) && (board[0][2] == 'x' || board[1][1] == 'x' || board[2][0] == 'x'))
else if (board[2][0] == board[1][1] && board[1][1] == 'x')
else if (board[0][2] == board[2][0] && board[0][2] == 'x')
}//非主對角線堵棋
if (count == 0)
} }}
char checkwin(char board[row][col], int row, int col)
}for (i = 0; i < row; i++)
}if (board[0][0] == board[1][1] && board[1][1] == board[2][2] && board[0][0] != ' ')
if (isfull(board, row, col) == 1)
return ' ';
}
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