做网站的公司挣钱吗网络广告营销的典型案例

下面是一个简单的井字棋（Tic-Tac-Toe）游戏的C语言实现。这个游戏是在命令行界面上运行的，支持两名玩家轮流输入他们的步骤。

#include <stdio.h>
#include <stdbool.h>#define SIZE 3void drawBoard(char board[SIZE][SIZE]) {for (int i = 0; i < SIZE; ++i) {for (int j = 0; j < SIZE; ++j) {printf(" %c ", board[i][j]);if (j < SIZE - 1) printf("|");}printf("\n");if (i < SIZE - 1) printf("---|---|---\n");}
}bool checkWin(char board[SIZE][SIZE], char player) {// Check rows, columns and diagonals for a winfor (int i = 0; i < SIZE; ++i) {if ((board[i][0] == player && board[i][1] == player && board[i][2] == player) ||(board[0][i] == player && board[1][i] == player && board[2][i] == player)) {return true;}}if ((board[0][0] == player && board[1][1] == player && board[2][2] == player) ||(board[0][2] == player && board[1][1] == player && board[2][0] == player)) {return true;}return false;
}bool isBoardFull(char board[SIZE][SIZE]) {for (int i = 0; i < SIZE; ++i)for (int j = 0; j < SIZE; ++j)if (board[i][j] == ' ') return false;return true;
}int main() {char board[SIZE][SIZE];for (int i = 0; i < SIZE; ++i)for (int j = 0; j < SIZE; ++j)board[i][j] = ' ';bool playerOneTurn = true;int row, col;while (true) {drawBoard(board);if (playerOneTurn) printf("Player X's turn.\n");else printf("Player O's turn.\n");printf("Enter row and column: ");scanf("%d %d", &row, &col);if (row >= 0 && row < SIZE && col >= 0 && col < SIZE && board[row][col] == ' ') {board[row][col] = playerOneTurn ? 'X' : 'O';if (checkWin(board, playerOneTurn ? 'X' : 'O')) {drawBoard(board);printf(playerOneTurn ? "Player X wins!\n" : "Player O wins!\n");break;}if (isBoardFull(board)) {drawBoard(board);printf("It's a draw!\n");break;}playerOneTurn = !playerOneTurn;} else {printf("Invalid move. Try again.\n");}}return 0;
}

程序说明：

• drawBoard 函数：用于打印当前的游戏板状态。
• checkWin 函数：检查当前是否有玩家获胜。它会检查所有可能的胜利条件：行、列以及两条对角线。
• isBoardFull 函数：判断游戏板是否已满，如果满了且没有玩家获胜，则游戏为平局。
• 主函数 (main)：初始化游戏板，循环处理玩家的输入，并根据玩家的输入更新游戏板。每轮都会检查是否有玩家获胜或游戏是否结束于平局。

这个程序提供了一个基础版本的井字棋游戏。你可以在此基础上添加更多功能，比如更友好的用户界面、与电脑进行对战等。