首先我们先来看一个稍微简单些的实现方式：
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#include <stdio.h>
#include <stdlib.h>
#define N 15
int chessboard[N + 1][N + 1] = { 0 };
int whoseTurn = 0;
void initGame(void);
void printChessboard(void);
void playChess(void);
int judge(int, int);
int main(void)
{
initGame();
while (1)
{
whoseTurn++;
playChess();
}
return 0;
}
void initGame(void)
{
char c;
printf("Please input \\'y\\' to enter the game:");
c = getchar();
if ('y' != c && 'Y' != c)
exit(0);
system("cls");
printChessboard();
}
void printChessboard(void)
{
int i, j;
for (i = 0; i <= N; i++)
{
for (j = 0; j <= N; j++)
{
if (0 == i)
printf("%3d", j);
else if (j == 0)
printf("%3d", i);
else if (1 == chessboard[i][j])
printf(" O");
else if (2 == chessboard[i][j])
printf(" X");
else
printf(" *");
}
printf("\\n");
}
}
void playChess(void)
{
int i, j, winner;
if (1 == whoseTurn % 2)
{
printf("Turn to player 1, please input the position:");
scanf("%d %d", &i, &j);
while (chessboard[i][j] != 0)
{
printf("This position has been occupied, please input the position again:");
scanf("%d %d", &i, &j);
}
chessboard[i][j] = 1;
}
else
{
printf("Turn to player 1, please input the position:");
scanf("%d %d", &i, &j);
while (chessboard[i][j] != 0)
{
printf("This position has been occupied, please input the position again:");
scanf("%d %d", &i, &j);
}
chessboard[i][j] = 2;
}
system("cls");
printChessboard();
if (judge(i, j))
{
if (1 == whoseTurn % 2)
{
printf("Winner is player 1!\\n");
exit(0);
}
else
{
printf("Winner is player 2!\\n");
exit(0);
}
}
}
int judge(int x, int y)
{
int i, j;
int t = 2 - whoseTurn % 2;
for (i = x - 4, j = y; i <= x; i++)
{
if (i >= 1 && i <= N - 4 && t == chessboard[i][j] && t == chessboard[i + 1][j] && t == chessboard[i + 2][j] && t == chessboard[i + 3][j] && t == chessboard[i + 4][j])
return 1;
}
for (i = x, j = y - 4; j <= y; j++)
{
if (j >= 1 && j <= N - 4 && t == chessboard[i][j] && t == chessboard[i][j + 1] && t == chessboard[i][j + 1] && t == chessboard[i][j + 3] && t == chessboard[i][j + 4])
return 1;
}
for (i = x - 4, j = y - 4; i <= x, j <= y; i++, j++)
{
if (i >= 1 && i <= N - 4 && j >= 1 && j <= N - 4 && t == chessboard[i][j] && t == chessboard[i + 1][j + 1] && t == chessboard[i + 2][j + 2] && t == chessboard[i + 3][j + 3] && t == chessboard[i + 4][j + 4])
return 1;
}
for (i = x + 4, j = y - 4; i >= 1, j <= y; i--, j++)
{
if (i >= 1 && i <= N - 4 && j >= 1 && j <= N - 4 && t == chessboard[i][j] && t == chessboard[i - 1][j + 1] && t == chessboard[i - 2][j + 2] && t == chessboard[i - 3][j + 3] && t == chessboard[i - 4][j + 4])
return 1;
}
return 0;
}
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演示截图
我们再来看一个更复杂些的